What does SFINAE not work correctly with following has_member function?

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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.

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ecatmur On BEST ANSWER

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].”