Clean ways to do multiple undos in C











up vote
12
down vote

favorite
4












Someone will probably say something about exceptions... but in C, what are other ways to do the following cleanly/clearly and without repeating so much code?



if (Do1()) { printf("Failed 1"); return 1; }
if (Do2()) { Undo1(); printf("Failed 2"); return 2; }
if (Do3()) { Undo2(); Undo1(); printf("Failed 3"); return 3; }
if (Do4()) { Undo3(); Undo2(); Undo1(); printf("Failed 4"); return 4; }
if (Do5()) { Undo4(); Undo3(); Undo2(); Undo1(); printf("Failed 5"); return 5; }
Etc...


This might be one case for using gotos. Or maybe multiple inner functions...










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  • 1




    Just to point out with regard to my previous comment: in these cases, it is generally the case that these cleanup actions need to be also performed when there are no early abortions (hence the mention of free/fclose); that makes the structure with goto & labels fairly straightforward and easy to read. This may not be the case that you are thinking of.
    – 9769953
    2 days ago






  • 2




    I believe you need to clarify the types of the functions, as we are getting all kinds of mixed answers. Are they the same type, or are all functions of different types?
    – Lundin
    2 days ago






  • 1




    @Lundin It is clear the code is an example of the code patterns that appear in many C functions, typically those that allocate resources. Precisely because the types do not matter here, they are not specified. It is pseudocode, if you want.
    – Acorn
    2 days ago








  • 5




    @Lundin No, I haven't said anything like it. And no, pseudo-code can be perfectly on topic.
    – Acorn
    2 days ago






  • 3




    @9769953 - I'd say the problem wasn't goto fail; so much as avoiding curly braces. And it's not like it's a new sort of bug. The offending line didn't have to be a goto, but could just as easily have been a exit(EXIT_FAILURE). Still the same bug, despite the different manifestation.
    – StoryTeller
    2 days ago

















up vote
12
down vote

favorite
4












Someone will probably say something about exceptions... but in C, what are other ways to do the following cleanly/clearly and without repeating so much code?



if (Do1()) { printf("Failed 1"); return 1; }
if (Do2()) { Undo1(); printf("Failed 2"); return 2; }
if (Do3()) { Undo2(); Undo1(); printf("Failed 3"); return 3; }
if (Do4()) { Undo3(); Undo2(); Undo1(); printf("Failed 4"); return 4; }
if (Do5()) { Undo4(); Undo3(); Undo2(); Undo1(); printf("Failed 5"); return 5; }
Etc...


This might be one case for using gotos. Or maybe multiple inner functions...










share|improve this question




















  • 1




    Just to point out with regard to my previous comment: in these cases, it is generally the case that these cleanup actions need to be also performed when there are no early abortions (hence the mention of free/fclose); that makes the structure with goto & labels fairly straightforward and easy to read. This may not be the case that you are thinking of.
    – 9769953
    2 days ago






  • 2




    I believe you need to clarify the types of the functions, as we are getting all kinds of mixed answers. Are they the same type, or are all functions of different types?
    – Lundin
    2 days ago






  • 1




    @Lundin It is clear the code is an example of the code patterns that appear in many C functions, typically those that allocate resources. Precisely because the types do not matter here, they are not specified. It is pseudocode, if you want.
    – Acorn
    2 days ago








  • 5




    @Lundin No, I haven't said anything like it. And no, pseudo-code can be perfectly on topic.
    – Acorn
    2 days ago






  • 3




    @9769953 - I'd say the problem wasn't goto fail; so much as avoiding curly braces. And it's not like it's a new sort of bug. The offending line didn't have to be a goto, but could just as easily have been a exit(EXIT_FAILURE). Still the same bug, despite the different manifestation.
    – StoryTeller
    2 days ago















up vote
12
down vote

favorite
4









up vote
12
down vote

favorite
4






4





Someone will probably say something about exceptions... but in C, what are other ways to do the following cleanly/clearly and without repeating so much code?



if (Do1()) { printf("Failed 1"); return 1; }
if (Do2()) { Undo1(); printf("Failed 2"); return 2; }
if (Do3()) { Undo2(); Undo1(); printf("Failed 3"); return 3; }
if (Do4()) { Undo3(); Undo2(); Undo1(); printf("Failed 4"); return 4; }
if (Do5()) { Undo4(); Undo3(); Undo2(); Undo1(); printf("Failed 5"); return 5; }
Etc...


This might be one case for using gotos. Or maybe multiple inner functions...










share|improve this question















Someone will probably say something about exceptions... but in C, what are other ways to do the following cleanly/clearly and without repeating so much code?



if (Do1()) { printf("Failed 1"); return 1; }
if (Do2()) { Undo1(); printf("Failed 2"); return 2; }
if (Do3()) { Undo2(); Undo1(); printf("Failed 3"); return 3; }
if (Do4()) { Undo3(); Undo2(); Undo1(); printf("Failed 4"); return 4; }
if (Do5()) { Undo4(); Undo3(); Undo2(); Undo1(); printf("Failed 5"); return 5; }
Etc...


This might be one case for using gotos. Or maybe multiple inner functions...







c






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edited 2 days ago









Peter Mortensen

13.3k1983111




13.3k1983111










asked 2 days ago









dargaud

73511024




73511024








  • 1




    Just to point out with regard to my previous comment: in these cases, it is generally the case that these cleanup actions need to be also performed when there are no early abortions (hence the mention of free/fclose); that makes the structure with goto & labels fairly straightforward and easy to read. This may not be the case that you are thinking of.
    – 9769953
    2 days ago






  • 2




    I believe you need to clarify the types of the functions, as we are getting all kinds of mixed answers. Are they the same type, or are all functions of different types?
    – Lundin
    2 days ago






  • 1




    @Lundin It is clear the code is an example of the code patterns that appear in many C functions, typically those that allocate resources. Precisely because the types do not matter here, they are not specified. It is pseudocode, if you want.
    – Acorn
    2 days ago








  • 5




    @Lundin No, I haven't said anything like it. And no, pseudo-code can be perfectly on topic.
    – Acorn
    2 days ago






  • 3




    @9769953 - I'd say the problem wasn't goto fail; so much as avoiding curly braces. And it's not like it's a new sort of bug. The offending line didn't have to be a goto, but could just as easily have been a exit(EXIT_FAILURE). Still the same bug, despite the different manifestation.
    – StoryTeller
    2 days ago
















  • 1




    Just to point out with regard to my previous comment: in these cases, it is generally the case that these cleanup actions need to be also performed when there are no early abortions (hence the mention of free/fclose); that makes the structure with goto & labels fairly straightforward and easy to read. This may not be the case that you are thinking of.
    – 9769953
    2 days ago






  • 2




    I believe you need to clarify the types of the functions, as we are getting all kinds of mixed answers. Are they the same type, or are all functions of different types?
    – Lundin
    2 days ago






  • 1




    @Lundin It is clear the code is an example of the code patterns that appear in many C functions, typically those that allocate resources. Precisely because the types do not matter here, they are not specified. It is pseudocode, if you want.
    – Acorn
    2 days ago








  • 5




    @Lundin No, I haven't said anything like it. And no, pseudo-code can be perfectly on topic.
    – Acorn
    2 days ago






  • 3




    @9769953 - I'd say the problem wasn't goto fail; so much as avoiding curly braces. And it's not like it's a new sort of bug. The offending line didn't have to be a goto, but could just as easily have been a exit(EXIT_FAILURE). Still the same bug, despite the different manifestation.
    – StoryTeller
    2 days ago










1




1




Just to point out with regard to my previous comment: in these cases, it is generally the case that these cleanup actions need to be also performed when there are no early abortions (hence the mention of free/fclose); that makes the structure with goto & labels fairly straightforward and easy to read. This may not be the case that you are thinking of.
– 9769953
2 days ago




Just to point out with regard to my previous comment: in these cases, it is generally the case that these cleanup actions need to be also performed when there are no early abortions (hence the mention of free/fclose); that makes the structure with goto & labels fairly straightforward and easy to read. This may not be the case that you are thinking of.
– 9769953
2 days ago




2




2




I believe you need to clarify the types of the functions, as we are getting all kinds of mixed answers. Are they the same type, or are all functions of different types?
– Lundin
2 days ago




I believe you need to clarify the types of the functions, as we are getting all kinds of mixed answers. Are they the same type, or are all functions of different types?
– Lundin
2 days ago




1




1




@Lundin It is clear the code is an example of the code patterns that appear in many C functions, typically those that allocate resources. Precisely because the types do not matter here, they are not specified. It is pseudocode, if you want.
– Acorn
2 days ago






@Lundin It is clear the code is an example of the code patterns that appear in many C functions, typically those that allocate resources. Precisely because the types do not matter here, they are not specified. It is pseudocode, if you want.
– Acorn
2 days ago






5




5




@Lundin No, I haven't said anything like it. And no, pseudo-code can be perfectly on topic.
– Acorn
2 days ago




@Lundin No, I haven't said anything like it. And no, pseudo-code can be perfectly on topic.
– Acorn
2 days ago




3




3




@9769953 - I'd say the problem wasn't goto fail; so much as avoiding curly braces. And it's not like it's a new sort of bug. The offending line didn't have to be a goto, but could just as easily have been a exit(EXIT_FAILURE). Still the same bug, despite the different manifestation.
– StoryTeller
2 days ago






@9769953 - I'd say the problem wasn't goto fail; so much as avoiding curly braces. And it's not like it's a new sort of bug. The offending line didn't have to be a goto, but could just as easily have been a exit(EXIT_FAILURE). Still the same bug, despite the different manifestation.
– StoryTeller
2 days ago














14 Answers
14






active

oldest

votes

















up vote
26
down vote













Yes, it's quite common to use goto in such cases to avoid repeating yourself.



An example:



int hello() {
int result;

if (Do1()) { result = 1; goto err_one; }
if (Do2()) { result = 2; goto err_two; }
if (Do3()) { result = 3; goto err_three; }
if (Do4()) { result = 4; goto err_four; }
if (Do5()) { result = 5; goto err_five; }

// Assuming you'd like to return 0 on success.
return 0;

err_five:
Undo4();
err_four:
Undo3();
err_three:
Undo2();
err_two:
Undo1();
err_one:
printf("Failed %i", result);
return result;
}


As always you probably also want to keep your functions small and batch together the operations in a meaningful manner to avoid a large "undo-code".






share|improve this answer



















  • 1




    Note: return 0 may or may not be needed, depending on what the function is supposed to do.
    – Acorn
    2 days ago






  • 1




    @Acorn Without, code wouldn't be equivalent to code presented in question, which undoes only on error...
    – Aconcagua
    2 days ago






  • 3




    @Lundin I agree. This answer assumes that the Do and Undo functions might have different signatures in reality which is probably more common in practice.
    – likle
    2 days ago






  • 10




    @Lundin There is no code repetition here. Please, don't misunderstand generic examples with concrete ones.
    – Acorn
    2 days ago






  • 10




    @Lundin It seems you don't understand how this approach works. If you have to add another case, there is no need to change anything else. That is the entire point.
    – Acorn
    2 days ago




















up vote
9
down vote













If you have the same signature for your function you can do something like this:



bool Do1(void) { printf("function %sn", __func__); return true;}
bool Do2(void) { printf("function %sn", __func__); return true;}
bool Do3(void) { printf("function %sn", __func__); return false;}
bool Do4(void) { printf("function %sn", __func__); return true;}
bool Do5(void) { printf("function %sn", __func__); return true;}

void Undo1(void) { printf("function %sn", __func__);}
void Undo2(void) { printf("function %sn", __func__);}
void Undo3(void) { printf("function %sn", __func__);}
void Undo4(void) { printf("function %sn", __func__);}
void Undo5(void) { printf("function %sn", __func__);}


typedef struct action {
bool (*Do)(void);
void (*Undo)(void);
} action_s;


int main(void)
{
action_s actions = {{Do1, Undo1},
{Do2, Undo2},
{Do3, Undo3},
{Do4, Undo4},
{Do5, Undo5},
{NULL, NULL}};

for (size_t i = 0; actions[i].Do; ++i) {
if (!actions[i].Do()) {
printf("Failed %zu.n", i + 1);
for (int j = i - 1; j >= 0; --j) {
actions[j].Undo();
}
return (i);
}
}

return (0);
}


You can change the return of one of Do functions to see how it react :)






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  • Comments are not for extended discussion; this conversation has been moved to chat.
    – Samuel Liew
    yesterday


















up vote
8
down vote














This might be one case for using gotos.




Sure, let's try that. Here's a possible implementation:



#include "stdio.h"
int main(int argc, char **argv) {
int errorCode = 0;
if (Do1()) { errorCode = 1; goto undo_0; }
if (Do2()) { errorCode = 2; goto undo_1; }
if (Do3()) { errorCode = 3; goto undo_2; }
if (Do4()) { errorCode = 4; goto undo_3; }
if (Do5()) { errorCode = 5; goto undo_4; }

undo_5: Undo5(); /* deliberate fallthrough */
undo_4: Undo4();
undo_3: Undo3();
undo_2: Undo2();
undo_1: Undo1();
undo_0: /* nothing to undo in this case */

if (errorCode != 0) {
printf("Failed %dn", errorCode);
return errorCode;
}
return 0;
}





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  • 2




    The -1 is probablky from someone who considers goto as absolutely evil, which it is not as long as it's abused.
    – Jabberwocky
    2 days ago






  • 1




    @Aconcagua It is rare to be able to factorize the printf, and even if you can, it is not typically done. In a "generic" example like this, you shouldn't write it like that, because beginners will think it is the common way of doing it, in my opinion. Also, there is the point about the initialization to -1 and also not talking about cases where you may have to return early (i.e. before the Undos). Finally, better answers are currently lower than this one.
    – Acorn
    2 days ago








  • 2




    @Lundin It isn't.
    – Acorn
    2 days ago






  • 1




    @Acorn The error behavior of every single error code is tightly coupled with the error behavior of every other error code. So if you add something in the middle during maintenance, it all comes crashing down.
    – Lundin
    2 days ago






  • 3




    @Lundin It isn't tied, at all. You are making assumptions, and they do not even correspond to actual code out there.
    – Acorn
    2 days ago


















up vote
7
down vote













For completeness a bit of obfuscation:



int foo(void)
{
int rc;

if (0
|| (rc = 1, do1())
|| (rc = 2, do2())
|| (rc = 3, do3())
|| (rc = 4, do4())
|| (rc = 5, do5())
|| (rc = 0)
)
{
/* More or less stolen from Chris' answer:
https://stackoverflow.com/a/53444967/694576) */
switch(rc - 1)
{
case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
undo5();

case 4:
undo4();

case 3:
undo3();

case 2:
undo2();

case 1:
undo1();

default:
break;
}
}

return rc;
}





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  • 5




    Please, stop this madness.
    – Acorn
    2 days ago






  • 3




    @Acorn: Why? Nice example for how to use the comma-operator ... ;-)
    – alk
    2 days ago






  • 1




    @Acorn: Also nicely symmetrical and compact, perfect to auto-generate.
    – alk
    2 days ago








  • 2




    The other solutions are just as easy to auto-generate (if you are talking about code generation).
    – Acorn
    2 days ago








  • 13




    Now all we are missing is a version with Duff's device and we'll be ready to submit this thread to IOCCC :)
    – Lundin
    2 days ago


















up vote
3
down vote













Use goto to manage cleanup in C.



For instance, check the Linux kernel coding style:




The rationale for using gotos is:




  • unconditional statements are easier to understand and follow nesting is reduced

  • errors by not updating individual exit points when making modifications are prevented

  • saves the compiler work to optimize redundant code away ;)


Example:



int fun(int a)
{
int result = 0;
char *buffer;

buffer = kmalloc(SIZE, GFP_KERNEL);
if (!buffer)
return -ENOMEM;

if (condition1) {
while (loop1) {
...
}
result = 1;
goto out_free_buffer;
}

...

out_free_buffer:
kfree(buffer);
return result;
}





In your particular case, it could look like:



int f(...)
{
int ret;

if (Do1()) {
printf("Failed 1");
ret = 1;
goto undo1;
}

...

if (Do5()) {
printf("Failed 5");
ret = 5;
goto undo5;
}

// all good, return here if you need to keep the resources
// (or not, if you want them deallocated; in that case initialize `ret`)
return 0;

undo5:
Undo4();
...
undo1:
return ret;
}





share|improve this answer



















  • 2




    This is somewhat acceptable use of goto - it is a pattern from BASIC known as "on error goto". I wish people wouldn't stop thinking there, but think one step further still. The better alternative to "on error goto" is to use a wrapper function and from the inner function return code; upon error. Leave resource allocation and clean-up to the outer wrapper. Thus separate resource allocation and algorithm in 2 different functions. Much cleaner design, easier to read and no goto debate. In general, I would recommend staying away from "the Linux kernel coding style" document.
    – Lundin
    2 days ago












  • @Lundin Not sure, are you now referring Tom's function pointer solution (your description sounds somehow different to me...)? If not, how would then these wrapper functions look like? Cannot think of anything better than int callDo2(void) { if(do2()) { undo1(); return 2; } return callDo3(); } at the moment, but cannot imagine either that you really meant such ones...
    – Aconcagua
    2 days ago








  • 1




    Basically replace all your goto with return, then in the outer wrapper function call "undo" based on what the function returned.
    – Lundin
    2 days ago










  • @Lundin Please show an example, because it does not sound like a good idea at all.
    – Acorn
    2 days ago


















up vote
2
down vote













There are probably many ways to do this, but one idea is since you won't call one function unless the preceeding one succeeded, you could chain your function calls using else if like this. And using a variable to track where it fails you can use a switch statement to roll back easily too.



int ret=0;
if(Do1()) {
ret=1;
} else if(Do2()) {
ret=2;
} else if(Do3()) {
ret=3;
} else if(Do4()) {
ret=4;
} else if(Do5()) {
ret=5;
}

switch(ret) {
case 5:
Undo4();
case 4:
Undo3();
case 3:
Undo2();
case 2:
Undo1();
case 1:
printf("Failed %dn",ret);
break;
}
return ret;





share|improve this answer



















  • 8




    Don't do this. The code is harder to read and doing more branches compared to simple goto.
    – Acorn
    2 days ago






  • 4




    Also, note it is wrong: if Do5() fails, we don't want to run Undo5() (typically).
    – Acorn
    2 days ago






  • 1




    switch(ret) should be switch(ret-1). Also an (emtpy) default case would be nice. All in all I like this approach.
    – alk
    2 days ago












  • @alk Writing the actual correct values would be better -- if you are keen on using this solution, which you should not ;) As for the default case, what for?
    – Acorn
    2 days ago






  • 6




    The lengths people will go to avoid uncoditional jump which was basically kept in C language exactly for such cases because it makes it more readable than arrow antipattern or switch/if ladder is amazing... Take my downvote.
    – Purple Ice
    2 days ago




















up vote
1
down vote













Yes, as explained by other answers, using goto for error-handling is often appropriate in C.



