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?
Void as a Function Return Type Void functions, also called nonvalue-returning functions, are used just like value-returning functions except void return types do not return a value when the function is executed. The void function accomplishes its task and then returns control to the caller.
Any method declared void doesn't return a value. It does not need to contain a return statement, but it may do so.
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
See @Shafik Yaghmour's answer for the full info.
The following paragraph forbids this for non-templates (7.1.5(3)):
The definition of a
constexpr
function shall satisfy the following constraints:
[...]
its return type shall be a literal type or a reference to literal type
To elaborate, a literal type is defined in 3.9(10) as a scalar type or a composition of literal type objects in an array or struct. void
is not a scalar type by 3.9(9).
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