Unsigned arithmetic in C++











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I just observed a strange phenomenon when doing unsigned arithmetic. It's expected that b and -a have the same number 4294967286 due to wraparound, but the actual output for b and -a is -10 and 4294967286 respectively. Could anyone help give a hint?



#include <iostream>



int main() {

  unsigned int a = 10;

  int b = -a;

  std::cout << b << ", " << -a << std::endl;

}


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c++







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  • 2




    Why do you expect that an int would have the value 4294967286, which is much larger than the maximum int?
    – molbdnilo
    2 days ago








  • 1




    int b = -a; -> unsigned int b = -a;
    – Paul R
    2 days ago






  • 1




    @molbdnilo My fault. I forgot that b overflows, too.
    – Zhe Chen
    2 days ago

















up vote
11
down vote

favorite
2












I just observed a strange phenomenon when doing unsigned arithmetic. It's expected that b and -a have the same number 4294967286 due to wraparound, but the actual output for b and -a is -10 and 4294967286 respectively. Could anyone help give a hint?



#include <iostream>



int main() {

  unsigned int a = 10;

  int b = -a;

  std::cout << b << ", " << -a << std::endl;

}


https://repl.it/repls/ExpertDrabOrganization






c++







share|improve this question




















  • 2




    Why do you expect that an int would have the value 4294967286, which is much larger than the maximum int?
    – molbdnilo
    2 days ago








  • 1




    int b = -a; -> unsigned int b = -a;
    – Paul R
    2 days ago






  • 1




    @molbdnilo My fault. I forgot that b overflows, too.
    – Zhe Chen
    2 days ago















up vote
11
down vote

favorite
2









up vote
11
down vote

favorite
2






2





I just observed a strange phenomenon when doing unsigned arithmetic. It's expected that b and -a have the same number 4294967286 due to wraparound, but the actual output for b and -a is -10 and 4294967286 respectively. Could anyone help give a hint?



#include <iostream>



int main() {

  unsigned int a = 10;

  int b = -a;

  std::cout << b << ", " << -a << std::endl;

}


https://repl.it/repls/ExpertDrabOrganization






c++







share|improve this question















I just observed a strange phenomenon when doing unsigned arithmetic. It's expected that b and -a have the same number 4294967286 due to wraparound, but the actual output for b and -a is -10 and 4294967286 respectively. Could anyone help give a hint?



#include <iostream>



int main() {

  unsigned int a = 10;

  int b = -a;

  std::cout << b << ", " << -a << std::endl;

}


https://repl.it/repls/ExpertDrabOrganization






c++




c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 days ago

























asked 2 days ago









Zhe Chen

1,42331528




1,42331528








  • 2




    Why do you expect that an int would have the value 4294967286, which is much larger than the maximum int?
    – molbdnilo
    2 days ago








  • 1




    int b = -a; -> unsigned int b = -a;
    – Paul R
    2 days ago






  • 1




    @molbdnilo My fault. I forgot that b overflows, too.
    – Zhe Chen
    2 days ago
















  • 2




    Why do you expect that an int would have the value 4294967286, which is much larger than the maximum int?
    – molbdnilo
    2 days ago








  • 1




    int b = -a; -> unsigned int b = -a;
    – Paul R
    2 days ago






  • 1




    @molbdnilo My fault. I forgot that b overflows, too.
    – Zhe Chen
    2 days ago










2




2




Why do you expect that an int would have the value 4294967286, which is much larger than the maximum int?
– molbdnilo
2 days ago






Why do you expect that an int would have the value 4294967286, which is much larger than the maximum int?
– molbdnilo
2 days ago






1




1




int b = -a; -> unsigned int b = -a;
– Paul R
2 days ago




int b = -a; -> unsigned int b = -a;
– Paul R
2 days ago




1




1




@molbdnilo My fault. I forgot that b overflows, too.
– Zhe Chen
2 days ago






@molbdnilo My fault. I forgot that b overflows, too.
– Zhe Chen
2 days ago














1 Answer
1






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oldest

votes

















up vote
17
down vote



accepted










-a is evaluated in unsigned arithmetic, and will be a number larger than std::numeric_limits<int>::max(). The unary operator - when applied to an unsigned type acts more like a modulus operator.



Therefore the behaviour of your program is implementation defined due to an out-of-range assignment to an int.






share|improve this answer



















  • 3




    implementation defined - I.e. look at the docs, and shout at your implementer if they don't do what they promise.
    – StoryTeller
    2 days ago












  • @StoryTeller: Since which flashy new standard?
    – Bathsheba
    2 days ago






  • 1




    @StoryTeller It was implementation-defined in C++03 too.
    – Angew
    2 days ago






  • 2




    @Angew - Interestingly enough, it's implementation defined in a C99 draft I have lying around. It would seem that a conversion was always with implementation defined semantics. Overflow in arithmetic with signed types is UB, however, wherever I checked. Curious.
    – StoryTeller
    2 days ago








  • 1




    gcc implementation defined behavior: The value is reduced module 2^N. So the same for clang. For MSVC this behavior is not ensured: docs.microsoft.com/en-us/cpp/c-language/…
    – Oliv
    2 days ago











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up vote
17
down vote



accepted










-a is evaluated in unsigned arithmetic, and will be a number larger than std::numeric_limits<int>::max(). The unary operator - when applied to an unsigned type acts more like a modulus operator.