That said, if possible, you probably should make your cleanup code safe to call even if the corresponding action was never performed. For example, instead of:



void foo()
{
int result;
int* p = malloc(...);
if (p == NULL) { result = 1; goto err1; }

int* p2 = malloc(...);
if (p2 == NULL) { result = 2; goto err2; }

int* p3 = malloc(...);
if (p3 == NULL) { result = 3; goto err3; }

// Do something with p, p2, and p3.
bar(p, p2, p3);

// Maybe we don't need p3 anymore.
free(p3);

return 0;

err3:
free(p3);
err2:
free(p2);
err1:
free(p1);
return result;
}


I'd advocate:



void foo()
{
int result = -1; // Or some generic error code for unknown errors.

int* p = NULL;
int* p2 = NULL;
int* p3 = NULL;

p = malloc(...);
if (p == NULL) { result = 1; goto exit; }

p2 = malloc(...);
if (p2 == NULL) { result = 2; goto exit; }

p3 = malloc(...);
if (p3 == NULL) { result = 3; goto exit; }

// Do something with p, p2, and p3.
bar(p, p2, p3);

// Set success *only* on the successful path.
result = 0;

exit:
// free(NULL) is a no-op, so this is safe even if p3 was never allocated.
free(p3);

if (result != 0)
{
free(p2);
free(p1);
}
return result;
}


It's slightly less efficient since it requires initializing variables to NULL, but it's more maintainable since you don't need extra labels. There's less stuff to get wrong when making changes to the code. Also, if there's cleanup code that you need on both success and failure paths, you can avoid code duplication.






share|improve this answer




























    up vote
    0
    down vote













    I typically approach this kind of problem by nesting the conditionals:



    int rval = 1;
    if (!Do1()) {
    if (!Do2()) {
    if (!Do3()) {
    if (!Do4()) {
    if (!Do5()) {
    return 0;
    // or "goto succeeded", or ...;
    } else {
    printf("Failed 5");
    rval = 5;
    }
    Undo4();
    } else {
    printf("Failed 4");
    rval = 4;
    }
    Undo3();
    } else {
    printf("Failed 3");
    rval = 3;
    }
    Undo2();
    } else {
    printf("Failed 2");
    rval = 2;
    }
    Undo1();
    } else {
    printf("Failed 1");
    rval = 1;
    }
    return rval;


    Usually, for me, the DoX() are some kind of resource acquisition, such as malloc(), and the UndoX() are corresponding resource releases that should be performed only in the event of failure. The nesting clearly shows the association between corresponding acquisitions and releases, and avoids the need for repetition of the code for undo operations. It's also very easy to write -- you don't need to create or maintain labels, and it's easy to put the resource release in the right place as soon as you write the acquisition.



    This approach does sometimes produce deeply nested code. That doesn't bother me much, but you might consider it an issue.






    share|improve this answer

















    • 2




      'deeply nested' - plus error and error handling get separated by quite some distance, the more for the less deeply nested ones...
      – Aconcagua
      yesterday










    • @Aconcagua, separation of error and error-handling is a pretty consistent characteristic of all of these approaches. It follows directly from avoiding repetition of the error-handling code.
      – John Bollinger
      yesterday




















    up vote
    0
    down vote













    Here is an answer that I have found resilient to bugs.



    Yes. It uses goto. I firmly believe you should use what gives you most clarity, rather than just blindly following the advice of those before you (goto as a construct can make spaghetti code, but in this instance every other error handling method ususally ends up more spaghetti-like than using this method of goto, so IMO it's superior).



    Some people may not like the form of this code, but I contest that when used to the style it is cleaner, easier to read (when everything's lined up, of course), and much more resilient to errors. If you have the properly linter/static analysis setup, and you're working with POSIX, it pretty much requires you to code in this fashion to allow for good error handling.



    static char *readbuf(char *path)
    {
    struct stat st;
    char *s = NULL;
    size_t size = 0;
    int fd = -1;

    if (!path) { return NULL; }

    if ((stat(path, &st)) < 0) { perror(path); goto _throw; }

    size = st.st_size;
    if (size == 0) { printf("%s is empty!n", path); goto _throw; }

    if (!(s = calloc(size, 1))) { perror("calloc"); goto _throw; }

    fd = open(path, O_RDONLY);
    if (fd < -1) { perror(path); goto _throw; }
    if ((read(fd, s, size)) < 0) { perror("read"); goto _throw; }
    close(fd); /* There's really no point checking close for errors */

    return s;

    _throw:
    if (fd > 0) close(fd);
    if (s) free(s);
    return NULL;
    }





    share|improve this answer




























      up vote
      -1
      down vote













      If the functions return some kind of state pointer or handle (like most allocation & initialization functions would), you can quite cleanly do this without goto by giving initial values to variables. Then you can have a single deallocation function that can handle the case where only part of the resources has been allocated.



      For example:




      void *mymemoryblock = NULL;
      FILE *myfile = NULL;
      int mysocket = -1;

      bool allocate_everything()
      {
      mymemoryblock = malloc(1000);
      if (!mymemoryblock)
      {
      return false;
      }

      myfile = fopen("/file", "r");
      if (!myfile)
      {
      return false;
      }

      mysocket = socket(AF_INET, SOCK_STREAM, 0);
      if (mysocket < 0)
      {
      return false;
      }

      return true;
      }

      void deallocate_everything()
      {
      if (mysocket >= 0)
      {
      close(mysocket);
      mysocket = -1;
      }

      if (myfile)
      {
      fclose(myfile);
      myfile = NULL;
      }

      if (mymemoryblock)
      {
      free(mymemoryblock);
      mymemoryblock = NULL;
      }
      }


      And then just do:




      if (allocate_everything())
      {
      do_the_deed();
      }
      deallocate_everything();





      share|improve this answer





















      • "If the functions return some kind of state pointer or handle..." Yes, but it is not the general case. Further, your solution requires 3 function for each function that allocates resources, plus global variables (or passing things around).
        – Acorn
        2 days ago




















      up vote
      -1
      down vote













      TL;DR:



      I believe it should be written as:



      int main (void)
      {
      int result = do_func();
      printf("Failed %dn", result);
      }




      Detailed explanation:



      If nothing can be assumed what-so-ever about the function types, we can't easily use an array of function pointers, which would otherwise be the correct answer.



      Assuming all function types are incompatible, then we would have to wrap in the original obscure design containing all those non-compatible functions, inside something else.



      We should make something that is readable, maintainable, fast. We should avoid tight coupling, so that the undo behavior of "Do_x" doesn't depend on the undo behavior of "Do_y".



      int main (void)
      {
      int result = do_func();
      printf("Failed %dn", result);
      }


      Where do_func is the function doing all the calls required by the algorithm, and the printf is the UI output, separated from the algorithm logic.



      do_func would be implemented like a wrapper function around the actual function calls, handling the outcome depending on the result:



      (With gcc -O3, do_func is inlined in the caller, so there is no overhead for having 2 separate functions)



      int do_it (void)
      {
      if(Do1()) { return 1; };
      if(Do2()) { return 2; };
      if(Do3()) { return 3; };
      if(Do4()) { return 4; };
      if(Do5()) { return 5; };
      return 0;
      }

      int do_func (void)
      {
      int result = do_it();
      if(result != 0)
      {
      undo[result-1]();
      }
      return result;
      }


      Here the specific behavior is controlled by the array undo, which is a wrapper around the various non-compatible functions. Which functions to to call, in which order, is all part of the specific behavior tied to each result code.



      We need to tidy it all up, so that we can couple a certain behavior to a certain result code. Then when needed, we only change the code in one single place if the behavior should be changed during maintenance:



      void Undo_stuff1 (void) { }
      void Undo_stuff2 (void) { Undo1(); }
      void Undo_stuff3 (void) { Undo2(); Undo1(); }
      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

      typedef void Undo_stuff_t (void);
      static Undo_stuff_t* undo[5] =
      {
      Undo_stuff1,
      Undo_stuff2,
      Undo_stuff3,
      Undo_stuff4,
      Undo_stuff5,
      };




      MCVE:



      #include <stdbool.h>
      #include <stdio.h>

      // some nonsense functions:
      bool Do1 (void) { puts(__func__); return false; }
      bool Do2 (void) { puts(__func__); return false; }
      bool Do3 (void) { puts(__func__); return false; }
      bool Do4 (void) { puts(__func__); return false; }
      bool Do5 (void) { puts(__func__); return true; }
      void Undo1 (void) { puts(__func__); }
      void Undo2 (void) { puts(__func__); }
      void Undo3 (void) { puts(__func__); }
      void Undo4 (void) { puts(__func__); }
      void Undo5 (void) { puts(__func__); }

      // wrappers specifying undo behavior:
      void Undo_stuff1 (void) { }
      void Undo_stuff2 (void) { Undo1(); }
      void Undo_stuff3 (void) { Undo2(); Undo1(); }
      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

      typedef void Undo_stuff_t (void);
      static Undo_stuff_t* undo[5] =
      {
      Undo_stuff1,
      Undo_stuff2,
      Undo_stuff3,
      Undo_stuff4,
      Undo_stuff5,
      };

      int do_it (void)
      {
      if(Do1()) { return 1; };
      if(Do2()) { return 2; };
      if(Do3()) { return 3; };
      if(Do4()) { return 4; };
      if(Do5()) { return 5; };
      return 0;
      }

      int do_func (void)
      {
      int result = do_it();
      if(result != 0)
      {
      undo[result-1]();
      }
      return result;
      }

      int main (void)
      {
      int result = do_func();
      printf("Failed %dn", result);
      }


      Output:



      Do1
      Do2
      Do3
      Do4
      Do5
      Undo4
      Undo3
      Undo2
      Undo1
      Failed 5





      share|improve this answer



















      • 12




        So, for each single function in your code that allocates resources, you are going to write 4 functions instead (which, some of them duplicate UndoN() calls in turn). Plus a wrapper. Plus a type. Plus a global array of that type. No further comments.
        – Acorn
        2 days ago












      • @Acorn No, for every ADT in my code, I'm going to write a user-friend API, a clarity of who's responsible for allocation/deallocation, private encapsulation, modular, maintainable code, without tight coupling, where everything is autonomous. In this particular case, it looks like we are patching up some old, broken design, in which case you need wrappers - which is the fault of the original (lack of) design. Of course you don't need that level of abstraction if it's just some quick & dirty hack, but in that case all that is best practices is thrown out the window anyway.
        – Lundin
        14 hours ago


















      up vote
      -2
      down vote













      typedef void(*undoer)();
      int undo( undoer*const* list ) {
      while(*list) {
      (*list)();
      ++list;
      }
      }
      void undo_push( undoer** list, undoer* undo ) {
      if (!undo) return;
      // swap
      undoer* tmp = *list;
      *list = undo;
      undo = tmp;
      undo_push( list+1, undo );
      }
      int func() {
      undoer undoers[6]={0};

      if (Do1()) { printf("Failed 1"); return 1; }
      undo_push( undoers, Undo1 );
      if (Do2()) { undo(undoers); printf("Failed 2"); return 2; }
      undo_push( undoers, Undo2 );
      if (Do3()) { undo(undoers); printf("Failed 3"); return 3; }
      undo_push( undoers, Undo3 );
      if (Do4()) { undo(undoers); printf("Failed 4"); return 4; }
      undo_push( undoers, Undo4 );
      if (Do5()) { undo(undoers); printf("Failed 5"); return 5; }
      return 6;
      }


      I made undo_push do the O(n) work. This is less efficient than having undo do the O(n) work, as we expect more push's than undos. But this version was a touch simpler.



      A more industrial strength version would have head and tail pointers and even capacity.



      The basic idea is to keep a queue of undo actions in a stack, then execute them if you need to clean up.



      Everything is local here, so we don't pollute global state.





      struct undoer {
      void(*action)(void*);
      void(*cleanup)(void*);
      void* state;
      };

      struct undoers {
      undoer* top;
      undoer buff[5];
      };
      void undo( undoers u ) {
      while (u.top != buff)
      {
      (u.top->action)(u.top->state);
      if (u.top->cleanup)
      (u.top->cleanup)(u.top->state);
      --u.top;
      }
      }
      void pundo(void* pu) {
      undo( *(undoers*)pu );
      free(pu);
      }
      void cleanup_undoers(undoers u) {
      while (u.top != buff)
      {
      if (u.top->cleanup)
      (u.top->cleanup)(u.top->state);
      --u.top;
      }
      }
      void pcleanup_undoers(void* pu) {
      cleanup_undoers(*(undoers*)pu);
      free(pu);
      }
      void push_undoer( undoers* to_here, undoer u ) {
      if (to_here->top != (to_here->buff+5))
      {
      to_here->top = u;
      ++(to_here->top)
      return;
      }
      undoers* chain = (undoers*)malloc( sizeof(undoers) );
      memcpy(chain, to_here, sizeof(undoers));
      memset(to_here, 0, sizeof(undoers));
      undoer chainer;
      chainer.action = pundo;
      chainer.cleanup = pcleanup_undoers;
      chainer.data = chain;
      push_undoer( to_here, chainer );
      push_undoer( to_here, u );
      }
      void paction( void* p ) {
      (void)(*a)() = ((void)(*)());
      a();
      }
      void push_undo( undoers* to_here, void(*action)() ) {
      undor u;
      u.action = paction;
      u.cleanup = 0;
      u.data = (void*)action;
      push_undoer(to_here, u);
      }


      then you get:



      int func() {
      undoers u={0};

      if (Do1()) { printf("Failed 1"); return 1; }
      push_undo( &u, Undo1 );
      if (Do2()) { undo(u); printf("Failed 2"); return 2; }
      push_undo( &u, Undo2 );
      if (Do3()) { undo(u); printf("Failed 3"); return 3; }
      push_undo( &u, Undo3 );
      if (Do4()) { undo(u); printf("Failed 4"); return 4; }
      push_undo( &u, Undo4 );
      if (Do5()) { undo(u); printf("Failed 5"); return 5; }
      cleanup_undoers(u);
      return 6;
      }


      but that is getting ridiculous.






      share|improve this answer



















      • 1




        More complex, requires that DoN/UndoN are actual functions (and the same signature), requires stack space, slower.
        – Acorn
        2 days ago




















      up vote
      -4
      down vote













      Let's try for something with zero curly braces:



      int result;
      result = Do1() ? 1 : 0;
      result = result ? result : Do2() ? 2 : 0;
      result = result ? result : Do3() ? 3 : 0;
      result = result ? result : Do4() ? 4 : 0;
      result = result ? result : Do5() ? 5 : 0;

      result > 4 ? (Undo5(),0) : 0;
      result > 3 ? Undo4() : 0;
      result > 2 ? Undo3() : 0;
      result > 1 ? Undo2() : 0;
      result > 0 ? Undo1() : 0;

      result ? printf("Failed %drn", result) : 0;


      Yes. 0; is a valid statement in C (and C++). In the case that some of the functions return something that is incompatible with this syntax (e.g. void perhaps) then the Undo5() style can be used.






      share|improve this answer























      • This assumes the UndoN functions return values, when in fact they may be (and most probably are) declared void (or aren't even functions at all).
        – Cássio Renan
        2 days ago












      • msvc is never a particularly standards compliant compiler, but without thinking about it I did actually develop this with void Undo functions. No idea if its actually valid. If it isn't one could just go with: `result > 4 ? (Undo5(), 0) : 0; Doesn't help of course. if 'UndoX' isn't actually a function.
        – Chris Becke
        2 days ago










      • yeah, MSVC is a bad, bad boy, for C++ at least. In C, this seems to be valid. My bad.
        – Cássio Renan
        2 days ago








      • 2




        I would argue that if (result > 4) Undo5(); is easier to understand than a ternary conditional with no false action and a discarded result. (if statements don't need curly braces)
        – pizzapants184
        2 days ago










      • True. I really was avoiding any kind of explicit flow control. ternary conditions are cheating I know.
        – Chris Becke
        yesterday


















      up vote
      -6
      down vote













      A sane (no gotos, no nested or chained ifs) approach would be



      int bar(void)
      {
      int rc = 0;

      do
      {
      if (do1())
      {
      rc = 1;
      break;
      }

      if (do2())
      {
      rc = 2;
      break;
      }

      ...

      if (do5())
      {
      rc = 5;
      break;
      }
      } while (0);

      if (rc)
      {
      /* More or less stolen from Chris' answer:
      https://stackoverflow.com/a/53444967/694576) */
      switch(rc - 1)
      {
      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
      undo5();

      case 4:
      undo4();

      case 3:
      undo3();

      case 2:
      undo2();

      case 1:
      undo1();

      default:
      break;
      }
      }

      return rc;
      }





      share|improve this answer

















      • 10




        If your definition of "sane" is "no goto" then you already failed because sanest way to handle this is indeed to use goto.
        – Purple Ice
        2 days ago










      • @PurpleIce: You are probably right for simple cases like the OP's. But the moment we have several such things woven into each other using goto is far to error prone. And yes, this latter case could be considered a design issue.
        – alk
        2 days ago








      • 3




        @alk The goto solution scales linearly to any complexity, as shown in other answers, just like this one, but with less clutter and extraneous loops and branches.
        – Acorn
        2 days ago






      • 3




        The do { ... } while (0) used here is just an obfuscated way of writing a goto. There’s no advantage at all compared to using goto, and it’s quite a bit harder to read.
        – NobodyNada
        2 days ago











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      14 Answers
      14






      active

      oldest

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      14 Answers
      14






      active

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      active

      oldest

      votes








      up vote
      26
      down vote













      Yes, it's quite common to use goto in such cases to avoid repeating yourself.



      An example:



      int hello() {
      int result;

      if (Do1()) { result = 1; goto err_one; }
      if (Do2()) { result = 2; goto err_two; }
      if (Do3()) { result = 3; goto err_three; }
      if (Do4()) { result = 4; goto err_four; }
      if (Do5()) { result = 5; goto err_five; }

      // Assuming you'd like to return 0 on success.
      return 0;

      err_five:
      Undo4();
      err_four:
      Undo3();
      err_three:
      Undo2();
      err_two:
      Undo1();
      err_one:
      printf("Failed %i", result);
      return result;
      }


      As always you probably also want to keep your functions small and batch together the operations in a meaningful manner to avoid a large "undo-code".






      share|improve this answer



















      • 1




        Note: return 0 may or may not be needed, depending on what the function is supposed to do.
        – Acorn
        2 days ago






      • 1




        @Acorn Without, code wouldn't be equivalent to code presented in question, which undoes only on error...
        – Aconcagua
        2 days ago






      • 3




        @Lundin I agree. This answer assumes that the Do and Undo functions might have different signatures in reality which is probably more common in practice.
        – likle
        2 days ago






      • 10




        @Lundin There is no code repetition here. Please, don't misunderstand generic examples with concrete ones.
        – Acorn
        2 days ago






      • 10




        @Lundin It seems you don't understand how this approach works. If you have to add another case, there is no need to change anything else. That is the entire point.
        – Acorn
        2 days ago

















      up vote
      26
      down vote













      Yes, it's quite common to use goto in such cases to avoid repeating yourself.



      An example:



      int hello() {
      int result;

      if (Do1()) { result = 1; goto err_one; }
      if (Do2()) { result = 2; goto err_two; }
      if (Do3()) { result = 3; goto err_three; }
      if (Do4()) { result = 4; goto err_four; }
      if (Do5()) { result = 5; goto err_five; }

      // Assuming you'd like to return 0 on success.
      return 0;

      err_five:
      Undo4();
      err_four:
      Undo3();
      err_three:
      Undo2();
      err_two:
      Undo1();
      err_one:
      printf("Failed %i", result);
      return result;
      }


      As always you probably also want to keep your functions small and batch together the operations in a meaningful manner to avoid a large "undo-code".






      share|improve this answer



















      • 1




        Note: return 0 may or may not be needed, depending on what the function is supposed to do.
        – Acorn
        2 days ago






      • 1




        @Acorn Without, code wouldn't be equivalent to code presented in question, which undoes only on error...
        – Aconcagua
        2 days ago






      • 3




        @Lundin I agree. This answer assumes that the Do and Undo functions might have different signatures in reality which is probably more common in practice.
        – likle
        2 days ago






      • 10




        @Lundin There is no code repetition here. Please, don't misunderstand generic examples with concrete ones.
        – Acorn
        2 days ago






      • 10




        @Lundin It seems you don't understand how this approach works. If you have to add another case, there is no need to change anything else. That is the entire point.
        – Acorn
        2 days ago















      up vote
      26
      down vote










      up vote
      26
      down vote









      Yes, it's quite common to use goto in such cases to avoid repeating yourself.



