c++ - 使用以下 has_member 函数时 SFINAE 无法正常工作的原因是什么？

Question

我正在尝试来自 Walter Brown's TMP talk 的示例我正试图让他的 has_member 实现正常工作。

然而，该实现似乎错误地返回 true，这让我相信我不理解 SFINAE 的一些细节。

#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;                                                                                                                 
}      

输出是:

Does FooWithType have type member? YES
Does FooNoType have type member? YES

我在 Ubuntu 上使用 gcc 4.8.2。

Answer 1

问题是 gcc 4.8.2(和 gcc 5.0 之前的版本)不认为别名模板中未使用的参数适用于 SFINAE。解决方法是转发到 voider 类模板:

template <class ... T> struct voider { using type = void; };
template <class ... T>
using void_t = typename voider<T...>::type;

来自 http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2014/n3911.pdf第 2.3 节:

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].”