问题
I'm trying out examples from Walter Brown's TMP talk and I'm trying to get his has_member
implementation working.
However the implementation seems to falsely return true which leads me to believe there is some detail of SFINAE that I am not understanding.
#include <iostream>
#include <type_traits>
template <class ...>
using void_t = void;
template <class, class = void>
struct has_type_member: std::false_type {};
template <class T>
struct has_type_member<T, void_t<typename T::type> >: std::true_type {};
struct FooWithType
{
typedef int type;
};
struct FooNoType
{
};
int main()
{
std::cout << "Does FooWithType have type member? " <<
(has_type_member<FooWithType>() ? "YES" : "NO") << "\n";
std::cout << "Does FooNoType have type member? " <<
(has_type_member<FooNoType>() ? "YES" : "NO") << "\n";
return 1;
}
Output is:
Does FooWithType have type member? YES
Does FooNoType have type member? YES
I am on gcc 4.8.2 on Ubuntu.
回答1:
The problem is that gcc 4.8.2 (and prior to gcc 5.0) does not regard unused arguments in alias templates as suitable for SFINAE. The workaround is to forward to a voider
class template:
template <class ... T> struct voider { using type = void; };
template <class ... T>
using void_t = typename voider<T...>::type;
From http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n3911.pdf section 2.3:
Alas, we have encountered implementation divergence (Clang vs. GCC) while working with the above very simple definition. We (continue to) conjecture that this is because of CWG issue 1558: “The treatment of unused arguments in an alias template specialization is not specified by the current wording of 14.5.7 [temp.alias].”
来源:https://stackoverflow.com/questions/30781523/what-does-sfinae-not-work-correctly-with-following-has-member-function