      An example:



      int hello() {
      int result;

      if (Do1()) { result = 1; goto err_one; }
      if (Do2()) { result = 2; goto err_two; }
      if (Do3()) { result = 3; goto err_three; }
      if (Do4()) { result = 4; goto err_four; }
      if (Do5()) { result = 5; goto err_five; }

      // Assuming you'd like to return 0 on success.
      return 0;

      err_five:
      Undo4();
      err_four:
      Undo3();
      err_three:
      Undo2();
      err_two:
      Undo1();
      err_one:
      printf("Failed %i", result);
      return result;
      }


      As always you probably also want to keep your functions small and batch together the operations in a meaningful manner to avoid a large "undo-code".






      share|improve this answer














      Yes, it's quite common to use goto in such cases to avoid repeating yourself.



      An example:



      int hello() {
      int result;

      if (Do1()) { result = 1; goto err_one; }
      if (Do2()) { result = 2; goto err_two; }
      if (Do3()) { result = 3; goto err_three; }
      if (Do4()) { result = 4; goto err_four; }
      if (Do5()) { result = 5; goto err_five; }

      // Assuming you'd like to return 0 on success.
      return 0;

      err_five:
      Undo4();
      err_four:
      Undo3();
      err_three:
      Undo2();
      err_two:
      Undo1();
      err_one:
      printf("Failed %i", result);
      return result;
      }


      As always you probably also want to keep your functions small and batch together the operations in a meaningful manner to avoid a large "undo-code".







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 days ago

























      answered 2 days ago









      likle

      91519




      91519








      • 1




        Note: return 0 may or may not be needed, depending on what the function is supposed to do.
        – Acorn
        2 days ago






      • 1




        @Acorn Without, code wouldn't be equivalent to code presented in question, which undoes only on error...
        – Aconcagua
        2 days ago






      • 3




        @Lundin I agree. This answer assumes that the Do and Undo functions might have different signatures in reality which is probably more common in practice.
        – likle
        2 days ago






      • 10




        @Lundin There is no code repetition here. Please, don't misunderstand generic examples with concrete ones.
        – Acorn
        2 days ago






      • 10




        @Lundin It seems you don't understand how this approach works. If you have to add another case, there is no need to change anything else. That is the entire point.
        – Acorn
        2 days ago
















      • 1




        Note: return 0 may or may not be needed, depending on what the function is supposed to do.
        – Acorn
        2 days ago






      • 1




        @Acorn Without, code wouldn't be equivalent to code presented in question, which undoes only on error...
        – Aconcagua
        2 days ago






      • 3




        @Lundin I agree. This answer assumes that the Do and Undo functions might have different signatures in reality which is probably more common in practice.
        – likle
        2 days ago






      • 10




        @Lundin There is no code repetition here. Please, don't misunderstand generic examples with concrete ones.
        – Acorn
        2 days ago






      • 10




        @Lundin It seems you don't understand how this approach works. If you have to add another case, there is no need to change anything else. That is the entire point.
        – Acorn
        2 days ago










      1




      1




      Note: return 0 may or may not be needed, depending on what the function is supposed to do.
      – Acorn
      2 days ago




      Note: return 0 may or may not be needed, depending on what the function is supposed to do.
      – Acorn
      2 days ago




      1




      1




      @Acorn Without, code wouldn't be equivalent to code presented in question, which undoes only on error...
      – Aconcagua
      2 days ago




      @Acorn Without, code wouldn't be equivalent to code presented in question, which undoes only on error...
      – Aconcagua
      2 days ago




      3




      3




      @Lundin I agree. This answer assumes that the Do and Undo functions might have different signatures in reality which is probably more common in practice.
      – likle
      2 days ago




      @Lundin I agree. This answer assumes that the Do and Undo functions might have different signatures in reality which is probably more common in practice.
      – likle
      2 days ago




      10




      10




      @Lundin There is no code repetition here. Please, don't misunderstand generic examples with concrete ones.
      – Acorn
      2 days ago




      @Lundin There is no code repetition here. Please, don't misunderstand generic examples with concrete ones.
      – Acorn
      2 days ago




      10




      10




      @Lundin It seems you don't understand how this approach works. If you have to add another case, there is no need to change anything else. That is the entire point.
      – Acorn
      2 days ago






      @Lundin It seems you don't understand how this approach works. If you have to add another case, there is no need to change anything else. That is the entire point.
      – Acorn
      2 days ago














      up vote
      9
      down vote













      If you have the same signature for your function you can do something like this:



      bool Do1(void) { printf("function %sn", __func__); return true;}
      bool Do2(void) { printf("function %sn", __func__); return true;}
      bool Do3(void) { printf("function %sn", __func__); return false;}
      bool Do4(void) { printf("function %sn", __func__); return true;}
      bool Do5(void) { printf("function %sn", __func__); return true;}

      void Undo1(void) { printf("function %sn", __func__);}
      void Undo2(void) { printf("function %sn", __func__);}
      void Undo3(void) { printf("function %sn", __func__);}
      void Undo4(void) { printf("function %sn", __func__);}
      void Undo5(void) { printf("function %sn", __func__);}


      typedef struct action {
      bool (*Do)(void);
      void (*Undo)(void);
      } action_s;


      int main(void)
      {
      action_s actions = {{Do1, Undo1},
      {Do2, Undo2},
      {Do3, Undo3},
      {Do4, Undo4},
      {Do5, Undo5},
      {NULL, NULL}};

      for (size_t i = 0; actions[i].Do; ++i) {
      if (!actions[i].Do()) {
      printf("Failed %zu.n", i + 1);
      for (int j = i - 1; j >= 0; --j) {
      actions[j].Undo();
      }
      return (i);
      }
      }

      return (0);
      }


      You can change the return of one of Do functions to see how it react :)






      share|improve this answer























      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Samuel Liew
        yesterday















      up vote
      9
      down vote













      If you have the same signature for your function you can do something like this:



      bool Do1(void) { printf("function %sn", __func__); return true;}
      bool Do2(void) { printf("function %sn", __func__); return true;}
      bool Do3(void) { printf("function %sn", __func__); return false;}
      bool Do4(void) { printf("function %sn", __func__); return true;}
      bool Do5(void) { printf("function %sn", __func__); return true;}

      void Undo1(void) { printf("function %sn", __func__);}
      void Undo2(void) { printf("function %sn", __func__);}
      void Undo3(void) { printf("function %sn", __func__);}
      void Undo4(void) { printf("function %sn", __func__);}
      void Undo5(void) { printf("function %sn", __func__);}


      typedef struct action {
      bool (*Do)(void);
      void (*Undo)(void);
      } action_s;


      int main(void)
      {
      action_s actions = {{Do1, Undo1},
      {Do2, Undo2},
      {Do3, Undo3},
      {Do4, Undo4},
      {Do5, Undo5},
      {NULL, NULL}};

      for (size_t i = 0; actions[i].Do; ++i) {
      if (!actions[i].Do()) {
      printf("Failed %zu.n", i + 1);
      for (int j = i - 1; j >= 0; --j) {
      actions[j].Undo();
      }
      return (i);
      }
      }

      return (0);
      }


      You can change the return of one of Do functions to see how it react :)






      share|improve this answer























      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Samuel Liew
        yesterday













      up vote
      9
      down vote










      up vote
      9
      down vote









      If you have the same signature for your function you can do something like this:



      bool Do1(void) { printf("function %sn", __func__); return true;}
      bool Do2(void) { printf("function %sn", __func__); return true;}
      bool Do3(void) { printf("function %sn", __func__); return false;}
      bool Do4(void) { printf("function %sn", __func__); return true;}
      bool Do5(void) { printf("function %sn", __func__); return true;}

      void Undo1(void) { printf("function %sn", __func__);}
      void Undo2(void) { printf("function %sn", __func__);}
      void Undo3(void) { printf("function %sn", __func__);}
      void Undo4(void) { printf("function %sn", __func__);}
      void Undo5(void) { printf("function %sn", __func__);}


      typedef struct action {
      bool (*Do)(void);
      void (*Undo)(void);
      } action_s;


      int main(void)
      {
      action_s actions = {{Do1, Undo1},
      {Do2, Undo2},
      {Do3, Undo3},
      {Do4, Undo4},
      {Do5, Undo5},
      {NULL, NULL}};

      for (size_t i = 0; actions[i].Do; ++i) {
      if (!actions[i].Do()) {
      printf("Failed %zu.n", i + 1);
      for (int j = i - 1; j >= 0; --j) {
      actions[j].Undo();
      }
      return (i);
      }
      }

      return (0);
      }


      You can change the return of one of Do functions to see how it react :)






      share|improve this answer














      If you have the same signature for your function you can do something like this:



      bool Do1(void) { printf("function %sn", __func__); return true;}
      bool Do2(void) { printf("function %sn", __func__); return true;}
      bool Do3(void) { printf("function %sn", __func__); return false;}
      bool Do4(void) { printf("function %sn", __func__); return true;}
      bool Do5(void) { printf("function %sn", __func__); return true;}

      void Undo1(void) { printf("function %sn", __func__);}
      void Undo2(void) { printf("function %sn", __func__);}
      void Undo3(void) { printf("function %sn", __func__);}
      void Undo4(void) { printf("function %sn", __func__);}
      void Undo5(void) { printf("function %sn", __func__);}


      typedef struct action {
      bool (*Do)(void);
      void (*Undo)(void);
      } action_s;


      int main(void)
      {
      action_s actions = {{Do1, Undo1},
      {Do2, Undo2},
      {Do3, Undo3},
      {Do4, Undo4},
      {Do5, Undo5},
      {NULL, NULL}};

      for (size_t i = 0; actions[i].Do; ++i) {
      if (!actions[i].Do()) {
      printf("Failed %zu.n", i + 1);
      for (int j = i - 1; j >= 0; --j) {
      actions[j].Undo();
      }
      return (i);
      }
      }

      return (0);
      }


      You can change the return of one of Do functions to see how it react :)







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 days ago









      Peter Mortensen

      13.3k1983111




      13.3k1983111










      answered 2 days ago









      Tom's

      1,842419




      1,842419












      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Samuel Liew
        yesterday


















      • Comments are not for extended discussion; this conversation has been moved to chat.
        – Samuel Liew
        yesterday
















      Comments are not for extended discussion; this conversation has been moved to chat.
      – Samuel Liew
      yesterday




      Comments are not for extended discussion; this conversation has been moved to chat.
      – Samuel Liew
      yesterday










      up vote
      8
      down vote














      This might be one case for using gotos.




      Sure, let's try that. Here's a possible implementation:



      #include "stdio.h"
      int main(int argc, char **argv) {
      int errorCode = 0;
      if (Do1()) { errorCode = 1; goto undo_0; }
      if (Do2()) { errorCode = 2; goto undo_1; }
      if (Do3()) { errorCode = 3; goto undo_2; }
      if (Do4()) { errorCode = 4; goto undo_3; }
      if (Do5()) { errorCode = 5; goto undo_4; }

      undo_5: Undo5(); /* deliberate fallthrough */
      undo_4: Undo4();
      undo_3: Undo3();
      undo_2: Undo2();
      undo_1: Undo1();
      undo_0: /* nothing to undo in this case */

      if (errorCode != 0) {
      printf("Failed %dn", errorCode);
      return errorCode;
      }
      return 0;
      }





      share|improve this answer



















      • 2




        The -1 is probablky from someone who considers goto as absolutely evil, which it is not as long as it's abused.
        – Jabberwocky
        2 days ago






      • 1




        @Aconcagua It is rare to be able to factorize the printf, and even if you can, it is not typically done. In a "generic" example like this, you shouldn't write it like that, because beginners will think it is the common way of doing it, in my opinion. Also, there is the point about the initialization to -1 and also not talking about cases where you may have to return early (i.e. before the Undos). Finally, better answers are currently lower than this one.
        – Acorn
        2 days ago








      • 2




        @Lundin It isn't.
        – Acorn
        2 days ago






      • 1




        @Acorn The error behavior of every single error code is tightly coupled with the error behavior of every other error code. So if you add something in the middle during maintenance, it all comes crashing down.
        – Lundin
        2 days ago






      • 3




        @Lundin It isn't tied, at all. You are making assumptions, and they do not even correspond to actual code out there.
        – Acorn
        2 days ago















      up vote
      8
      down vote














      This might be one case for using gotos.




      Sure, let's try that. Here's a possible implementation:



      #include "stdio.h"
      int main(int argc, char **argv) {
      int errorCode = 0;
      if (Do1()) { errorCode = 1; goto undo_0; }
      if (Do2()) { errorCode = 2; goto undo_1; }
      if (Do3()) { errorCode = 3; goto undo_2; }
      if (Do4()) { errorCode = 4; goto undo_3; }
      if (Do5()) { errorCode = 5; goto undo_4; }

      undo_5: Undo5(); /* deliberate fallthrough */
      undo_4: Undo4();
      undo_3: Undo3();
      undo_2: Undo2();
      undo_1: Undo1();
      undo_0: /* nothing to undo in this case */

      if (errorCode != 0) {
      printf("Failed %dn", errorCode);
      return errorCode;
      }
      return 0;
      }





      share|improve this answer



















      • 2




        The -1 is probablky from someone who considers goto as absolutely evil, which it is not as long as it's abused.
        – Jabberwocky
        2 days ago






      • 1




        @Aconcagua It is rare to be able to factorize the printf, and even if you can, it is not typically done. In a "generic" example like this, you shouldn't write it like that, because beginners will think it is the common way of doing it, in my opinion. Also, there is the point about the initialization to -1 and also not talking about cases where you may have to return early (i.e. before the Undos). Finally, better answers are currently lower than this one.
        – Acorn
        2 days ago








      • 2




        @Lundin It isn't.
        – Acorn
        2 days ago






      • 1




        @Acorn The error behavior of every single error code is tightly coupled with the error behavior of every other error code. So if you add something in the middle during maintenance, it all comes crashing down.
        – Lundin
        2 days ago






      • 3




        @Lundin It isn't tied, at all. You are making assumptions, and they do not even correspond to actual code out there.
        – Acorn
        2 days ago













      up vote
      8
      down vote










      up vote
      8
      down vote










      This might be one case for using gotos.




      Sure, let's try that. Here's a possible implementation:



      #include "stdio.h"
      int main(int argc, char **argv) {
      int errorCode = 0;
      if (Do1()) { errorCode = 1; goto undo_0; }
      if (Do2()) { errorCode = 2; goto undo_1; }
      if (Do3()) { errorCode = 3; goto undo_2; }
      if (Do4()) { errorCode = 4; goto undo_3; }
      if (Do5()) { errorCode = 5; goto undo_4; }

      undo_5: Undo5(); /* deliberate fallthrough */
      undo_4: Undo4();
      undo_3: Undo3();
      undo_2: Undo2();
      undo_1: Undo1();
      undo_0: /* nothing to undo in this case */

      if (errorCode != 0) {
      printf("Failed %dn", errorCode);
      return errorCode;
      }
      return 0;
      }





      share|improve this answer















      This might be one case for using gotos.




      Sure, let's try that. Here's a possible implementation:



      #include "stdio.h"
      int main(int argc, char **argv) {
      int errorCode = 0;
      if (Do1()) { errorCode = 1; goto undo_0; }
      if (Do2()) { errorCode = 2; goto undo_1; }
      if (Do3()) { errorCode = 3; goto undo_2; }
      if (Do4()) { errorCode = 4; goto undo_3; }
      if (Do5()) { errorCode = 5; goto undo_4; }

      undo_5: Undo5(); /* deliberate fallthrough */
      undo_4: Undo4();
      undo_3: Undo3();
      undo_2: Undo2();
      undo_1: Undo1();
      undo_0: /* nothing to undo in this case */

      if (errorCode != 0) {
      printf("Failed %dn", errorCode);
      return errorCode;
      }
      return 0;
      }






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 days ago

























      answered 2 days ago









      Blaze

      2,8911524




      2,8911524








      • 2




        The -1 is probablky from someone who considers goto as absolutely evil, which it is not as long as it's abused.
        – Jabberwocky
        2 days ago






      • 1




        @Aconcagua It is rare to be able to factorize the printf, and even if you can, it is not typically done. In a "generic" example like this, you shouldn't write it like that, because beginners will think it is the common way of doing it, in my opinion. Also, there is the point about the initialization to -1 and also not talking about cases where you may have to return early (i.e. before the Undos). Finally, better answers are currently lower than this one.
        – Acorn
        2 days ago








      • 2




        @Lundin It isn't.
        – Acorn
        2 days ago






      • 1




        @Acorn The error behavior of every single error code is tightly coupled with the error behavior of every other error code. So if you add something in the middle during maintenance, it all comes crashing down.
        – Lundin
        2 days ago






      • 3




        @Lundin It isn't tied, at all. You are making assumptions, and they do not even correspond to actual code out there.
        – Acorn
        2 days ago














      • 2




        The -1 is probablky from someone who considers goto as absolutely evil, which it is not as long as it's abused.
        – Jabberwocky
        2 days ago






      • 1




        @Aconcagua It is rare to be able to factorize the printf, and even if you can, it is not typically done. In a "generic" example like this, you shouldn't write it like that, because beginners will think it is the common way of doing it, in my opinion. Also, there is the point about the initialization to -1 and also not talking about cases where you may have to return early (i.e. before the Undos). Finally, better answers are currently lower than this one.
        – Acorn
        2 days ago








      • 2




        @Lundin It isn't.
        – Acorn
        2 days ago






      • 1




        @Acorn The error behavior of every single error code is tightly coupled with the error behavior of every other error code. So if you add something in the middle during maintenance, it all comes crashing down.
        – Lundin
        2 days ago






      • 3




        @Lundin It isn't tied, at all. You are making assumptions, and they do not even correspond to actual code out there.
        – Acorn
        2 days ago








      2




      2




      The -1 is probablky from someone who considers goto as absolutely evil, which it is not as long as it's abused.
      – Jabberwocky
      2 days ago




      The -1 is probablky from someone who considers goto as absolutely evil, which it is not as long as it's abused.
      – Jabberwocky
      2 days ago




      1




      1




      @Aconcagua It is rare to be able to factorize the printf, and even if you can, it is not typically done. In a "generic" example like this, you shouldn't write it like that, because beginners will think it is the common way of doing it, in my opinion. Also, there is the point about the initialization to -1 and also not talking about cases where you may have to return early (i.e. before the Undos). Finally, better answers are currently lower than this one.
      – Acorn
      2 days ago






      @Aconcagua It is rare to be able to factorize the printf, and even if you can, it is not typically done. In a "generic" example like this, you shouldn't write it like that, because beginners will think it is the common way of doing it, in my opinion. Also, there is the point about the initialization to -1 and also not talking about cases where you may have to return early (i.e. before the Undos). Finally, better answers are currently lower than this one.
      – Acorn
      2 days ago






      2




      2




      @Lundin It isn't.
      – Acorn
      2 days ago




      @Lundin It isn't.
      – Acorn
      2 days ago




      1




      1




      @Acorn The error behavior of every single error code is tightly coupled with the error behavior of every other error code. So if you add something in the middle during maintenance, it all comes crashing down.
      – Lundin
      2 days ago




      @Acorn The error behavior of every single error code is tightly coupled with the error behavior of every other error code. So if you add something in the middle during maintenance, it all comes crashing down.
      – Lundin
      2 days ago




