Why does the compiler prefer f(const void*) to f(const std::string &)?
up vote
32
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Consider the following piece of code:
#include <iostream>
#include <string>
// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose
void f(const std::string &) { std::cout << "const std::string &"; }
void f(const void *) { std::cout << "const void *"; }
int main()
{
f("hello");
std::cout << std::endl;
}
I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:
$ g++ -std=c++11 strings_1.cpp -Wall
$ ./a.out
const void *
Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).
So, why does the compiler pick f(const void*) over f(const std::string &)?
c++ stdstring string-literals function-overloading overload-resolution
edited Nov 17 at 19:00
curiousguy
4,47722940
asked Nov 17 at 18:27
omar
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up vote
32
down vote
favorite
6
Consider the following piece of code:
#include <iostream>
#include <string>
// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose
void f(const std::string &) { std::cout << "const std::string &"; }
void f(const void *) { std::cout << "const void *"; }
int main()
{
f("hello");
std::cout << std::endl;
}
I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:
$ g++ -std=c++11 strings_1.cpp -Wall
$ ./a.out
const void *
Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).
So, why does the compiler pick f(const void*) over f(const std::string &)?
c++ stdstring string-literals function-overloading overload-resolution
edited Nov 17 at 19:00
curiousguy
4,47722940
asked Nov 17 at 18:27
omar
21918
New contributor
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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4
Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
Nov 17 at 18:53
@geza awesome. I was looking for it, thanks.
– omar
Nov 17 at 18:56
The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
Nov 17 at 19:01
3
Well, a string literal is not anstd::string, it's a static array ofchars, which decays to a pointer to its first character. This behavior is inherited from C which never had something likestd::string, but ample amounts of code handling strings nonetheless.
– cmaster
Nov 17 at 21:49
1
If you specifically want astd::stringliteral you can achieve that by adding asbehind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
2 days ago
|
show 2 more comments
up vote
32
down vote
favorite
6
up vote
32
down vote
favorite
6
6
Consider the following piece of code:
#include <iostream>
#include <string>
// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose
void f(const std::string &) { std::cout << "const std::string &"; }
void f(const void *) { std::cout << "const void *"; }
int main()
{
f("hello");
std::cout << std::endl;
}
I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:
$ g++ -std=c++11 strings_1.cpp -Wall
$ ./a.out
const void *
Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).
So, why does the compiler pick f(const void*) over f(const std::string &)?
c++ stdstring string-literals function-overloading overload-resolution
edited Nov 17 at 19:00
curiousguy
4,47722940
asked Nov 17 at 18:27
omar
21918
New contributor
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Consider the following piece of code:
#include <iostream>
#include <string>
// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose
void f(const std::string &) { std::cout << "const std::string &"; }
void f(const void *) { std::cout << "const void *"; }
int main()
{
f("hello");
std::cout << std::endl;
}
I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:
$ g++ -std=c++11 strings_1.cpp -Wall
$ ./a.out
const void *
Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).
So, why does the compiler pick f(const void*) over f(const std::string &)?
c++ stdstring string-literals function-overloading overload-resolution
c++ stdstring string-literals function-overloading overload-resolution
edited Nov 17 at 19:00
curiousguy
4,47722940
asked Nov 17 at 18:27
omar
21918
New contributor
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 17 at 19:00
curiousguy
4,47722940
asked Nov 17 at 18:27
omar
21918
New contributor
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Nov 17 at 19:00
curiousguy
4,47722940
edited Nov 17 at 19:00
curiousguy
4,47722940
edited Nov 17 at 19:00
curiousguy
4,47722940
4,47722940
asked Nov 17 at 18:27
omar
21918
New contributor
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 17 at 18:27
omar
21918
asked Nov 17 at 18:27
omar
21918
21918
New contributor
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
Nov 17 at 18:53
@geza awesome. I was looking for it, thanks.
– omar
Nov 17 at 18:56
The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
Nov 17 at 19:01
3
Well, a string literal is not anstd::string, it's a static array ofchars, which decays to a pointer to its first character. This behavior is inherited from C which never had something likestd::string, but ample amounts of code handling strings nonetheless.
– cmaster
Nov 17 at 21:49
1
If you specifically want astd::stringliteral you can achieve that by adding asbehind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
2 days ago
|
show 2 more comments
4
Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
Nov 17 at 18:53
@geza awesome. I was looking for it, thanks.
– omar
Nov 17 at 18:56
The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
Nov 17 at 19:01
3
Well, a string literal is not anstd::string, it's a static array ofchars, which decays to a pointer to its first character. This behavior is inherited from C which never had something likestd::string, but ample amounts of code handling strings nonetheless.
– cmaster
Nov 17 at 21:49
1
If you specifically want astd::stringliteral you can achieve that by adding asbehind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
2 days ago
4
4
Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
Nov 17 at 18:53
Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
Nov 17 at 18:53
@geza awesome. I was looking for it, thanks.
– omar
Nov 17 at 18:56
@geza awesome. I was looking for it, thanks.
– omar
Nov 17 at 18:56
The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
Nov 17 at 19:01
The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
Nov 17 at 19:01
3
3
Well, a string literal is not an
std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.– cmaster
Nov 17 at 21:49
Well, a string literal is not an
std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.– cmaster
Nov 17 at 21:49
1
1
If you specifically want a
std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s– henje
2 days ago
If you specifically want a
std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s– henje
2 days ago
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
40
down vote
accepted
Converting to a std::string requires a "user defined conversion".
Converting to void const* does not.
User defined conversions are ordered behind built in ones.
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
40
down vote
accepted
Converting to a std::string requires a "user defined conversion".
Converting to void const* does not.
User defined conversions are ordered behind built in ones.
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
add a comment |
up vote
40
down vote
accepted
Converting to a std::string requires a "user defined conversion".
Converting to void const* does not.
User defined conversions are ordered behind built in ones.
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
add a comment |
up vote
40
down vote
accepted
up vote
40
down vote
accepted
Converting to a std::string requires a "user defined conversion".
Converting to void const* does not.
User defined conversions are ordered behind built in ones.
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
Converting to a std::string requires a "user defined conversion".
Converting to void const* does not.
User defined conversions are ordered behind built in ones.
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
answered Nov 17 at 18:40
Yakk - Adam Nevraumont
178k19185364
178k19185364
add a comment |
add a comment |
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4
Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
Nov 17 at 18:53
@geza awesome. I was looking for it, thanks.
– omar
Nov 17 at 18:56
The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
Nov 17 at 19:01
3
Well, a string literal is not an
std::string, it's a static array ofchars, which decays to a pointer to its first character. This behavior is inherited from C which never had something likestd::string, but ample amounts of code handling strings nonetheless.– cmaster
Nov 17 at 21:49
1
If you specifically want a
std::stringliteral you can achieve that by adding asbehind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s– henje
2 days ago