什么是SFINAE不能正常工作与下面的has_member函数？

问题描述

我正在尝试的示例，我想让他的 has_member 实现工作。

但是实现似乎错误地返回true，这导致我相信有一些细节的SFINAE，我不明白。

  #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< FooWithType有类型成员吗？<< 
（has_type_member< FooWithType>（）？YES：NO）< \\\
; 
 
 std :: cout<< FooNoType有类型成员吗？<< 
（has_type_member< FooNoType>（）？YES：NO）< \\\
; 
 
 return 1; 
} 
  

输出是：

  FooWithType有类型成员吗？ YES 
 FooNoType有类型成员吗？ YES 
  

我在Ubuntu上的gcc 4.8.2。

voider 类模板：

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

从第2.3节：

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.

解决方案

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:

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