      3




      3




      @Lundin It isn't tied, at all. You are making assumptions, and they do not even correspond to actual code out there.
      – Acorn
      2 days ago




      @Lundin It isn't tied, at all. You are making assumptions, and they do not even correspond to actual code out there.
      – Acorn
      2 days ago










      up vote
      7
      down vote













      For completeness a bit of obfuscation:



      int foo(void)
      {
      int rc;

      if (0
      || (rc = 1, do1())
      || (rc = 2, do2())
      || (rc = 3, do3())
      || (rc = 4, do4())
      || (rc = 5, do5())
      || (rc = 0)
      )
      {
      /* More or less stolen from Chris' answer:
      https://stackoverflow.com/a/53444967/694576) */
      switch(rc - 1)
      {
      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
      undo5();

      case 4:
      undo4();

      case 3:
      undo3();

      case 2:
      undo2();

      case 1:
      undo1();

      default:
      break;
      }
      }

      return rc;
      }





      share|improve this answer



















      • 5




        Please, stop this madness.
        – Acorn
        2 days ago






      • 3




        @Acorn: Why? Nice example for how to use the comma-operator ... ;-)
        – alk
        2 days ago






      • 1




        @Acorn: Also nicely symmetrical and compact, perfect to auto-generate.
        – alk
        2 days ago








      • 2




        The other solutions are just as easy to auto-generate (if you are talking about code generation).
        – Acorn
        2 days ago








      • 13




        Now all we are missing is a version with Duff's device and we'll be ready to submit this thread to IOCCC :)
        – Lundin
        2 days ago















      up vote
      7
      down vote













      For completeness a bit of obfuscation:



      int foo(void)
      {
      int rc;

      if (0
      || (rc = 1, do1())
      || (rc = 2, do2())
      || (rc = 3, do3())
      || (rc = 4, do4())
      || (rc = 5, do5())
      || (rc = 0)
      )
      {
      /* More or less stolen from Chris' answer:
      https://stackoverflow.com/a/53444967/694576) */
      switch(rc - 1)
      {
      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
      undo5();

      case 4:
      undo4();

      case 3:
      undo3();

      case 2:
      undo2();

      case 1:
      undo1();

      default:
      break;
      }
      }

      return rc;
      }





      share|improve this answer



















      • 5




        Please, stop this madness.
        – Acorn
        2 days ago






      • 3




        @Acorn: Why? Nice example for how to use the comma-operator ... ;-)
        – alk
        2 days ago






      • 1




        @Acorn: Also nicely symmetrical and compact, perfect to auto-generate.
        – alk
        2 days ago








      • 2




        The other solutions are just as easy to auto-generate (if you are talking about code generation).
        – Acorn
        2 days ago








      • 13




        Now all we are missing is a version with Duff's device and we'll be ready to submit this thread to IOCCC :)
        – Lundin
        2 days ago













      up vote
      7
      down vote










      up vote
      7
      down vote









      For completeness a bit of obfuscation:



      int foo(void)
      {
      int rc;

      if (0
      || (rc = 1, do1())
      || (rc = 2, do2())
      || (rc = 3, do3())
      || (rc = 4, do4())
      || (rc = 5, do5())
      || (rc = 0)
      )
      {
      /* More or less stolen from Chris' answer:
      https://stackoverflow.com/a/53444967/694576) */
      switch(rc - 1)
      {
      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
      undo5();

      case 4:
      undo4();

      case 3:
      undo3();

      case 2:
      undo2();

      case 1:
      undo1();

      default:
      break;
      }
      }

      return rc;
      }





      share|improve this answer














      For completeness a bit of obfuscation:



      int foo(void)
      {
      int rc;

      if (0
      || (rc = 1, do1())
      || (rc = 2, do2())
      || (rc = 3, do3())
      || (rc = 4, do4())
      || (rc = 5, do5())
      || (rc = 0)
      )
      {
      /* More or less stolen from Chris' answer:
      https://stackoverflow.com/a/53444967/694576) */
      switch(rc - 1)
      {
      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
      undo5();

      case 4:
      undo4();

      case 3:
      undo3();

      case 2:
      undo2();

      case 1:
      undo1();

      default:
      break;
      }
      }

      return rc;
      }






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 days ago

























      answered 2 days ago









      alk

      57.7k759167




      57.7k759167








      • 5




        Please, stop this madness.
        – Acorn
        2 days ago






      • 3




        @Acorn: Why? Nice example for how to use the comma-operator ... ;-)
        – alk
        2 days ago






      • 1




        @Acorn: Also nicely symmetrical and compact, perfect to auto-generate.
        – alk
        2 days ago








      • 2




        The other solutions are just as easy to auto-generate (if you are talking about code generation).
        – Acorn
        2 days ago








      • 13




        Now all we are missing is a version with Duff's device and we'll be ready to submit this thread to IOCCC :)
        – Lundin
        2 days ago














      • 5




        Please, stop this madness.
        – Acorn
        2 days ago






      • 3




        @Acorn: Why? Nice example for how to use the comma-operator ... ;-)
        – alk
        2 days ago






      • 1




        @Acorn: Also nicely symmetrical and compact, perfect to auto-generate.
        – alk
        2 days ago








      • 2




        The other solutions are just as easy to auto-generate (if you are talking about code generation).
        – Acorn
        2 days ago








      • 13




        Now all we are missing is a version with Duff's device and we'll be ready to submit this thread to IOCCC :)
        – Lundin
        2 days ago








      5




      5




      Please, stop this madness.
      – Acorn
      2 days ago




      Please, stop this madness.
      – Acorn
      2 days ago




      3




      3




      @Acorn: Why? Nice example for how to use the comma-operator ... ;-)
      – alk
      2 days ago




      @Acorn: Why? Nice example for how to use the comma-operator ... ;-)
      – alk
      2 days ago




      1




      1




      @Acorn: Also nicely symmetrical and compact, perfect to auto-generate.
      – alk
      2 days ago






      @Acorn: Also nicely symmetrical and compact, perfect to auto-generate.
      – alk
      2 days ago






      2




      2




      The other solutions are just as easy to auto-generate (if you are talking about code generation).
      – Acorn
      2 days ago






      The other solutions are just as easy to auto-generate (if you are talking about code generation).
      – Acorn
      2 days ago






      13




      13




      Now all we are missing is a version with Duff's device and we'll be ready to submit this thread to IOCCC :)
      – Lundin
      2 days ago




      Now all we are missing is a version with Duff's device and we'll be ready to submit this thread to IOCCC :)
      – Lundin
      2 days ago










      up vote
      3
      down vote













      Use goto to manage cleanup in C.



      For instance, check the Linux kernel coding style:




      The rationale for using gotos is:




      • unconditional statements are easier to understand and follow nesting is reduced

      • errors by not updating individual exit points when making modifications are prevented

      • saves the compiler work to optimize redundant code away ;)


      Example:



      int fun(int a)
      {
      int result = 0;
      char *buffer;

      buffer = kmalloc(SIZE, GFP_KERNEL);
      if (!buffer)
      return -ENOMEM;

      if (condition1) {
      while (loop1) {
      ...
      }
      result = 1;
      goto out_free_buffer;
      }

      ...

      out_free_buffer:
      kfree(buffer);
      return result;
      }





      In your particular case, it could look like:



      int f(...)
      {
      int ret;

      if (Do1()) {
      printf("Failed 1");
      ret = 1;
      goto undo1;
      }

      ...

      if (Do5()) {
      printf("Failed 5");
      ret = 5;
      goto undo5;
      }

      // all good, return here if you need to keep the resources
      // (or not, if you want them deallocated; in that case initialize `ret`)
      return 0;

      undo5:
      Undo4();
      ...
      undo1:
      return ret;
      }





      share|improve this answer



















      • 2




        This is somewhat acceptable use of goto - it is a pattern from BASIC known as "on error goto". I wish people wouldn't stop thinking there, but think one step further still. The better alternative to "on error goto" is to use a wrapper function and from the inner function return code; upon error. Leave resource allocation and clean-up to the outer wrapper. Thus separate resource allocation and algorithm in 2 different functions. Much cleaner design, easier to read and no goto debate. In general, I would recommend staying away from "the Linux kernel coding style" document.
        – Lundin
        2 days ago












      • @Lundin Not sure, are you now referring Tom's function pointer solution (your description sounds somehow different to me...)? If not, how would then these wrapper functions look like? Cannot think of anything better than int callDo2(void) { if(do2()) { undo1(); return 2; } return callDo3(); } at the moment, but cannot imagine either that you really meant such ones...
        – Aconcagua
        2 days ago








      • 1




        Basically replace all your goto with return, then in the outer wrapper function call "undo" based on what the function returned.
        – Lundin
        2 days ago










      • @Lundin Please show an example, because it does not sound like a good idea at all.
        – Acorn
        2 days ago















      up vote
      3
      down vote













      Use goto to manage cleanup in C.



      For instance, check the Linux kernel coding style:




      The rationale for using gotos is:




      • unconditional statements are easier to understand and follow nesting is reduced

      • errors by not updating individual exit points when making modifications are prevented

      • saves the compiler work to optimize redundant code away ;)


      Example:



      int fun(int a)
      {
      int result = 0;
      char *buffer;

      buffer = kmalloc(SIZE, GFP_KERNEL);
      if (!buffer)
      return -ENOMEM;

      if (condition1) {
      while (loop1) {
      ...
      }
      result = 1;
      goto out_free_buffer;
      }

      ...

      out_free_buffer:
      kfree(buffer);
      return result;
      }





      In your particular case, it could look like:



      int f(...)
      {
      int ret;

      if (Do1()) {
      printf("Failed 1");
      ret = 1;
      goto undo1;
      }

      ...

      if (Do5()) {
      printf("Failed 5");
      ret = 5;
      goto undo5;
      }

      // all good, return here if you need to keep the resources
      // (or not, if you want them deallocated; in that case initialize `ret`)
      return 0;

      undo5:
      Undo4();
      ...
      undo1:
      return ret;
      }





      share|improve this answer



















      • 2




        This is somewhat acceptable use of goto - it is a pattern from BASIC known as "on error goto". I wish people wouldn't stop thinking there, but think one step further still. The better alternative to "on error goto" is to use a wrapper function and from the inner function return code; upon error. Leave resource allocation and clean-up to the outer wrapper. Thus separate resource allocation and algorithm in 2 different functions. Much cleaner design, easier to read and no goto debate. In general, I would recommend staying away from "the Linux kernel coding style" document.
        – Lundin
        2 days ago












      • @Lundin Not sure, are you now referring Tom's function pointer solution (your description sounds somehow different to me...)? If not, how would then these wrapper functions look like? Cannot think of anything better than int callDo2(void) { if(do2()) { undo1(); return 2; } return callDo3(); } at the moment, but cannot imagine either that you really meant such ones...
        – Aconcagua
        2 days ago








      • 1




        Basically replace all your goto with return, then in the outer wrapper function call "undo" based on what the function returned.
        – Lundin
        2 days ago










      • @Lundin Please show an example, because it does not sound like a good idea at all.
        – Acorn
        2 days ago













      up vote
      3
      down vote










      up vote
      3
      down vote









      Use goto to manage cleanup in C.



      For instance, check the Linux kernel coding style:




      The rationale for using gotos is:




      • unconditional statements are easier to understand and follow nesting is reduced

      • errors by not updating individual exit points when making modifications are prevented

      • saves the compiler work to optimize redundant code away ;)


      Example:



      int fun(int a)
      {
      int result = 0;
      char *buffer;

      buffer = kmalloc(SIZE, GFP_KERNEL);
      if (!buffer)
      return -ENOMEM;

      if (condition1) {
      while (loop1) {
      ...
      }
      result = 1;
      goto out_free_buffer;
      }

      ...

      out_free_buffer:
      kfree(buffer);
      return result;
      }





      In your particular case, it could look like:



      int f(...)
      {
      int ret;

      if (Do1()) {
      printf("Failed 1");
      ret = 1;
      goto undo1;
      }

      ...

      if (Do5()) {
      printf("Failed 5");
      ret = 5;
      goto undo5;
      }

      // all good, return here if you need to keep the resources
      // (or not, if you want them deallocated; in that case initialize `ret`)
      return 0;

      undo5:
      Undo4();
      ...
      undo1:
      return ret;
      }





      share|improve this answer














      Use goto to manage cleanup in C.



      For instance, check the Linux kernel coding style:




      The rationale for using gotos is:




      • unconditional statements are easier to understand and follow nesting is reduced

      • errors by not updating individual exit points when making modifications are prevented

      • saves the compiler work to optimize redundant code away ;)


      Example:



      int fun(int a)
      {
      int result = 0;
      char *buffer;

      buffer = kmalloc(SIZE, GFP_KERNEL);
      if (!buffer)
      return -ENOMEM;

      if (condition1) {
      while (loop1) {
      ...
      }
      result = 1;
      goto out_free_buffer;
      }

      ...

      out_free_buffer:
      kfree(buffer);
      return result;
      }





      In your particular case, it could look like:



      int f(...)
      {
      int ret;

      if (Do1()) {
      printf("Failed 1");
      ret = 1;
      goto undo1;
      }

      ...

      if (Do5()) {
      printf("Failed 5");
      ret = 5;
      goto undo5;
      }

      // all good, return here if you need to keep the resources
      // (or not, if you want them deallocated; in that case initialize `ret`)
      return 0;

      undo5:
      Undo4();
      ...
      undo1:
      return ret;
      }






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 days ago

























      answered 2 days ago









      Acorn

      4,67311135




      4,67311135








      • 2




        This is somewhat acceptable use of goto - it is a pattern from BASIC known as "on error goto". I wish people wouldn't stop thinking there, but think one step further still. The better alternative to "on error goto" is to use a wrapper function and from the inner function return code; upon error. Leave resource allocation and clean-up to the outer wrapper. Thus separate resource allocation and algorithm in 2 different functions. Much cleaner design, easier to read and no goto debate. In general, I would recommend staying away from "the Linux kernel coding style" document.
        – Lundin
        2 days ago












      • @Lundin Not sure, are you now referring Tom's function pointer solution (your description sounds somehow different to me...)? If not, how would then these wrapper functions look like? Cannot think of anything better than int callDo2(void) { if(do2()) { undo1(); return 2; } return callDo3(); } at the moment, but cannot imagine either that you really meant such ones...
        – Aconcagua
        2 days ago








      • 1




        Basically replace all your goto with return, then in the outer wrapper function call "undo" based on what the function returned.
        – Lundin
        2 days ago










      • @Lundin Please show an example, because it does not sound like a good idea at all.
        – Acorn
        2 days ago














      • 2




        This is somewhat acceptable use of goto - it is a pattern from BASIC known as "on error goto". I wish people wouldn't stop thinking there, but think one step further still. The better alternative to "on error goto" is to use a wrapper function and from the inner function return code; upon error. Leave resource allocation and clean-up to the outer wrapper. Thus separate resource allocation and algorithm in 2 different functions. Much cleaner design, easier to read and no goto debate. In general, I would recommend staying away from "the Linux kernel coding style" document.
        – Lundin
        2 days ago












      • @Lundin Not sure, are you now referring Tom's function pointer solution (your description sounds somehow different to me...)? If not, how would then these wrapper functions look like? Cannot think of anything better than int callDo2(void) { if(do2()) { undo1(); return 2; } return callDo3(); } at the moment, but cannot imagine either that you really meant such ones...
        – Aconcagua
        2 days ago








      • 1




        Basically replace all your goto with return, then in the outer wrapper function call "undo" based on what the function returned.
        – Lundin
        2 days ago










      • @Lundin Please show an example, because it does not sound like a good idea at all.
        – Acorn
        2 days ago








      2




      2




      This is somewhat acceptable use of goto - it is a pattern from BASIC known as "on error goto". I wish people wouldn't stop thinking there, but think one step further still. The better alternative to "on error goto" is to use a wrapper function and from the inner function return code; upon error. Leave resource allocation and clean-up to the outer wrapper. Thus separate resource allocation and algorithm in 2 different functions. Much cleaner design, easier to read and no goto debate. In general, I would recommend staying away from "the Linux kernel coding style" document.
      – Lundin
      2 days ago






      This is somewhat acceptable use of goto - it is a pattern from BASIC known as "on error goto". I wish people wouldn't stop thinking there, but think one step further still. The better alternative to "on error goto" is to use a wrapper function and from the inner function return code; upon error. Leave resource allocation and clean-up to the outer wrapper. Thus separate resource allocation and algorithm in 2 different functions. Much cleaner design, easier to read and no goto debate. In general, I would recommend staying away from "the Linux kernel coding style" document.
      – Lundin
      2 days ago














      @Lundin Not sure, are you now referring Tom's function pointer solution (your description sounds somehow different to me...)? If not, how would then these wrapper functions look like? Cannot think of anything better than int callDo2(void) { if(do2()) { undo1(); return 2; } return callDo3(); } at the moment, but cannot imagine either that you really meant such ones...
      – Aconcagua
      2 days ago






      @Lundin Not sure, are you now referring Tom's function pointer solution (your description sounds somehow different to me...)? If not, how would then these wrapper functions look like? Cannot think of anything better than int callDo2(void) { if(do2()) { undo1(); return 2; } return callDo3(); } at the moment, but cannot imagine either that you really meant such ones...
      – Aconcagua
      2 days ago






      1




      1




      Basically replace all your goto with return, then in the outer wrapper function call "undo" based on what the function returned.
      – Lundin
      2 days ago




      Basically replace all your goto with return, then in the outer wrapper function call "undo" based on what the function returned.
      – Lundin
      2 days ago












      @Lundin Please show an example, because it does not sound like a good idea at all.
      – Acorn
      2 days ago




      @Lundin Please show an example, because it does not sound like a good idea at all.
      – Acorn
      2 days ago










      up vote
      2
      down vote













      There are probably many ways to do this, but one idea is since you won't call one function unless the preceeding one succeeded, you could chain your function calls using else if like this. And using a variable to track where it fails you can use a switch statement to roll back easily too.



      int ret=0;
      if(Do1()) {
      ret=1;
      } else if(Do2()) {
      ret=2;
      } else if(Do3()) {
      ret=3;
      } else if(Do4()) {
      ret=4;
      } else if(Do5()) {
      ret=5;
      }

      switch(ret) {
      case 5:
      Undo4();
      case 4:
      Undo3();
      case 3:
      Undo2();
      case 2:
      Undo1();
      case 1:
      printf("Failed %dn",ret);
      break;
      }
      return ret;





      share|improve this answer



















      • 8




        Don't do this. The code is harder to read and doing more branches compared to simple goto.
        – Acorn
        2 days ago






      • 4




        Also, note it is wrong: if Do5() fails, we don't want to run Undo5() (typically).
        – Acorn
        2 days ago






      • 1




        switch(ret) should be switch(ret-1). Also an (emtpy) default case would be nice. All in all I like this approach.
        – alk
        2 days ago












      • @alk Writing the actual correct values would be better -- if you are keen on using this solution, which you should not ;) As for the default case, what for?
        – Acorn
        2 days ago






      • 6




        The lengths people will go to avoid uncoditional jump which was basically kept in C language exactly for such cases because it makes it more readable than arrow antipattern or switch/if ladder is amazing... Take my downvote.
        – Purple Ice
        2 days ago

















      up vote
      2
      down vote













      There are probably many ways to do this, but one idea is since you won't call one function unless the preceeding one succeeded, you could chain your function calls using else if like this. And using a variable to track where it fails you can use a switch statement to roll back easily too.



