Using `std::function` to call non-void function

Question

A while ago I used std::function pretty much like this:

std::function func = [](int i) -> int { return i; };

Answer 1

Your code has undefined behavior. It may or may not work as you expect. The reason it has undefined behavior is because of 20.8.11.2.1 [func.wrap.func.con]/p7:

Requires: F shall be CopyConstructible. f shall be Callable (20.8.11.2) for argument types ArgTypes and return type R.

For f to be Callable for return type R, f must return something implicitly convertible to the return type of the std::function (void in your case). And int is not implicitly convertible to void.

I would expect your code to work on most implementations. However on at least one implementation (libc++), it fails to compile:

test.cpp:7:30: error: no viable conversion from 'int (int)' to 'std::function'
    std::function ff = f;
                             ^    ~

Ironically the rationale for this behavior stems from another SO question.

The other question presented a problem with std::function usage. The solution to that problem involved having the implementation enforce the Requires: clause at compile time. In contrast, the solution to this question's problem is forbidding the implementation from enforcing the Requires: clause.