Therefore the behaviour of your program is implementation defined due to an out-of-range assignment to an int.






share|improve this answer



















  • 3




    implementation defined - I.e. look at the docs, and shout at your implementer if they don't do what they promise.
    – StoryTeller
    2 days ago












  • @StoryTeller: Since which flashy new standard?
    – Bathsheba
    2 days ago






  • 1




    @StoryTeller It was implementation-defined in C++03 too.
    – Angew
    2 days ago






  • 2




    @Angew - Interestingly enough, it's implementation defined in a C99 draft I have lying around. It would seem that a conversion was always with implementation defined semantics. Overflow in arithmetic with signed types is UB, however, wherever I checked. Curious.
    – StoryTeller
    2 days ago








  • 1




    gcc implementation defined behavior: The value is reduced module 2^N. So the same for clang. For MSVC this behavior is not ensured: docs.microsoft.com/en-us/cpp/c-language/…
    – Oliv
    2 days ago















up vote
17
down vote



accepted










-a is evaluated in unsigned arithmetic, and will be a number larger than std::numeric_limits<int>::max(). The unary operator - when applied to an unsigned type acts more like a modulus operator.



Therefore the behaviour of your program is implementation defined due to an out-of-range assignment to an int.






share|improve this answer



















  • 3




    implementation defined - I.e. look at the docs, and shout at your implementer if they don't do what they promise.
    – StoryTeller
    2 days ago












  • @StoryTeller: Since which flashy new standard?
    – Bathsheba
    2 days ago






  • 1




    @StoryTeller It was implementation-defined in C++03 too.
    – Angew
    2 days ago






  • 2




    @Angew - Interestingly enough, it's implementation defined in a C99 draft I have lying around. It would seem that a conversion was always with implementation defined semantics. Overflow in arithmetic with signed types is UB, however, wherever I checked. Curious.
    – StoryTeller
    2 days ago








  • 1




    gcc implementation defined behavior: The value is reduced module 2^N. So the same for clang. For MSVC this behavior is not ensured: docs.microsoft.com/en-us/cpp/c-language/…
    – Oliv
    2 days ago













up vote
17
down vote



accepted







up vote
17
down vote



accepted






-a is evaluated in unsigned arithmetic, and will be a number larger than std::numeric_limits<int>::max(). The unary operator - when applied to an unsigned type acts more like a modulus operator.



Therefore the behaviour of your program is implementation defined due to an out-of-range assignment to an int.






share|improve this answer














-a is evaluated in unsigned arithmetic, and will be a number larger than std::numeric_limits<int>::max(). The unary operator - when applied to an unsigned type acts more like a modulus operator.



Therefore the behaviour of your program is implementation defined due to an out-of-range assignment to an int.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 days ago









Bakudan

13.3k84264




13.3k84264










answered 2 days ago









Bathsheba

172k27244366




172k27244366








  • 3




    implementation defined - I.e. look at the docs, and shout at your implementer if they don't do what they promise.
    – StoryTeller
    2 days ago












  • @StoryTeller: Since which flashy new standard?
    – Bathsheba
    2 days ago






  • 1




    @StoryTeller It was implementation-defined in C++03 too.
    – Angew
    2 days ago






  • 2




    @Angew - Interestingly enough, it's implementation defined in a C99 draft I have lying around. It would seem that a conversion was always with implementation defined semantics. Overflow in arithmetic with signed types is UB, however, wherever I checked. Curious.
    – StoryTeller
    2 days ago








  • 1




    gcc implementation defined behavior: The value is reduced module 2^N. So the same for clang. For MSVC this behavior is not ensured: docs.microsoft.com/en-us/cpp/c-language/…
    – Oliv
    2 days ago














  • 3




    implementation defined - I.e. look at the docs, and shout at your implementer if they don't do what they promise.
    – StoryTeller
    2 days ago












  • @StoryTeller: Since which flashy new standard?
    – Bathsheba
    2 days ago






  • 1




    @StoryTeller It was implementation-defined in C++03 too.
    – Angew
    2 days ago






  • 2




    @Angew - Interestingly enough, it's implementation defined in a C99 draft I have lying around. It would seem that a conversion was always with implementation defined semantics. Overflow in arithmetic with signed types is UB, however, wherever I checked. Curious.
    – StoryTeller
    2 days ago








  • 1




    gcc implementation defined behavior: The value is reduced module 2^N. So the same for clang. For MSVC this behavior is not ensured: docs.microsoft.com/en-us/cpp/c-language/…
    – Oliv
    2 days ago








3




3




implementation defined - I.e. look at the docs, and shout at your implementer if they don't do what they promise.
– StoryTeller
2 days ago






implementation defined - I.e. look at the docs, and shout at your implementer if they don't do what they promise.
– StoryTeller
2 days ago














@StoryTeller: Since which flashy new standard?
– Bathsheba
2 days ago




@StoryTeller: Since which flashy new standard?
– Bathsheba
2 days ago




1




1




@StoryTeller It was implementation-defined in C++03 too.
– Angew
2 days ago




@StoryTeller It was implementation-defined in C++03 too.
– Angew
2 days ago




2




2




@Angew - Interestingly enough, it's implementation defined in a C99 draft I have lying around. It would seem that a conversion was always with implementation defined semantics. Overflow in arithmetic with signed types is UB, however, wherever I checked. Curious.
– StoryTeller
2 days ago






@Angew - Interestingly enough, it's implementation defined in a C99 draft I have lying around. It would seem that a conversion was always with implementation defined semantics. Overflow in arithmetic with signed types is UB, however, wherever I checked. Curious.
– StoryTeller
2 days ago






1




1




gcc implementation defined behavior: The value is reduced module 2^N. So the same for clang. For MSVC this behavior is not ensured: docs.microsoft.com/en-us/cpp/c-language/…
– Oliv
2 days ago




gcc implementation defined behavior: The value is reduced module 2^N. So the same for clang. For MSVC this behavior is not ensured: docs.microsoft.com/en-us/cpp/c-language/…
– Oliv
2 days ago


















 

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