      int ret=0;
      if(Do1()) {
      ret=1;
      } else if(Do2()) {
      ret=2;
      } else if(Do3()) {
      ret=3;
      } else if(Do4()) {
      ret=4;
      } else if(Do5()) {
      ret=5;
      }

      switch(ret) {
      case 5:
      Undo4();
      case 4:
      Undo3();
      case 3:
      Undo2();
      case 2:
      Undo1();
      case 1:
      printf("Failed %dn",ret);
      break;
      }
      return ret;





      share|improve this answer



















      • 8




        Don't do this. The code is harder to read and doing more branches compared to simple goto.
        – Acorn
        2 days ago






      • 4




        Also, note it is wrong: if Do5() fails, we don't want to run Undo5() (typically).
        – Acorn
        2 days ago






      • 1




        switch(ret) should be switch(ret-1). Also an (emtpy) default case would be nice. All in all I like this approach.
        – alk
        2 days ago












      • @alk Writing the actual correct values would be better -- if you are keen on using this solution, which you should not ;) As for the default case, what for?
        – Acorn
        2 days ago






      • 6




        The lengths people will go to avoid uncoditional jump which was basically kept in C language exactly for such cases because it makes it more readable than arrow antipattern or switch/if ladder is amazing... Take my downvote.
        – Purple Ice
        2 days ago















      up vote
      2
      down vote










      up vote
      2
      down vote









      There are probably many ways to do this, but one idea is since you won't call one function unless the preceeding one succeeded, you could chain your function calls using else if like this. And using a variable to track where it fails you can use a switch statement to roll back easily too.



      int ret=0;
      if(Do1()) {
      ret=1;
      } else if(Do2()) {
      ret=2;
      } else if(Do3()) {
      ret=3;
      } else if(Do4()) {
      ret=4;
      } else if(Do5()) {
      ret=5;
      }

      switch(ret) {
      case 5:
      Undo4();
      case 4:
      Undo3();
      case 3:
      Undo2();
      case 2:
      Undo1();
      case 1:
      printf("Failed %dn",ret);
      break;
      }
      return ret;





      share|improve this answer














      There are probably many ways to do this, but one idea is since you won't call one function unless the preceeding one succeeded, you could chain your function calls using else if like this. And using a variable to track where it fails you can use a switch statement to roll back easily too.



      int ret=0;
      if(Do1()) {
      ret=1;
      } else if(Do2()) {
      ret=2;
      } else if(Do3()) {
      ret=3;
      } else if(Do4()) {
      ret=4;
      } else if(Do5()) {
      ret=5;
      }

      switch(ret) {
      case 5:
      Undo4();
      case 4:
      Undo3();
      case 3:
      Undo2();
      case 2:
      Undo1();
      case 1:
      printf("Failed %dn",ret);
      break;
      }
      return ret;






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 2 days ago

























      answered 2 days ago









      Chris Turner

      6,2521917




      6,2521917








      • 8




        Don't do this. The code is harder to read and doing more branches compared to simple goto.
        – Acorn
        2 days ago






      • 4




        Also, note it is wrong: if Do5() fails, we don't want to run Undo5() (typically).
        – Acorn
        2 days ago






      • 1




        switch(ret) should be switch(ret-1). Also an (emtpy) default case would be nice. All in all I like this approach.
        – alk
        2 days ago












      • @alk Writing the actual correct values would be better -- if you are keen on using this solution, which you should not ;) As for the default case, what for?
        – Acorn
        2 days ago






      • 6




        The lengths people will go to avoid uncoditional jump which was basically kept in C language exactly for such cases because it makes it more readable than arrow antipattern or switch/if ladder is amazing... Take my downvote.
        – Purple Ice
        2 days ago
















      • 8




        Don't do this. The code is harder to read and doing more branches compared to simple goto.
        – Acorn
        2 days ago






      • 4




        Also, note it is wrong: if Do5() fails, we don't want to run Undo5() (typically).
        – Acorn
        2 days ago






      • 1




        switch(ret) should be switch(ret-1). Also an (emtpy) default case would be nice. All in all I like this approach.
        – alk
        2 days ago












      • @alk Writing the actual correct values would be better -- if you are keen on using this solution, which you should not ;) As for the default case, what for?
        – Acorn
        2 days ago






      • 6




        The lengths people will go to avoid uncoditional jump which was basically kept in C language exactly for such cases because it makes it more readable than arrow antipattern or switch/if ladder is amazing... Take my downvote.
        – Purple Ice
        2 days ago










      8




      8




      Don't do this. The code is harder to read and doing more branches compared to simple goto.
      – Acorn
      2 days ago




      Don't do this. The code is harder to read and doing more branches compared to simple goto.
      – Acorn
      2 days ago




      4




      4




      Also, note it is wrong: if Do5() fails, we don't want to run Undo5() (typically).
      – Acorn
      2 days ago




      Also, note it is wrong: if Do5() fails, we don't want to run Undo5() (typically).
      – Acorn
      2 days ago




      1




      1




      switch(ret) should be switch(ret-1). Also an (emtpy) default case would be nice. All in all I like this approach.
      – alk
      2 days ago






      switch(ret) should be switch(ret-1). Also an (emtpy) default case would be nice. All in all I like this approach.
      – alk
      2 days ago














      @alk Writing the actual correct values would be better -- if you are keen on using this solution, which you should not ;) As for the default case, what for?
      – Acorn
      2 days ago




      @alk Writing the actual correct values would be better -- if you are keen on using this solution, which you should not ;) As for the default case, what for?
      – Acorn
      2 days ago




      6




      6




      The lengths people will go to avoid uncoditional jump which was basically kept in C language exactly for such cases because it makes it more readable than arrow antipattern or switch/if ladder is amazing... Take my downvote.
      – Purple Ice
      2 days ago






      The lengths people will go to avoid uncoditional jump which was basically kept in C language exactly for such cases because it makes it more readable than arrow antipattern or switch/if ladder is amazing... Take my downvote.
      – Purple Ice
      2 days ago












      up vote
      1
      down vote













      Yes, as explained by other answers, using goto for error-handling is often appropriate in C.



      That said, if possible, you probably should make your cleanup code safe to call even if the corresponding action was never performed. For example, instead of:



      void foo()
      {
      int result;
      int* p = malloc(...);
      if (p == NULL) { result = 1; goto err1; }

      int* p2 = malloc(...);
      if (p2 == NULL) { result = 2; goto err2; }

      int* p3 = malloc(...);
      if (p3 == NULL) { result = 3; goto err3; }

      // Do something with p, p2, and p3.
      bar(p, p2, p3);

      // Maybe we don't need p3 anymore.
      free(p3);

      return 0;

      err3:
      free(p3);
      err2:
      free(p2);
      err1:
      free(p1);
      return result;
      }


      I'd advocate:



      void foo()
      {
      int result = -1; // Or some generic error code for unknown errors.

      int* p = NULL;
      int* p2 = NULL;
      int* p3 = NULL;

      p = malloc(...);
      if (p == NULL) { result = 1; goto exit; }

      p2 = malloc(...);
      if (p2 == NULL) { result = 2; goto exit; }

      p3 = malloc(...);
      if (p3 == NULL) { result = 3; goto exit; }

      // Do something with p, p2, and p3.
      bar(p, p2, p3);

      // Set success *only* on the successful path.
      result = 0;

      exit:
      // free(NULL) is a no-op, so this is safe even if p3 was never allocated.
      free(p3);

      if (result != 0)
      {
      free(p2);
      free(p1);
      }
      return result;
      }


      It's slightly less efficient since it requires initializing variables to NULL, but it's more maintainable since you don't need extra labels. There's less stuff to get wrong when making changes to the code. Also, if there's cleanup code that you need on both success and failure paths, you can avoid code duplication.






      share|improve this answer

























        up vote
        1
        down vote













        Yes, as explained by other answers, using goto for error-handling is often appropriate in C.



        That said, if possible, you probably should make your cleanup code safe to call even if the corresponding action was never performed. For example, instead of:



        void foo()
        {
        int result;
        int* p = malloc(...);
        if (p == NULL) { result = 1; goto err1; }

        int* p2 = malloc(...);
        if (p2 == NULL) { result = 2; goto err2; }

        int* p3 = malloc(...);
        if (p3 == NULL) { result = 3; goto err3; }

        // Do something with p, p2, and p3.
        bar(p, p2, p3);

        // Maybe we don't need p3 anymore.
        free(p3);

        return 0;

        err3:
        free(p3);
        err2:
        free(p2);
        err1:
        free(p1);
        return result;
        }


        I'd advocate:



        void foo()
        {
        int result = -1; // Or some generic error code for unknown errors.

        int* p = NULL;
        int* p2 = NULL;
        int* p3 = NULL;

        p = malloc(...);
        if (p == NULL) { result = 1; goto exit; }

        p2 = malloc(...);
        if (p2 == NULL) { result = 2; goto exit; }

        p3 = malloc(...);
        if (p3 == NULL) { result = 3; goto exit; }

        // Do something with p, p2, and p3.
        bar(p, p2, p3);

        // Set success *only* on the successful path.
        result = 0;

        exit:
        // free(NULL) is a no-op, so this is safe even if p3 was never allocated.
        free(p3);

        if (result != 0)
        {
        free(p2);
        free(p1);
        }
        return result;
        }


        It's slightly less efficient since it requires initializing variables to NULL, but it's more maintainable since you don't need extra labels. There's less stuff to get wrong when making changes to the code. Also, if there's cleanup code that you need on both success and failure paths, you can avoid code duplication.






        share|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Yes, as explained by other answers, using goto for error-handling is often appropriate in C.



          That said, if possible, you probably should make your cleanup code safe to call even if the corresponding action was never performed. For example, instead of:



          void foo()
          {
          int result;
          int* p = malloc(...);
          if (p == NULL) { result = 1; goto err1; }

          int* p2 = malloc(...);
          if (p2 == NULL) { result = 2; goto err2; }

          int* p3 = malloc(...);
          if (p3 == NULL) { result = 3; goto err3; }

          // Do something with p, p2, and p3.
          bar(p, p2, p3);

          // Maybe we don't need p3 anymore.
          free(p3);

          return 0;

          err3:
          free(p3);
          err2:
          free(p2);
          err1:
          free(p1);
          return result;
          }


          I'd advocate:



          void foo()
          {
          int result = -1; // Or some generic error code for unknown errors.

          int* p = NULL;
          int* p2 = NULL;
          int* p3 = NULL;

          p = malloc(...);
          if (p == NULL) { result = 1; goto exit; }

          p2 = malloc(...);
          if (p2 == NULL) { result = 2; goto exit; }

          p3 = malloc(...);
          if (p3 == NULL) { result = 3; goto exit; }

          // Do something with p, p2, and p3.
          bar(p, p2, p3);

          // Set success *only* on the successful path.
          result = 0;

          exit:
          // free(NULL) is a no-op, so this is safe even if p3 was never allocated.
          free(p3);

          if (result != 0)
          {
          free(p2);
          free(p1);
          }
          return result;
          }


          It's slightly less efficient since it requires initializing variables to NULL, but it's more maintainable since you don't need extra labels. There's less stuff to get wrong when making changes to the code. Also, if there's cleanup code that you need on both success and failure paths, you can avoid code duplication.






          share|improve this answer












          Yes, as explained by other answers, using goto for error-handling is often appropriate in C.



          That said, if possible, you probably should make your cleanup code safe to call even if the corresponding action was never performed. For example, instead of:



          void foo()
          {
          int result;
          int* p = malloc(...);
          if (p == NULL) { result = 1; goto err1; }

          int* p2 = malloc(...);
          if (p2 == NULL) { result = 2; goto err2; }

          int* p3 = malloc(...);
          if (p3 == NULL) { result = 3; goto err3; }

          // Do something with p, p2, and p3.
          bar(p, p2, p3);

          // Maybe we don't need p3 anymore.
          free(p3);

          return 0;

          err3:
          free(p3);
          err2:
          free(p2);
          err1:
          free(p1);
          return result;
          }


          I'd advocate:



          void foo()
          {
          int result = -1; // Or some generic error code for unknown errors.

          int* p = NULL;
          int* p2 = NULL;
          int* p3 = NULL;

          p = malloc(...);
          if (p == NULL) { result = 1; goto exit; }

          p2 = malloc(...);
          if (p2 == NULL) { result = 2; goto exit; }

          p3 = malloc(...);
          if (p3 == NULL) { result = 3; goto exit; }

          // Do something with p, p2, and p3.
          bar(p, p2, p3);

          // Set success *only* on the successful path.
          result = 0;

          exit:
          // free(NULL) is a no-op, so this is safe even if p3 was never allocated.
          free(p3);

          if (result != 0)
          {
          free(p2);
          free(p1);
          }
          return result;
          }


          It's slightly less efficient since it requires initializing variables to NULL, but it's more maintainable since you don't need extra labels. There's less stuff to get wrong when making changes to the code. Also, if there's cleanup code that you need on both success and failure paths, you can avoid code duplication.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 2 days ago









          jamesdlin

          25.8k65791




          25.8k65791






















              up vote
              0
              down vote













              I typically approach this kind of problem by nesting the conditionals:



              int rval = 1;
              if (!Do1()) {
              if (!Do2()) {
              if (!Do3()) {
              if (!Do4()) {
              if (!Do5()) {
              return 0;
              // or "goto succeeded", or ...;
              } else {
              printf("Failed 5");
              rval = 5;
              }
              Undo4();
              } else {
              printf("Failed 4");
              rval = 4;
              }
              Undo3();
              } else {
              printf("Failed 3");
              rval = 3;
              }
              Undo2();
              } else {
              printf("Failed 2");
              rval = 2;
              }
              Undo1();
              } else {
              printf("Failed 1");
              rval = 1;
              }
              return rval;


              Usually, for me, the DoX() are some kind of resource acquisition, such as malloc(), and the UndoX() are corresponding resource releases that should be performed only in the event of failure. The nesting clearly shows the association between corresponding acquisitions and releases, and avoids the need for repetition of the code for undo operations. It's also very easy to write -- you don't need to create or maintain labels, and it's easy to put the resource release in the right place as soon as you write the acquisition.



              This approach does sometimes produce deeply nested code. That doesn't bother me much, but you might consider it an issue.






              share|improve this answer

















              • 2




                'deeply nested' - plus error and error handling get separated by quite some distance, the more for the less deeply nested ones...
                – Aconcagua
                yesterday










              • @Aconcagua, separation of error and error-handling is a pretty consistent characteristic of all of these approaches. It follows directly from avoiding repetition of the error-handling code.
                – John Bollinger
                yesterday

















              up vote
              0
              down vote













              I typically approach this kind of problem by nesting the conditionals:



              int rval = 1;
              if (!Do1()) {
              if (!Do2()) {
              if (!Do3()) {
              if (!Do4()) {
              if (!Do5()) {
              return 0;
              // or "goto succeeded", or ...;
              } else {
              printf("Failed 5");
              rval = 5;
              }
              Undo4();
              } else {
              printf("Failed 4");
              rval = 4;
              }
              Undo3();
              } else {
              printf("Failed 3");
              rval = 3;
              }
              Undo2();
              } else {
              printf("Failed 2");
              rval = 2;
              }
              Undo1();
              } else {
              printf("Failed 1");
              rval = 1;
              }
              return rval;


              Usually, for me, the DoX() are some kind of resource acquisition, such as malloc(), and the UndoX() are corresponding resource releases that should be performed only in the event of failure. The nesting clearly shows the association between corresponding acquisitions and releases, and avoids the need for repetition of the code for undo operations. It's also very easy to write -- you don't need to create or maintain labels, and it's easy to put the resource release in the right place as soon as you write the acquisition.



              This approach does sometimes produce deeply nested code. That doesn't bother me much, but you might consider it an issue.






              share|improve this answer

















              • 2




                'deeply nested' - plus error and error handling get separated by quite some distance, the more for the less deeply nested ones...
                – Aconcagua
                yesterday










              • @Aconcagua, separation of error and error-handling is a pretty consistent characteristic of all of these approaches. It follows directly from avoiding repetition of the error-handling code.
                – John Bollinger
                yesterday















              up vote
              0
              down vote










              up vote
              0
              down vote









              I typically approach this kind of problem by nesting the conditionals:



              int rval = 1;
              if (!Do1()) {
              if (!Do2()) {
              if (!Do3()) {
              if (!Do4()) {
              if (!Do5()) {
              return 0;
              // or "goto succeeded", or ...;
              } else {
              printf("Failed 5");
              rval = 5;
              }
              Undo4();
              } else {
              printf("Failed 4");
              rval = 4;
              }
              Undo3();
              } else {
              printf("Failed 3");
              rval = 3;
              }
              Undo2();
              } else {
              printf("Failed 2");
              rval = 2;
              }
              Undo1();
              } else {
              printf("Failed 1");
              rval = 1;
              }
              return rval;


              Usually, for me, the DoX() are some kind of resource acquisition, such as malloc(), and the UndoX() are corresponding resource releases that should be performed only in the event of failure. The nesting clearly shows the association between corresponding acquisitions and releases, and avoids the need for repetition of the code for undo operations. It's also very easy to write -- you don't need to create or maintain labels, and it's easy to put the resource release in the right place as soon as you write the acquisition.



              This approach does sometimes produce deeply nested code. That doesn't bother me much, but you might consider it an issue.






              share|improve this answer












              I typically approach this kind of problem by nesting the conditionals:



              int rval = 1;
              if (!Do1()) {
              if (!Do2()) {
              if (!Do3()) {
              if (!Do4()) {
              if (!Do5()) {
              return 0;
              // or "goto succeeded", or ...;
              } else {
              printf("Failed 5");
              rval = 5;
              }
              Undo4();
              } else {
              printf("Failed 4");
              rval = 4;
              }
              Undo3();
              } else {
              printf("Failed 3");
              rval = 3;
              }
              Undo2();
              } else {
              printf("Failed 2");
              rval = 2;
              }
              Undo1();
              } else {
              printf("Failed 1");
              rval = 1;
              }
              return rval;


              Usually, for me, the DoX() are some kind of resource acquisition, such as malloc(), and the UndoX() are corresponding resource releases that should be performed only in the event of failure. The nesting clearly shows the association between corresponding acquisitions and releases, and avoids the need for repetition of the code for undo operations. It's also very easy to write -- you don't need to create or maintain labels, and it's easy to put the resource release in the right place as soon as you write the acquisition.



              This approach does sometimes produce deeply nested code. That doesn't bother me much, but you might consider it an issue.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 2 days ago









              John Bollinger

              76.6k63771




              76.6k63771








              • 2




                'deeply nested' - plus error and error handling get separated by quite some distance, the more for the less deeply nested ones...
                – Aconcagua
                yesterday










              • @Aconcagua, separation of error and error-handling is a pretty consistent characteristic of all of these approaches. It follows directly from avoiding repetition of the error-handling code.
                – John Bollinger
                yesterday
















              • 2




                'deeply nested' - plus error and error handling get separated by quite some distance, the more for the less deeply nested ones...
                – Aconcagua
                yesterday










              • @Aconcagua, separation of error and error-handling is a pretty consistent characteristic of all of these approaches. It follows directly from avoiding repetition of the error-handling code.
                – John Bollinger
                yesterday










              2




              2




              'deeply nested' - plus error and error handling get separated by quite some distance, the more for the less deeply nested ones...
              – Aconcagua
              yesterday




              'deeply nested' - plus error and error handling get separated by quite some distance, the more for the less deeply nested ones...
              – Aconcagua
              yesterday












              @Aconcagua, separation of error and error-handling is a pretty consistent characteristic of all of these approaches. It follows directly from avoiding repetition of the error-handling code.
              – John Bollinger
              yesterday






              @Aconcagua, separation of error and error-handling is a pretty consistent characteristic of all of these approaches. It follows directly from avoiding repetition of the error-handling code.
              – John Bollinger
              yesterday












              up vote
              0
              down vote













              Here is an answer that I have found resilient to bugs.



