Returning a void?
c++, c++11, return, templates, void
Solution
This looks valid by the draft C++11 standard, if we look at section `5.2.3` Explicit type conversion (functional notation) paragraph 2 says (emphasis mine):
The expression T(), where T is a simple-type-specifier or typename-specifier for a non-array complete object type or the (possibly cv-qualified) void type, creates a prvalue of the specified type, whose value is that produced by value-initializing (8.5) an object of type T; no initialization is done for the void() case.[...]
the wording is pretty similar pre C++11 as well.
This okay in a constexpr even though section `7.1.5` paragraph `3` says:
The definition of a constexpr function shall satisfy the following constraints:
and includes this bullet:
its return type shall be a literal type;
and void is not a literal in C++11 as per section `3.9` paragraph 10, but if we then look at paragraph 6 it gives an exception that fits this case, it says:
If the instantiated template specialization of a constexpr function template or member function of a class template would fail to satisfy the requirements for a constexpr function or constexpr constructor, that specialization is not a constexpr function or constexpr constructor. [ Note: If the function is a member function it will still be const as described below. —end note ] If no specialization of the template would yield a constexpr function or constexpr constructor, the program is ill-formed; no diagnostic required.
As Casey noted in the C++14 draft standard void is a literal, this is section `3.9` Types paragraph 10 says:
A type is a literal type if it is:
and includes:
— void; or
Problem
I do not understand why this code compiles without error: ``` #include <iostream> template <class T> struct Test { static constexpr T f() {return T();} }; int main() { Test<void> test; test.f(); // Why not an error? return 0; } ``` Is it ok according to the standard, or is it a compiler tolerance?