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It's more complicated than this. C++ compilers can't just "reject" UB code - they don't have enough information at all times to prove that code invokes UB, the

It's more complicated than this. C++ compilers can't just "reject" UB code - they don't have enough information at all times to prove that code invokes UB, the languages is not powerful enough to represent this in all cases. When people say "just fix this in compilers, don't compile when they run into UB" it's a fundamental misunderstanding of the problem. Yes, in some cases the compiler can prove UB, and uses it to write optimiations, but it doesn't prove that all code is not invoking UB.

gpderetta 9y ago
> a fundamental misunderstanding of the problem No kidding! There is one right here: > Yes, in some cases the compiler can prove UB, and uses it to write optimiation Compilers do not prove UB , on the contrary they postulate no-UB and then use this axiom to prove other properties of the code.
- staticassertion 9y ago
  I suppose that's true, yes. My point was more that your compiler is not hitting a line of code and saying "I know this is or is not UB" because it can not do that in the general case - it is definitely incorrect to say it is 'proving' this. Would you consider that an accurate representation?
  - masklinn 9y ago
    > My point was more that your compiler is not hitting a line of code and saying "I know this is or is not UB" because it can not do that in the general case I think it's still a (cause of) misunderstanding. For the compiler, UBs are situations which axiomatically can not occur. "This is UB" is not a concept, because the compiler assumes at all point that UBs can not occur. The compiler doesn't go "oh you're dereferencing a pointer which may be null, fuck you", it goes "you're dereferencing a pointer so it can't be null, and thus I can remove anything assuming possible nullability".
    - staticassertion 9y ago
      > The compiler doesn't go "oh you're dereferencing a pointer which may be null, fuck you", it goes "you're dereferencing a pointer so it can't be null, and thus I can remove anything assuming possible nullability". :) This is what I'm trying to say.

gpderetta 9y ago

> a fundamental misunderstanding of the problem No kidding! There is one right here: > Yes, in some cases the compiler can prove UB, and uses it to write optimiation Compilers do not prove UB , on the contrary they postulate no-UB and then use this axiom to prove other properties of the code.

staticassertion 9y ago
I suppose that's true, yes. My point was more that your compiler is not hitting a line of code and saying "I know this is or is not UB" because it can not do that in the general case - it is definitely incorrect to say it is 'proving' this. Would you consider that an accurate representation?
- masklinn 9y ago
  > My point was more that your compiler is not hitting a line of code and saying "I know this is or is not UB" because it can not do that in the general case I think it's still a (cause of) misunderstanding. For the compiler, UBs are situations which axiomatically can not occur. "This is UB" is not a concept, because the compiler assumes at all point that UBs can not occur. The compiler doesn't go "oh you're dereferencing a pointer which may be null, fuck you", it goes "you're dereferencing a pointer so it can't be null, and thus I can remove anything assuming possible nullability".
  - staticassertion 9y ago
    > The compiler doesn't go "oh you're dereferencing a pointer which may be null, fuck you", it goes "you're dereferencing a pointer so it can't be null, and thus I can remove anything assuming possible nullability". :) This is what I'm trying to say.