              Yes. It uses goto. I firmly believe you should use what gives you most clarity, rather than just blindly following the advice of those before you (goto as a construct can make spaghetti code, but in this instance every other error handling method ususally ends up more spaghetti-like than using this method of goto, so IMO it's superior).



              Some people may not like the form of this code, but I contest that when used to the style it is cleaner, easier to read (when everything's lined up, of course), and much more resilient to errors. If you have the properly linter/static analysis setup, and you're working with POSIX, it pretty much requires you to code in this fashion to allow for good error handling.



              static char *readbuf(char *path)
              {
              struct stat st;
              char *s = NULL;
              size_t size = 0;
              int fd = -1;

              if (!path) { return NULL; }

              if ((stat(path, &st)) < 0) { perror(path); goto _throw; }

              size = st.st_size;
              if (size == 0) { printf("%s is empty!n", path); goto _throw; }

              if (!(s = calloc(size, 1))) { perror("calloc"); goto _throw; }

              fd = open(path, O_RDONLY);
              if (fd < -1) { perror(path); goto _throw; }
              if ((read(fd, s, size)) < 0) { perror("read"); goto _throw; }
              close(fd); /* There's really no point checking close for errors */

              return s;

              _throw:
              if (fd > 0) close(fd);
              if (s) free(s);
              return NULL;
              }





              share|improve this answer

























                up vote
                0
                down vote













                Here is an answer that I have found resilient to bugs.



                Yes. It uses goto. I firmly believe you should use what gives you most clarity, rather than just blindly following the advice of those before you (goto as a construct can make spaghetti code, but in this instance every other error handling method ususally ends up more spaghetti-like than using this method of goto, so IMO it's superior).



                Some people may not like the form of this code, but I contest that when used to the style it is cleaner, easier to read (when everything's lined up, of course), and much more resilient to errors. If you have the properly linter/static analysis setup, and you're working with POSIX, it pretty much requires you to code in this fashion to allow for good error handling.



                static char *readbuf(char *path)
                {
                struct stat st;
                char *s = NULL;
                size_t size = 0;
                int fd = -1;

                if (!path) { return NULL; }

                if ((stat(path, &st)) < 0) { perror(path); goto _throw; }

                size = st.st_size;
                if (size == 0) { printf("%s is empty!n", path); goto _throw; }

                if (!(s = calloc(size, 1))) { perror("calloc"); goto _throw; }

                fd = open(path, O_RDONLY);
                if (fd < -1) { perror(path); goto _throw; }
                if ((read(fd, s, size)) < 0) { perror("read"); goto _throw; }
                close(fd); /* There's really no point checking close for errors */

                return s;

                _throw:
                if (fd > 0) close(fd);
                if (s) free(s);
                return NULL;
                }





                share|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Here is an answer that I have found resilient to bugs.



                  Yes. It uses goto. I firmly believe you should use what gives you most clarity, rather than just blindly following the advice of those before you (goto as a construct can make spaghetti code, but in this instance every other error handling method ususally ends up more spaghetti-like than using this method of goto, so IMO it's superior).



                  Some people may not like the form of this code, but I contest that when used to the style it is cleaner, easier to read (when everything's lined up, of course), and much more resilient to errors. If you have the properly linter/static analysis setup, and you're working with POSIX, it pretty much requires you to code in this fashion to allow for good error handling.



                  static char *readbuf(char *path)
                  {
                  struct stat st;
                  char *s = NULL;
                  size_t size = 0;
                  int fd = -1;

                  if (!path) { return NULL; }

                  if ((stat(path, &st)) < 0) { perror(path); goto _throw; }

                  size = st.st_size;
                  if (size == 0) { printf("%s is empty!n", path); goto _throw; }

                  if (!(s = calloc(size, 1))) { perror("calloc"); goto _throw; }

                  fd = open(path, O_RDONLY);
                  if (fd < -1) { perror(path); goto _throw; }
                  if ((read(fd, s, size)) < 0) { perror("read"); goto _throw; }
                  close(fd); /* There's really no point checking close for errors */

                  return s;

                  _throw:
                  if (fd > 0) close(fd);
                  if (s) free(s);
                  return NULL;
                  }





                  share|improve this answer












                  Here is an answer that I have found resilient to bugs.



                  Yes. It uses goto. I firmly believe you should use what gives you most clarity, rather than just blindly following the advice of those before you (goto as a construct can make spaghetti code, but in this instance every other error handling method ususally ends up more spaghetti-like than using this method of goto, so IMO it's superior).



                  Some people may not like the form of this code, but I contest that when used to the style it is cleaner, easier to read (when everything's lined up, of course), and much more resilient to errors. If you have the properly linter/static analysis setup, and you're working with POSIX, it pretty much requires you to code in this fashion to allow for good error handling.



                  static char *readbuf(char *path)
                  {
                  struct stat st;
                  char *s = NULL;
                  size_t size = 0;
                  int fd = -1;

                  if (!path) { return NULL; }

                  if ((stat(path, &st)) < 0) { perror(path); goto _throw; }

                  size = st.st_size;
                  if (size == 0) { printf("%s is empty!n", path); goto _throw; }

                  if (!(s = calloc(size, 1))) { perror("calloc"); goto _throw; }

                  fd = open(path, O_RDONLY);
                  if (fd < -1) { perror(path); goto _throw; }
                  if ((read(fd, s, size)) < 0) { perror("read"); goto _throw; }
                  close(fd); /* There's really no point checking close for errors */

                  return s;

                  _throw:
                  if (fd > 0) close(fd);
                  if (s) free(s);
                  return NULL;
                  }






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered yesterday









                  Finn O'leary

                  13912




                  13912






















                      up vote
                      -1
                      down vote













                      If the functions return some kind of state pointer or handle (like most allocation & initialization functions would), you can quite cleanly do this without goto by giving initial values to variables. Then you can have a single deallocation function that can handle the case where only part of the resources has been allocated.



                      For example:




                      void *mymemoryblock = NULL;
                      FILE *myfile = NULL;
                      int mysocket = -1;

                      bool allocate_everything()
                      {
                      mymemoryblock = malloc(1000);
                      if (!mymemoryblock)
                      {
                      return false;
                      }

                      myfile = fopen("/file", "r");
                      if (!myfile)
                      {
                      return false;
                      }

                      mysocket = socket(AF_INET, SOCK_STREAM, 0);
                      if (mysocket < 0)
                      {
                      return false;
                      }

                      return true;
                      }

                      void deallocate_everything()
                      {
                      if (mysocket >= 0)
                      {
                      close(mysocket);
                      mysocket = -1;
                      }

                      if (myfile)
                      {
                      fclose(myfile);
                      myfile = NULL;
                      }

                      if (mymemoryblock)
                      {
                      free(mymemoryblock);
                      mymemoryblock = NULL;
                      }
                      }


                      And then just do:




                      if (allocate_everything())
                      {
                      do_the_deed();
                      }
                      deallocate_everything();





                      share|improve this answer





















                      • "If the functions return some kind of state pointer or handle..." Yes, but it is not the general case. Further, your solution requires 3 function for each function that allocates resources, plus global variables (or passing things around).
                        – Acorn
                        2 days ago

















                      up vote
                      -1
                      down vote













                      If the functions return some kind of state pointer or handle (like most allocation & initialization functions would), you can quite cleanly do this without goto by giving initial values to variables. Then you can have a single deallocation function that can handle the case where only part of the resources has been allocated.



                      For example:




                      void *mymemoryblock = NULL;
                      FILE *myfile = NULL;
                      int mysocket = -1;

                      bool allocate_everything()
                      {
                      mymemoryblock = malloc(1000);
                      if (!mymemoryblock)
                      {
                      return false;
                      }

                      myfile = fopen("/file", "r");
                      if (!myfile)
                      {
                      return false;
                      }

                      mysocket = socket(AF_INET, SOCK_STREAM, 0);
                      if (mysocket < 0)
                      {
                      return false;
                      }

                      return true;
                      }

                      void deallocate_everything()
                      {
                      if (mysocket >= 0)
                      {
                      close(mysocket);
                      mysocket = -1;
                      }

                      if (myfile)
                      {
                      fclose(myfile);
                      myfile = NULL;
                      }

                      if (mymemoryblock)
                      {
                      free(mymemoryblock);
                      mymemoryblock = NULL;
                      }
                      }


                      And then just do:




                      if (allocate_everything())
                      {
                      do_the_deed();
                      }
                      deallocate_everything();





                      share|improve this answer





















                      • "If the functions return some kind of state pointer or handle..." Yes, but it is not the general case. Further, your solution requires 3 function for each function that allocates resources, plus global variables (or passing things around).
                        – Acorn
                        2 days ago















                      up vote
                      -1
                      down vote










                      up vote
                      -1
                      down vote









                      If the functions return some kind of state pointer or handle (like most allocation & initialization functions would), you can quite cleanly do this without goto by giving initial values to variables. Then you can have a single deallocation function that can handle the case where only part of the resources has been allocated.



                      For example:




                      void *mymemoryblock = NULL;
                      FILE *myfile = NULL;
                      int mysocket = -1;

                      bool allocate_everything()
                      {
                      mymemoryblock = malloc(1000);
                      if (!mymemoryblock)
                      {
                      return false;
                      }

                      myfile = fopen("/file", "r");
                      if (!myfile)
                      {
                      return false;
                      }

                      mysocket = socket(AF_INET, SOCK_STREAM, 0);
                      if (mysocket < 0)
                      {
                      return false;
                      }

                      return true;
                      }

                      void deallocate_everything()
                      {
                      if (mysocket >= 0)
                      {
                      close(mysocket);
                      mysocket = -1;
                      }

                      if (myfile)
                      {
                      fclose(myfile);
                      myfile = NULL;
                      }

                      if (mymemoryblock)
                      {
                      free(mymemoryblock);
                      mymemoryblock = NULL;
                      }
                      }


                      And then just do:




                      if (allocate_everything())
                      {
                      do_the_deed();
                      }
                      deallocate_everything();





                      share|improve this answer












                      If the functions return some kind of state pointer or handle (like most allocation & initialization functions would), you can quite cleanly do this without goto by giving initial values to variables. Then you can have a single deallocation function that can handle the case where only part of the resources has been allocated.



                      For example:




                      void *mymemoryblock = NULL;
                      FILE *myfile = NULL;
                      int mysocket = -1;

                      bool allocate_everything()
                      {
                      mymemoryblock = malloc(1000);
                      if (!mymemoryblock)
                      {
                      return false;
                      }

                      myfile = fopen("/file", "r");
                      if (!myfile)
                      {
                      return false;
                      }

                      mysocket = socket(AF_INET, SOCK_STREAM, 0);
                      if (mysocket < 0)
                      {
                      return false;
                      }

                      return true;
                      }

                      void deallocate_everything()
                      {
                      if (mysocket >= 0)
                      {
                      close(mysocket);
                      mysocket = -1;
                      }

                      if (myfile)
                      {
                      fclose(myfile);
                      myfile = NULL;
                      }

                      if (mymemoryblock)
                      {
                      free(mymemoryblock);
                      mymemoryblock = NULL;
                      }
                      }


                      And then just do:




                      if (allocate_everything())
                      {
                      do_the_deed();
                      }
                      deallocate_everything();






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 2 days ago









                      jpa

                      5,1391226




                      5,1391226












                      • "If the functions return some kind of state pointer or handle..." Yes, but it is not the general case. Further, your solution requires 3 function for each function that allocates resources, plus global variables (or passing things around).
                        – Acorn
                        2 days ago




















                      • "If the functions return some kind of state pointer or handle..." Yes, but it is not the general case. Further, your solution requires 3 function for each function that allocates resources, plus global variables (or passing things around).
                        – Acorn
                        2 days ago


















                      "If the functions return some kind of state pointer or handle..." Yes, but it is not the general case. Further, your solution requires 3 function for each function that allocates resources, plus global variables (or passing things around).
                      – Acorn
                      2 days ago






                      "If the functions return some kind of state pointer or handle..." Yes, but it is not the general case. Further, your solution requires 3 function for each function that allocates resources, plus global variables (or passing things around).
                      – Acorn
                      2 days ago












                      up vote
                      -1
                      down vote













                      TL;DR:



                      I believe it should be written as:



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }




                      Detailed explanation:



                      If nothing can be assumed what-so-ever about the function types, we can't easily use an array of function pointers, which would otherwise be the correct answer.



                      Assuming all function types are incompatible, then we would have to wrap in the original obscure design containing all those non-compatible functions, inside something else.



                      We should make something that is readable, maintainable, fast. We should avoid tight coupling, so that the undo behavior of "Do_x" doesn't depend on the undo behavior of "Do_y".



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Where do_func is the function doing all the calls required by the algorithm, and the printf is the UI output, separated from the algorithm logic.



                      do_func would be implemented like a wrapper function around the actual function calls, handling the outcome depending on the result:



                      (With gcc -O3, do_func is inlined in the caller, so there is no overhead for having 2 separate functions)



                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }


                      Here the specific behavior is controlled by the array undo, which is a wrapper around the various non-compatible functions. Which functions to to call, in which order, is all part of the specific behavior tied to each result code.



                      We need to tidy it all up, so that we can couple a certain behavior to a certain result code. Then when needed, we only change the code in one single place if the behavior should be changed during maintenance:



                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };




                      MCVE:



                      #include <stdbool.h>
                      #include <stdio.h>

                      // some nonsense functions:
                      bool Do1 (void) { puts(__func__); return false; }
                      bool Do2 (void) { puts(__func__); return false; }
                      bool Do3 (void) { puts(__func__); return false; }
                      bool Do4 (void) { puts(__func__); return false; }
                      bool Do5 (void) { puts(__func__); return true; }
                      void Undo1 (void) { puts(__func__); }
                      void Undo2 (void) { puts(__func__); }
                      void Undo3 (void) { puts(__func__); }
                      void Undo4 (void) { puts(__func__); }
                      void Undo5 (void) { puts(__func__); }

                      // wrappers specifying undo behavior:
                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };

                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }

                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Output:



                      Do1
                      Do2
                      Do3
                      Do4
                      Do5
                      Undo4
                      Undo3
                      Undo2
                      Undo1
                      Failed 5





                      share|improve this answer



















                      • 12




                        So, for each single function in your code that allocates resources, you are going to write 4 functions instead (which, some of them duplicate UndoN() calls in turn). Plus a wrapper. Plus a type. Plus a global array of that type. No further comments.
                        – Acorn
                        2 days ago












                      • @Acorn No, for every ADT in my code, I'm going to write a user-friend API, a clarity of who's responsible for allocation/deallocation, private encapsulation, modular, maintainable code, without tight coupling, where everything is autonomous. In this particular case, it looks like we are patching up some old, broken design, in which case you need wrappers - which is the fault of the original (lack of) design. Of course you don't need that level of abstraction if it's just some quick & dirty hack, but in that case all that is best practices is thrown out the window anyway.
                        – Lundin
                        14 hours ago















                      up vote
                      -1
                      down vote













                      TL;DR:



                      I believe it should be written as:



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }




                      Detailed explanation:



                      If nothing can be assumed what-so-ever about the function types, we can't easily use an array of function pointers, which would otherwise be the correct answer.



                      Assuming all function types are incompatible, then we would have to wrap in the original obscure design containing all those non-compatible functions, inside something else.



                      We should make something that is readable, maintainable, fast. We should avoid tight coupling, so that the undo behavior of "Do_x" doesn't depend on the undo behavior of "Do_y".



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Where do_func is the function doing all the calls required by the algorithm, and the printf is the UI output, separated from the algorithm logic.



                      do_func would be implemented like a wrapper function around the actual function calls, handling the outcome depending on the result:



                      (With gcc -O3, do_func is inlined in the caller, so there is no overhead for having 2 separate functions)



                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }


                      Here the specific behavior is controlled by the array undo, which is a wrapper around the various non-compatible functions. Which functions to to call, in which order, is all part of the specific behavior tied to each result code.



                      We need to tidy it all up, so that we can couple a certain behavior to a certain result code. Then when needed, we only change the code in one single place if the behavior should be changed during maintenance:



                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };




                      MCVE:



                      #include <stdbool.h>
                      #include <stdio.h>

                      // some nonsense functions:
                      bool Do1 (void) { puts(__func__); return false; }
                      bool Do2 (void) { puts(__func__); return false; }
                      bool Do3 (void) { puts(__func__); return false; }
                      bool Do4 (void) { puts(__func__); return false; }
                      bool Do5 (void) { puts(__func__); return true; }
                      void Undo1 (void) { puts(__func__); }
                      void Undo2 (void) { puts(__func__); }
                      void Undo3 (void) { puts(__func__); }
                      void Undo4 (void) { puts(__func__); }
                      void Undo5 (void) { puts(__func__); }

                      // wrappers specifying undo behavior:
                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };

                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }

                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Output:



                      Do1
                      Do2
                      Do3
                      Do4
                      Do5
                      Undo4
                      Undo3
                      Undo2
                      Undo1
                      Failed 5





                      share|improve this answer



















                      • 12




                        So, for each single function in your code that allocates resources, you are going to write 4 functions instead (which, some of them duplicate UndoN() calls in turn). Plus a wrapper. Plus a type. Plus a global array of that type. No further comments.
                        – Acorn
                        2 days ago












                      • @Acorn No, for every ADT in my code, I'm going to write a user-friend API, a clarity of who's responsible for allocation/deallocation, private encapsulation, modular, maintainable code, without tight coupling, where everything is autonomous. In this particular case, it looks like we are patching up some old, broken design, in which case you need wrappers - which is the fault of the original (lack of) design. Of course you don't need that level of abstraction if it's just some quick & dirty hack, but in that case all that is best practices is thrown out the window anyway.
                        – Lundin
                        14 hours ago













                      up vote
                      -1
                      down vote










                      up vote
                      -1
                      down vote









                      TL;DR:



                      I believe it should be written as:



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }




                      Detailed explanation:



                      If nothing can be assumed what-so-ever about the function types, we can't easily use an array of function pointers, which would otherwise be the correct answer.



                      Assuming all function types are incompatible, then we would have to wrap in the original obscure design containing all those non-compatible functions, inside something else.



                      We should make something that is readable, maintainable, fast. We should avoid tight coupling, so that the undo behavior of "Do_x" doesn't depend on the undo behavior of "Do_y".



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Where do_func is the function doing all the calls required by the algorithm, and the printf is the UI output, separated from the algorithm logic.



                      do_func would be implemented like a wrapper function around the actual function calls, handling the outcome depending on the result:



                      (With gcc -O3, do_func is inlined in the caller, so there is no overhead for having 2 separate functions)



                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }


                      Here the specific behavior is controlled by the array undo, which is a wrapper around the various non-compatible functions. Which functions to to call, in which order, is all part of the specific behavior tied to each result code.



                      We need to tidy it all up, so that we can couple a certain behavior to a certain result code. Then when needed, we only change the code in one single place if the behavior should be changed during maintenance:



                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };




                      MCVE:



                      #include <stdbool.h>
                      #include <stdio.h>

                      // some nonsense functions:
                      bool Do1 (void) { puts(__func__); return false; }
                      bool Do2 (void) { puts(__func__); return false; }
                      bool Do3 (void) { puts(__func__); return false; }
                      bool Do4 (void) { puts(__func__); return false; }
                      bool Do5 (void) { puts(__func__); return true; }
                      void Undo1 (void) { puts(__func__); }
                      void Undo2 (void) { puts(__func__); }
                      void Undo3 (void) { puts(__func__); }
                      void Undo4 (void) { puts(__func__); }
                      void Undo5 (void) { puts(__func__); }

                      // wrappers specifying undo behavior:
                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };

                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }

                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Output:



                      Do1
                      Do2
                      Do3
                      Do4
                      Do5
                      Undo4
                      Undo3
                      Undo2
                      Undo1
                      Failed 5





                      share|improve this answer














                      TL;DR:



                      I believe it should be written as:



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }




                      Detailed explanation:



                      If nothing can be assumed what-so-ever about the function types, we can't easily use an array of function pointers, which would otherwise be the correct answer.



