Lambda converted function pointer comparison

Question

Simply, will the following assert ever fire?

template<typename T>
auto destructor()
{
    return +[](void* p){
        ((T*)p)->~T();
    };
}

assert(destructor<int>() != destructor<char>());

The standard only seems to say that the lambda converted function pointer does the same thing as the lambda itself, then you realize trivially destructed types all have no-op destructors and are therefore identical.

For the exact snippet shown, comparison for hand rolled vtables in type erased contexts without RTTI. The general question might have much more use. — Passer By, Jan 28 '18 at 10:19
What is the role of `+` after return here? There should also be `;` after return, right? — user7860670, Jan 28 '18 at 10:35
It could be the same under the as-if rule, for certain. But I don't see any guarantee in the standard you can rely on here. — StoryTeller - Unslander Monica, Jan 28 '18 at 10:40
@VTT The `+` is just to force the conversion, the `;` is a typo :P — Passer By, Jan 28 '18 at 10:48
Conversion to what? Plus converting lambda is something new to me. — user7860670, Jan 28 '18 at 10:51
@VTT Conversion of a non-capturing lambda to an equivalent function pointer. https://stackoverflow.com/questions/18889028/a-positive-lambda-what-sorcery-is-this — Passer By, Jan 28 '18 at 10:53

score 4 · Answer 1 · answered Jan 28 '18 at 11:06

[expr.prim.lambda]/6:

The value returned by this conversion function is the address of a function F that, when invoked, has the same effect as invoking the closure type's function call operator.

The pointed to function is specified in term of its behavior, not in term of its identity. So it is unspecified if this assertion will fire.

Lambda converted function pointer comparison

1 Answers1