                      Assuming all function types are incompatible, then we would have to wrap in the original obscure design containing all those non-compatible functions, inside something else.



                      We should make something that is readable, maintainable, fast. We should avoid tight coupling, so that the undo behavior of "Do_x" doesn't depend on the undo behavior of "Do_y".



                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Where do_func is the function doing all the calls required by the algorithm, and the printf is the UI output, separated from the algorithm logic.



                      do_func would be implemented like a wrapper function around the actual function calls, handling the outcome depending on the result:



                      (With gcc -O3, do_func is inlined in the caller, so there is no overhead for having 2 separate functions)



                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }


                      Here the specific behavior is controlled by the array undo, which is a wrapper around the various non-compatible functions. Which functions to to call, in which order, is all part of the specific behavior tied to each result code.



                      We need to tidy it all up, so that we can couple a certain behavior to a certain result code. Then when needed, we only change the code in one single place if the behavior should be changed during maintenance:



                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };




                      MCVE:



                      #include <stdbool.h>
                      #include <stdio.h>

                      // some nonsense functions:
                      bool Do1 (void) { puts(__func__); return false; }
                      bool Do2 (void) { puts(__func__); return false; }
                      bool Do3 (void) { puts(__func__); return false; }
                      bool Do4 (void) { puts(__func__); return false; }
                      bool Do5 (void) { puts(__func__); return true; }
                      void Undo1 (void) { puts(__func__); }
                      void Undo2 (void) { puts(__func__); }
                      void Undo3 (void) { puts(__func__); }
                      void Undo4 (void) { puts(__func__); }
                      void Undo5 (void) { puts(__func__); }

                      // wrappers specifying undo behavior:
                      void Undo_stuff1 (void) { }
                      void Undo_stuff2 (void) { Undo1(); }
                      void Undo_stuff3 (void) { Undo2(); Undo1(); }
                      void Undo_stuff4 (void) { Undo3(); Undo2(); Undo1(); }
                      void Undo_stuff5 (void) { Undo4(); Undo3(); Undo2(); Undo1(); }

                      typedef void Undo_stuff_t (void);
                      static Undo_stuff_t* undo[5] =
                      {
                      Undo_stuff1,
                      Undo_stuff2,
                      Undo_stuff3,
                      Undo_stuff4,
                      Undo_stuff5,
                      };

                      int do_it (void)
                      {
                      if(Do1()) { return 1; };
                      if(Do2()) { return 2; };
                      if(Do3()) { return 3; };
                      if(Do4()) { return 4; };
                      if(Do5()) { return 5; };
                      return 0;
                      }

                      int do_func (void)
                      {
                      int result = do_it();
                      if(result != 0)
                      {
                      undo[result-1]();
                      }
                      return result;
                      }

                      int main (void)
                      {
                      int result = do_func();
                      printf("Failed %dn", result);
                      }


                      Output:



                      Do1
                      Do2
                      Do3
                      Do4
                      Do5
                      Undo4
                      Undo3
                      Undo2
                      Undo1
                      Failed 5






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 2 days ago

























                      answered 2 days ago









                      Lundin

                      105k16153257




                      105k16153257








                      • 12




                        So, for each single function in your code that allocates resources, you are going to write 4 functions instead (which, some of them duplicate UndoN() calls in turn). Plus a wrapper. Plus a type. Plus a global array of that type. No further comments.
                        – Acorn
                        2 days ago












                      • @Acorn No, for every ADT in my code, I'm going to write a user-friend API, a clarity of who's responsible for allocation/deallocation, private encapsulation, modular, maintainable code, without tight coupling, where everything is autonomous. In this particular case, it looks like we are patching up some old, broken design, in which case you need wrappers - which is the fault of the original (lack of) design. Of course you don't need that level of abstraction if it's just some quick & dirty hack, but in that case all that is best practices is thrown out the window anyway.
                        – Lundin
                        14 hours ago














                      • 12




                        So, for each single function in your code that allocates resources, you are going to write 4 functions instead (which, some of them duplicate UndoN() calls in turn). Plus a wrapper. Plus a type. Plus a global array of that type. No further comments.
                        – Acorn
                        2 days ago












                      • @Acorn No, for every ADT in my code, I'm going to write a user-friend API, a clarity of who's responsible for allocation/deallocation, private encapsulation, modular, maintainable code, without tight coupling, where everything is autonomous. In this particular case, it looks like we are patching up some old, broken design, in which case you need wrappers - which is the fault of the original (lack of) design. Of course you don't need that level of abstraction if it's just some quick & dirty hack, but in that case all that is best practices is thrown out the window anyway.
                        – Lundin
                        14 hours ago








                      12




                      12




                      So, for each single function in your code that allocates resources, you are going to write 4 functions instead (which, some of them duplicate UndoN() calls in turn). Plus a wrapper. Plus a type. Plus a global array of that type. No further comments.
                      – Acorn
                      2 days ago






                      So, for each single function in your code that allocates resources, you are going to write 4 functions instead (which, some of them duplicate UndoN() calls in turn). Plus a wrapper. Plus a type. Plus a global array of that type. No further comments.
                      – Acorn
                      2 days ago














                      @Acorn No, for every ADT in my code, I'm going to write a user-friend API, a clarity of who's responsible for allocation/deallocation, private encapsulation, modular, maintainable code, without tight coupling, where everything is autonomous. In this particular case, it looks like we are patching up some old, broken design, in which case you need wrappers - which is the fault of the original (lack of) design. Of course you don't need that level of abstraction if it's just some quick & dirty hack, but in that case all that is best practices is thrown out the window anyway.
                      – Lundin
                      14 hours ago




                      @Acorn No, for every ADT in my code, I'm going to write a user-friend API, a clarity of who's responsible for allocation/deallocation, private encapsulation, modular, maintainable code, without tight coupling, where everything is autonomous. In this particular case, it looks like we are patching up some old, broken design, in which case you need wrappers - which is the fault of the original (lack of) design. Of course you don't need that level of abstraction if it's just some quick & dirty hack, but in that case all that is best practices is thrown out the window anyway.
                      – Lundin
                      14 hours ago










                      up vote
                      -2
                      down vote













                      typedef void(*undoer)();
                      int undo( undoer*const* list ) {
                      while(*list) {
                      (*list)();
                      ++list;
                      }
                      }
                      void undo_push( undoer** list, undoer* undo ) {
                      if (!undo) return;
                      // swap
                      undoer* tmp = *list;
                      *list = undo;
                      undo = tmp;
                      undo_push( list+1, undo );
                      }
                      int func() {
                      undoer undoers[6]={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      undo_push( undoers, Undo1 );
                      if (Do2()) { undo(undoers); printf("Failed 2"); return 2; }
                      undo_push( undoers, Undo2 );
                      if (Do3()) { undo(undoers); printf("Failed 3"); return 3; }
                      undo_push( undoers, Undo3 );
                      if (Do4()) { undo(undoers); printf("Failed 4"); return 4; }
                      undo_push( undoers, Undo4 );
                      if (Do5()) { undo(undoers); printf("Failed 5"); return 5; }
                      return 6;
                      }


                      I made undo_push do the O(n) work. This is less efficient than having undo do the O(n) work, as we expect more push's than undos. But this version was a touch simpler.



                      A more industrial strength version would have head and tail pointers and even capacity.



                      The basic idea is to keep a queue of undo actions in a stack, then execute them if you need to clean up.



                      Everything is local here, so we don't pollute global state.





                      struct undoer {
                      void(*action)(void*);
                      void(*cleanup)(void*);
                      void* state;
                      };

                      struct undoers {
                      undoer* top;
                      undoer buff[5];
                      };
                      void undo( undoers u ) {
                      while (u.top != buff)
                      {
                      (u.top->action)(u.top->state);
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pundo(void* pu) {
                      undo( *(undoers*)pu );
                      free(pu);
                      }
                      void cleanup_undoers(undoers u) {
                      while (u.top != buff)
                      {
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pcleanup_undoers(void* pu) {
                      cleanup_undoers(*(undoers*)pu);
                      free(pu);
                      }
                      void push_undoer( undoers* to_here, undoer u ) {
                      if (to_here->top != (to_here->buff+5))
                      {
                      to_here->top = u;
                      ++(to_here->top)
                      return;
                      }
                      undoers* chain = (undoers*)malloc( sizeof(undoers) );
                      memcpy(chain, to_here, sizeof(undoers));
                      memset(to_here, 0, sizeof(undoers));
                      undoer chainer;
                      chainer.action = pundo;
                      chainer.cleanup = pcleanup_undoers;
                      chainer.data = chain;
                      push_undoer( to_here, chainer );
                      push_undoer( to_here, u );
                      }
                      void paction( void* p ) {
                      (void)(*a)() = ((void)(*)());
                      a();
                      }
                      void push_undo( undoers* to_here, void(*action)() ) {
                      undor u;
                      u.action = paction;
                      u.cleanup = 0;
                      u.data = (void*)action;
                      push_undoer(to_here, u);
                      }


                      then you get:



                      int func() {
                      undoers u={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      push_undo( &u, Undo1 );
                      if (Do2()) { undo(u); printf("Failed 2"); return 2; }
                      push_undo( &u, Undo2 );
                      if (Do3()) { undo(u); printf("Failed 3"); return 3; }
                      push_undo( &u, Undo3 );
                      if (Do4()) { undo(u); printf("Failed 4"); return 4; }
                      push_undo( &u, Undo4 );
                      if (Do5()) { undo(u); printf("Failed 5"); return 5; }
                      cleanup_undoers(u);
                      return 6;
                      }


                      but that is getting ridiculous.






                      share|improve this answer



















                      • 1




                        More complex, requires that DoN/UndoN are actual functions (and the same signature), requires stack space, slower.
                        – Acorn
                        2 days ago

















                      up vote
                      -2
                      down vote













                      typedef void(*undoer)();
                      int undo( undoer*const* list ) {
                      while(*list) {
                      (*list)();
                      ++list;
                      }
                      }
                      void undo_push( undoer** list, undoer* undo ) {
                      if (!undo) return;
                      // swap
                      undoer* tmp = *list;
                      *list = undo;
                      undo = tmp;
                      undo_push( list+1, undo );
                      }
                      int func() {
                      undoer undoers[6]={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      undo_push( undoers, Undo1 );
                      if (Do2()) { undo(undoers); printf("Failed 2"); return 2; }
                      undo_push( undoers, Undo2 );
                      if (Do3()) { undo(undoers); printf("Failed 3"); return 3; }
                      undo_push( undoers, Undo3 );
                      if (Do4()) { undo(undoers); printf("Failed 4"); return 4; }
                      undo_push( undoers, Undo4 );
                      if (Do5()) { undo(undoers); printf("Failed 5"); return 5; }
                      return 6;
                      }


                      I made undo_push do the O(n) work. This is less efficient than having undo do the O(n) work, as we expect more push's than undos. But this version was a touch simpler.



                      A more industrial strength version would have head and tail pointers and even capacity.



                      The basic idea is to keep a queue of undo actions in a stack, then execute them if you need to clean up.



                      Everything is local here, so we don't pollute global state.





                      struct undoer {
                      void(*action)(void*);
                      void(*cleanup)(void*);
                      void* state;
                      };

                      struct undoers {
                      undoer* top;
                      undoer buff[5];
                      };
                      void undo( undoers u ) {
                      while (u.top != buff)
                      {
                      (u.top->action)(u.top->state);
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pundo(void* pu) {
                      undo( *(undoers*)pu );
                      free(pu);
                      }
                      void cleanup_undoers(undoers u) {
                      while (u.top != buff)
                      {
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pcleanup_undoers(void* pu) {
                      cleanup_undoers(*(undoers*)pu);
                      free(pu);
                      }
                      void push_undoer( undoers* to_here, undoer u ) {
                      if (to_here->top != (to_here->buff+5))
                      {
                      to_here->top = u;
                      ++(to_here->top)
                      return;
                      }
                      undoers* chain = (undoers*)malloc( sizeof(undoers) );
                      memcpy(chain, to_here, sizeof(undoers));
                      memset(to_here, 0, sizeof(undoers));
                      undoer chainer;
                      chainer.action = pundo;
                      chainer.cleanup = pcleanup_undoers;
                      chainer.data = chain;
                      push_undoer( to_here, chainer );
                      push_undoer( to_here, u );
                      }
                      void paction( void* p ) {
                      (void)(*a)() = ((void)(*)());
                      a();
                      }
                      void push_undo( undoers* to_here, void(*action)() ) {
                      undor u;
                      u.action = paction;
                      u.cleanup = 0;
                      u.data = (void*)action;
                      push_undoer(to_here, u);
                      }


                      then you get:



                      int func() {
                      undoers u={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      push_undo( &u, Undo1 );
                      if (Do2()) { undo(u); printf("Failed 2"); return 2; }
                      push_undo( &u, Undo2 );
                      if (Do3()) { undo(u); printf("Failed 3"); return 3; }
                      push_undo( &u, Undo3 );
                      if (Do4()) { undo(u); printf("Failed 4"); return 4; }
                      push_undo( &u, Undo4 );
                      if (Do5()) { undo(u); printf("Failed 5"); return 5; }
                      cleanup_undoers(u);
                      return 6;
                      }


                      but that is getting ridiculous.






                      share|improve this answer



















                      • 1




                        More complex, requires that DoN/UndoN are actual functions (and the same signature), requires stack space, slower.
                        – Acorn
                        2 days ago















                      up vote
                      -2
                      down vote










                      up vote
                      -2
                      down vote









                      typedef void(*undoer)();
                      int undo( undoer*const* list ) {
                      while(*list) {
                      (*list)();
                      ++list;
                      }
                      }
                      void undo_push( undoer** list, undoer* undo ) {
                      if (!undo) return;
                      // swap
                      undoer* tmp = *list;
                      *list = undo;
                      undo = tmp;
                      undo_push( list+1, undo );
                      }
                      int func() {
                      undoer undoers[6]={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      undo_push( undoers, Undo1 );
                      if (Do2()) { undo(undoers); printf("Failed 2"); return 2; }
                      undo_push( undoers, Undo2 );
                      if (Do3()) { undo(undoers); printf("Failed 3"); return 3; }
                      undo_push( undoers, Undo3 );
                      if (Do4()) { undo(undoers); printf("Failed 4"); return 4; }
                      undo_push( undoers, Undo4 );
                      if (Do5()) { undo(undoers); printf("Failed 5"); return 5; }
                      return 6;
                      }


                      I made undo_push do the O(n) work. This is less efficient than having undo do the O(n) work, as we expect more push's than undos. But this version was a touch simpler.



                      A more industrial strength version would have head and tail pointers and even capacity.



                      The basic idea is to keep a queue of undo actions in a stack, then execute them if you need to clean up.



                      Everything is local here, so we don't pollute global state.





                      struct undoer {
                      void(*action)(void*);
                      void(*cleanup)(void*);
                      void* state;
                      };

                      struct undoers {
                      undoer* top;
                      undoer buff[5];
                      };
                      void undo( undoers u ) {
                      while (u.top != buff)
                      {
                      (u.top->action)(u.top->state);
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pundo(void* pu) {
                      undo( *(undoers*)pu );
                      free(pu);
                      }
                      void cleanup_undoers(undoers u) {
                      while (u.top != buff)
                      {
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pcleanup_undoers(void* pu) {
                      cleanup_undoers(*(undoers*)pu);
                      free(pu);
                      }
                      void push_undoer( undoers* to_here, undoer u ) {
                      if (to_here->top != (to_here->buff+5))
                      {
                      to_here->top = u;
                      ++(to_here->top)
                      return;
                      }
                      undoers* chain = (undoers*)malloc( sizeof(undoers) );
                      memcpy(chain, to_here, sizeof(undoers));
                      memset(to_here, 0, sizeof(undoers));
                      undoer chainer;
                      chainer.action = pundo;
                      chainer.cleanup = pcleanup_undoers;
                      chainer.data = chain;
                      push_undoer( to_here, chainer );
                      push_undoer( to_here, u );
                      }
                      void paction( void* p ) {
                      (void)(*a)() = ((void)(*)());
                      a();
                      }
                      void push_undo( undoers* to_here, void(*action)() ) {
                      undor u;
                      u.action = paction;
                      u.cleanup = 0;
                      u.data = (void*)action;
                      push_undoer(to_here, u);
                      }


                      then you get:



                      int func() {
                      undoers u={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      push_undo( &u, Undo1 );
                      if (Do2()) { undo(u); printf("Failed 2"); return 2; }
                      push_undo( &u, Undo2 );
                      if (Do3()) { undo(u); printf("Failed 3"); return 3; }
                      push_undo( &u, Undo3 );
                      if (Do4()) { undo(u); printf("Failed 4"); return 4; }
                      push_undo( &u, Undo4 );
                      if (Do5()) { undo(u); printf("Failed 5"); return 5; }
                      cleanup_undoers(u);
                      return 6;
                      }


                      but that is getting ridiculous.






                      share|improve this answer














                      typedef void(*undoer)();
                      int undo( undoer*const* list ) {
                      while(*list) {
                      (*list)();
                      ++list;
                      }
                      }
                      void undo_push( undoer** list, undoer* undo ) {
                      if (!undo) return;
                      // swap
                      undoer* tmp = *list;
                      *list = undo;
                      undo = tmp;
                      undo_push( list+1, undo );
                      }
                      int func() {
                      undoer undoers[6]={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      undo_push( undoers, Undo1 );
                      if (Do2()) { undo(undoers); printf("Failed 2"); return 2; }
                      undo_push( undoers, Undo2 );
                      if (Do3()) { undo(undoers); printf("Failed 3"); return 3; }
                      undo_push( undoers, Undo3 );
                      if (Do4()) { undo(undoers); printf("Failed 4"); return 4; }
                      undo_push( undoers, Undo4 );
                      if (Do5()) { undo(undoers); printf("Failed 5"); return 5; }
                      return 6;
                      }


                      I made undo_push do the O(n) work. This is less efficient than having undo do the O(n) work, as we expect more push's than undos. But this version was a touch simpler.



                      A more industrial strength version would have head and tail pointers and even capacity.



                      The basic idea is to keep a queue of undo actions in a stack, then execute them if you need to clean up.



                      Everything is local here, so we don't pollute global state.





                      struct undoer {
                      void(*action)(void*);
                      void(*cleanup)(void*);
                      void* state;
                      };

                      struct undoers {
                      undoer* top;
                      undoer buff[5];
                      };
                      void undo( undoers u ) {
                      while (u.top != buff)
                      {
                      (u.top->action)(u.top->state);
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pundo(void* pu) {
                      undo( *(undoers*)pu );
                      free(pu);
                      }
                      void cleanup_undoers(undoers u) {
                      while (u.top != buff)
                      {
                      if (u.top->cleanup)
                      (u.top->cleanup)(u.top->state);
                      --u.top;
                      }
                      }
                      void pcleanup_undoers(void* pu) {
                      cleanup_undoers(*(undoers*)pu);
                      free(pu);
                      }
                      void push_undoer( undoers* to_here, undoer u ) {
                      if (to_here->top != (to_here->buff+5))
                      {
                      to_here->top = u;
                      ++(to_here->top)
                      return;
                      }
                      undoers* chain = (undoers*)malloc( sizeof(undoers) );
                      memcpy(chain, to_here, sizeof(undoers));
                      memset(to_here, 0, sizeof(undoers));
                      undoer chainer;
                      chainer.action = pundo;
                      chainer.cleanup = pcleanup_undoers;
                      chainer.data = chain;
                      push_undoer( to_here, chainer );
                      push_undoer( to_here, u );
                      }
                      void paction( void* p ) {
                      (void)(*a)() = ((void)(*)());
                      a();
                      }
                      void push_undo( undoers* to_here, void(*action)() ) {
                      undor u;
                      u.action = paction;
                      u.cleanup = 0;
                      u.data = (void*)action;
                      push_undoer(to_here, u);
                      }


                      then you get:



                      int func() {
                      undoers u={0};

                      if (Do1()) { printf("Failed 1"); return 1; }
                      push_undo( &u, Undo1 );
                      if (Do2()) { undo(u); printf("Failed 2"); return 2; }
                      push_undo( &u, Undo2 );
                      if (Do3()) { undo(u); printf("Failed 3"); return 3; }
                      push_undo( &u, Undo3 );
                      if (Do4()) { undo(u); printf("Failed 4"); return 4; }
                      push_undo( &u, Undo4 );
                      if (Do5()) { undo(u); printf("Failed 5"); return 5; }
                      cleanup_undoers(u);
                      return 6;
                      }


                      but that is getting ridiculous.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 2 days ago

























                      answered 2 days ago









                      Yakk - Adam Nevraumont

                      179k19186365




                      179k19186365








                      • 1




                        More complex, requires that DoN/UndoN are actual functions (and the same signature), requires stack space, slower.
                        – Acorn
                        2 days ago
















                      • 1




                        More complex, requires that DoN/UndoN are actual functions (and the same signature), requires stack space, slower.
                        – Acorn
                        2 days ago










                      1




                      1




                      More complex, requires that DoN/UndoN are actual functions (and the same signature), requires stack space, slower.
                      – Acorn
                      2 days ago






                      More complex, requires that DoN/UndoN are actual functions (and the same signature), requires stack space, slower.
                      – Acorn
                      2 days ago












                      up vote
                      -4
                      down vote













                      Let's try for something with zero curly braces:



                      int result;
                      result = Do1() ? 1 : 0;
                      result = result ? result : Do2() ? 2 : 0;
                      result = result ? result : Do3() ? 3 : 0;
                      result = result ? result : Do4() ? 4 : 0;
                      result = result ? result : Do5() ? 5 : 0;

                      result > 4 ? (Undo5(),0) : 0;
                      result > 3 ? Undo4() : 0;
                      result > 2 ? Undo3() : 0;
                      result > 1 ? Undo2() : 0;
                      result > 0 ? Undo1() : 0;

                      result ? printf("Failed %drn", result) : 0;


                      Yes. 0; is a valid statement in C (and C++). In the case that some of the functions return something that is incompatible with this syntax (e.g. void perhaps) then the Undo5() style can be used.






                      share|improve this answer























                      • This assumes the UndoN functions return values, when in fact they may be (and most probably are) declared void (or aren't even functions at all).
                        – Cássio Renan
                        2 days ago












                      • msvc is never a particularly standards compliant compiler, but without thinking about it I did actually develop this with void Undo functions. No idea if its actually valid. If it isn't one could just go with: `result > 4 ? (Undo5(), 0) : 0; Doesn't help of course. if 'UndoX' isn't actually a function.
                        – Chris Becke
                        2 days ago










                      • yeah, MSVC is a bad, bad boy, for C++ at least. In C, this seems to be valid. My bad.
                        – Cássio Renan
                        2 days ago








                      • 2




                        I would argue that if (result > 4) Undo5(); is easier to understand than a ternary conditional with no false action and a discarded result. (if statements don't need curly braces)
                        – pizzapants184
                        2 days ago










                      • True. I really was avoiding any kind of explicit flow control. ternary conditions are cheating I know.
                        – Chris Becke
                        yesterday















                      up vote
                      -4
                      down vote













                      Let's try for something with zero curly braces:



                      int result;
                      result = Do1() ? 1 : 0;
                      result = result ? result : Do2() ? 2 : 0;
                      result = result ? result : Do3() ? 3 : 0;
                      result = result ? result : Do4() ? 4 : 0;
                      result = result ? result : Do5() ? 5 : 0;

                      result > 4 ? (Undo5(),0) : 0;
                      result > 3 ? Undo4() : 0;
                      result > 2 ? Undo3() : 0;
                      result > 1 ? Undo2() : 0;
                      result > 0 ? Undo1() : 0;

                      result ? printf("Failed %drn", result) : 0;


                      Yes. 0; is a valid statement in C (and C++). In the case that some of the functions return something that is incompatible with this syntax (e.g. void perhaps) then the Undo5() style can be used.






                      share|improve this answer























                      • This assumes the UndoN functions return values, when in fact they may be (and most probably are) declared void (or aren't even functions at all).
                        – Cássio Renan
                        2 days ago












                      • msvc is never a particularly standards compliant compiler, but without thinking about it I did actually develop this with void Undo functions. No idea if its actually valid. If it isn't one could just go with: `result > 4 ? (Undo5(), 0) : 0; Doesn't help of course. if 'UndoX' isn't actually a function.
                        – Chris Becke
                        2 days ago










                      • yeah, MSVC is a bad, bad boy, for C++ at least. In C, this seems to be valid. My bad.
                        – Cássio Renan
                        2 days ago








                      • 2




                        I would argue that if (result > 4) Undo5(); is easier to understand than a ternary conditional with no false action and a discarded result. (if statements don't need curly braces)
                        – pizzapants184
                        2 days ago










                      • True. I really was avoiding any kind of explicit flow control. ternary conditions are cheating I know.
                        – Chris Becke
                        yesterday













                      up vote
                      -4
                      down vote










                      up vote
                      -4
                      down vote









                      Let's try for something with zero curly braces:



                      int result;
                      result = Do1() ? 1 : 0;
                      result = result ? result : Do2() ? 2 : 0;
                      result = result ? result : Do3() ? 3 : 0;
                      result = result ? result : Do4() ? 4 : 0;
                      result = result ? result : Do5() ? 5 : 0;

                      result > 4 ? (Undo5(),0) : 0;
                      result > 3 ? Undo4() : 0;
                      result > 2 ? Undo3() : 0;
                      result > 1 ? Undo2() : 0;
                      result > 0 ? Undo1() : 0;

                      result ? printf("Failed %drn", result) : 0;


                      Yes. 0; is a valid statement in C (and C++). In the case that some of the functions return something that is incompatible with this syntax (e.g. void perhaps) then the Undo5() style can be used.






                      share|improve this answer














                      Let's try for something with zero curly braces:



                      int result;
                      result = Do1() ? 1 : 0;
                      result = result ? result : Do2() ? 2 : 0;
                      result = result ? result : Do3() ? 3 : 0;
                      result = result ? result : Do4() ? 4 : 0;
                      result = result ? result : Do5() ? 5 : 0;

                      result > 4 ? (Undo5(),0) : 0;
                      result > 3 ? Undo4() : 0;
                      result > 2 ? Undo3() : 0;
                      result > 1 ? Undo2() : 0;
                      result > 0 ? Undo1() : 0;

                      result ? printf("Failed %drn", result) : 0;


                      Yes. 0; is a valid statement in C (and C++). In the case that some of the functions return something that is incompatible with this syntax (e.g. void perhaps) then the Undo5() style can be used.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 2 days ago









                      Peter Mortensen

                      13.3k1983111




                      13.3k1983111










                      answered 2 days ago









                      Chris Becke

                      26.3k656120




                      26.3k656120












                      • This assumes the UndoN functions return values, when in fact they may be (and most probably are) declared void (or aren't even functions at all).
                        – Cássio Renan
                        2 days ago












                      • msvc is never a particularly standards compliant compiler, but without thinking about it I did actually develop this with void Undo functions. No idea if its actually valid. If it isn't one could just go with: `result > 4 ? (Undo5(), 0) : 0; Doesn't help of course. if 'UndoX' isn't actually a function.
                        – Chris Becke
                        2 days ago










                      • yeah, MSVC is a bad, bad boy, for C++ at least. In C, this seems to be valid. My bad.
                        – Cássio Renan
                        2 days ago








                      • 2




                        I would argue that if (result > 4) Undo5(); is easier to understand than a ternary conditional with no false action and a discarded result. (if statements don't need curly braces)
                        – pizzapants184
                        2 days ago










                      • True. I really was avoiding any kind of explicit flow control. ternary conditions are cheating I know.
                        – Chris Becke
                        yesterday


















                      • This assumes the UndoN functions return values, when in fact they may be (and most probably are) declared void (or aren't even functions at all).
                        – Cássio Renan
                        2 days ago












                      • msvc is never a particularly standards compliant compiler, but without thinking about it I did actually develop this with void Undo functions. No idea if its actually valid. If it isn't one could just go with: `result > 4 ? (Undo5(), 0) : 0; Doesn't help of course. if 'UndoX' isn't actually a function.
                        – Chris Becke
                        2 days ago










                      • yeah, MSVC is a bad, bad boy, for C++ at least. In C, this seems to be valid. My bad.
                        – Cássio Renan
                        2 days ago








                      • 2




                        I would argue that if (result > 4) Undo5(); is easier to understand than a ternary conditional with no false action and a discarded result. (if statements don't need curly braces)
                        – pizzapants184
                        2 days ago










                      • True. I really was avoiding any kind of explicit flow control. ternary conditions are cheating I know.
                        – Chris Becke
                        yesterday
















                      This assumes the UndoN functions return values, when in fact they may be (and most probably are) declared void (or aren't even functions at all).
                      – Cássio Renan
                      2 days ago






                      This assumes the UndoN functions return values, when in fact they may be (and most probably are) declared void (or aren't even functions at all).
                      – Cássio Renan
                      2 days ago














                      msvc is never a particularly standards compliant compiler, but without thinking about it I did actually develop this with void Undo functions. No idea if its actually valid. If it isn't one could just go with: `result > 4 ? (Undo5(), 0) : 0; Doesn't help of course. if 'UndoX' isn't actually a function.
                      – Chris Becke
                      2 days ago




                      msvc is never a particularly standards compliant compiler, but without thinking about it I did actually develop this with void Undo functions. No idea if its actually valid. If it isn't one could just go with: `result > 4 ? (Undo5(), 0) : 0; Doesn't help of course. if 'UndoX' isn't actually a function.
                      – Chris Becke
                      2 days ago












                      yeah, MSVC is a bad, bad boy, for C++ at least. In C, this seems to be valid. My bad.
                      – Cássio Renan
                      2 days ago






                      yeah, MSVC is a bad, bad boy, for C++ at least. In C, this seems to be valid. My bad.
                      – Cássio Renan
                      2 days ago






                      2




                      2




                      I would argue that if (result > 4) Undo5(); is easier to understand than a ternary conditional with no false action and a discarded result. (if statements don't need curly braces)
                      – pizzapants184
                      2 days ago




                      I would argue that if (result > 4) Undo5(); is easier to understand than a ternary conditional with no false action and a discarded result. (if statements don't need curly braces)
                      – pizzapants184
                      2 days ago












                      True. I really was avoiding any kind of explicit flow control. ternary conditions are cheating I know.
                      – Chris Becke
                      yesterday




                      True. I really was avoiding any kind of explicit flow control. ternary conditions are cheating I know.
                      – Chris Becke
                      yesterday










                      up vote
                      -6
                      down vote













                      A sane (no gotos, no nested or chained ifs) approach would be



                      int bar(void)
                      {
                      int rc = 0;

                      do
                      {
                      if (do1())
                      {
                      rc = 1;
                      break;
                      }

                      if (do2())
                      {
                      rc = 2;
                      break;
                      }

                      ...

                      if (do5())
                      {
                      rc = 5;
                      break;
                      }
                      } while (0);

                      if (rc)
                      {
                      /* More or less stolen from Chris' answer:
                      https://stackoverflow.com/a/53444967/694576) */
                      switch(rc - 1)
                      {
                      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
                      undo5();

                      case 4:
                      undo4();

                      case 3:
                      undo3();

                      case 2:
                      undo2();

                      case 1:
                      undo1();

                      default:
                      break;
                      }
                      }

                      return rc;
                      }





                      share|improve this answer

















                      • 10




                        If your definition of "sane" is "no goto" then you already failed because sanest way to handle this is indeed to use goto.
                        – Purple Ice
                        2 days ago










                      • @PurpleIce: You are probably right for simple cases like the OP's. But the moment we have several such things woven into each other using goto is far to error prone. And yes, this latter case could be considered a design issue.
                        – alk
                        2 days ago








                      • 3




                        @alk The goto solution scales linearly to any complexity, as shown in other answers, just like this one, but with less clutter and extraneous loops and branches.
                        – Acorn
                        2 days ago






                      • 3




                        The do { ... } while (0) used here is just an obfuscated way of writing a goto. There’s no advantage at all compared to using goto, and it’s quite a bit harder to read.
                        – NobodyNada
                        2 days ago















                      up vote
                      -6
                      down vote













                      A sane (no gotos, no nested or chained ifs) approach would be



                      int bar(void)
                      {
                      int rc = 0;

                      do
                      {
                      if (do1())
                      {
                      rc = 1;
                      break;
                      }

                      if (do2())
                      {
                      rc = 2;
                      break;
                      }

                      ...

                      if (do5())
                      {
                      rc = 5;
                      break;
                      }
                      } while (0);

                      if (rc)
                      {
                      /* More or less stolen from Chris' answer:
                      https://stackoverflow.com/a/53444967/694576) */
                      switch(rc - 1)
                      {
                      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
                      undo5();

                      case 4:
                      undo4();

                      case 3:
                      undo3();

                      case 2:
                      undo2();

                      case 1:
                      undo1();

                      default:
                      break;
                      }
                      }

                      return rc;
                      }





                      share|improve this answer

















                      • 10




                        If your definition of "sane" is "no goto" then you already failed because sanest way to handle this is indeed to use goto.
                        – Purple Ice
                        2 days ago










                      • @PurpleIce: You are probably right for simple cases like the OP's. But the moment we have several such things woven into each other using goto is far to error prone. And yes, this latter case could be considered a design issue.
                        – alk
                        2 days ago








                      • 3




                        @alk The goto solution scales linearly to any complexity, as shown in other answers, just like this one, but with less clutter and extraneous loops and branches.
                        – Acorn
                        2 days ago






                      • 3




                        The do { ... } while (0) used here is just an obfuscated way of writing a goto. There’s no advantage at all compared to using goto, and it’s quite a bit harder to read.
                        – NobodyNada
                        2 days ago













                      up vote
                      -6
                      down vote










                      up vote
                      -6
                      down vote









                      A sane (no gotos, no nested or chained ifs) approach would be



                      int bar(void)
                      {
                      int rc = 0;

                      do
                      {
                      if (do1())
                      {
                      rc = 1;
                      break;
                      }

                      if (do2())
                      {
                      rc = 2;
                      break;
                      }

                      ...

                      if (do5())
                      {
                      rc = 5;
                      break;
                      }
                      } while (0);

                      if (rc)
                      {
                      /* More or less stolen from Chris' answer:
                      https://stackoverflow.com/a/53444967/694576) */
                      switch(rc - 1)
                      {
                      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
                      undo5();

                      case 4:
                      undo4();

                      case 3:
                      undo3();

                      case 2:
                      undo2();

                      case 1:
                      undo1();

                      default:
                      break;
                      }
                      }

                      return rc;
                      }





                      share|improve this answer












                      A sane (no gotos, no nested or chained ifs) approach would be



                      int bar(void)
                      {
                      int rc = 0;

                      do
                      {
                      if (do1())
                      {
                      rc = 1;
                      break;
                      }

                      if (do2())
                      {
                      rc = 2;
                      break;
                      }

                      ...

                      if (do5())
                      {
                      rc = 5;
                      break;
                      }
                      } while (0);

                      if (rc)
                      {
                      /* More or less stolen from Chris' answer:
                      https://stackoverflow.com/a/53444967/694576) */
                      switch(rc - 1)
                      {
                      case 5: /* Not needed for this example, but left in in case we'd add do6() ... */
                      undo5();

                      case 4:
                      undo4();

                      case 3:
                      undo3();

                      case 2:
                      undo2();

                      case 1:
                      undo1();

                      default:
                      break;
                      }
                      }

                      return rc;
                      }






                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 2 days ago









                      alk

                      57.7k759167




                      57.7k759167








                      • 10




                        If your definition of "sane" is "no goto" then you already failed because sanest way to handle this is indeed to use goto.
                        – Purple Ice
                        2 days ago










                      • @PurpleIce: You are probably right for simple cases like the OP's. But the moment we have several such things woven into each other using goto is far to error prone. And yes, this latter case could be considered a design issue.
                        – alk
                        2 days ago








                      • 3




                        @alk The goto solution scales linearly to any complexity, as shown in other answers, just like this one, but with less clutter and extraneous loops and branches.
                        – Acorn
                        2 days ago






                      • 3




                        The do { ... } while (0) used here is just an obfuscated way of writing a goto. There’s no advantage at all compared to using goto, and it’s quite a bit harder to read.
                        – NobodyNada
                        2 days ago














                      • 10




                        If your definition of "sane" is "no goto" then you already failed because sanest way to handle this is indeed to use goto.
                        – Purple Ice
                        2 days ago










                      • @PurpleIce: You are probably right for simple cases like the OP's. But the moment we have several such things woven into each other using goto is far to error prone. And yes, this latter case could be considered a design issue.
                        – alk
                        2 days ago








                      • 3




                        @alk The goto solution scales linearly to any complexity, as shown in other answers, just like this one, but with less clutter and extraneous loops and branches.
                        – Acorn
                        2 days ago






                      • 3




                        The do { ... } while (0) used here is just an obfuscated way of writing a goto. There’s no advantage at all compared to using goto, and it’s quite a bit harder to read.
                        – NobodyNada
                        2 days ago








                      10




                      10




                      If your definition of "sane" is "no goto" then you already failed because sanest way to handle this is indeed to use goto.
                      – Purple Ice
                      2 days ago




                      If your definition of "sane" is "no goto" then you already failed because sanest way to handle this is indeed to use goto.
                      – Purple Ice
                      2 days ago












                      @PurpleIce: You are probably right for simple cases like the OP's. But the moment we have several such things woven into each other using goto is far to error prone. And yes, this latter case could be considered a design issue.
                      – alk
                      2 days ago






                      @PurpleIce: You are probably right for simple cases like the OP's. But the moment we have several such things woven into each other using goto is far to error prone. And yes, this latter case could be considered a design issue.
                      – alk
                      2 days ago






                      3




                      3




                      @alk The goto solution scales linearly to any complexity, as shown in other answers, just like this one, but with less clutter and extraneous loops and branches.
                      – Acorn
                      2 days ago




                      @alk The goto solution scales linearly to any complexity, as shown in other answers, just like this one, but with less clutter and extraneous loops and branches.
                      – Acorn
                      2 days ago




                      3




                      3




                      The do { ... } while (0) used here is just an obfuscated way of writing a goto. There’s no advantage at all compared to using goto, and it’s quite a bit harder to read.
                      – NobodyNada
                      2 days ago




                      The do { ... } while (0) used here is just an obfuscated way of writing a goto. There’s no advantage at all compared to using goto, and it’s quite a bit harder to read.
                      – NobodyNada
                      2 days ago


















                       

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