SFINAE 检查继承的成员函数

问题描述

限时删除！！

使用 SFINAE，我 可以检测给定的类是否具有某个成员函数.但是如果我想测试继承的成员函数怎么办?

Using SFINAE, i can detect wether a given class has a certain member function. But what if i want to test for inherited member functions?

以下在 VC8 和 GCC4 中不起作用(即检测到 A 具有成员函数 foo()，而不是 B继承一个):

The following does not work in VC8 and GCC4 (i.e. detects that A has a member function foo(), but not that B inherits one):

#include <iostream>

template<typename T, typename Sig>
struct has_foo {
    template <typename U, U> struct type_check;
    template <typename V> static char (& chk(type_check<Sig, &V::foo>*))[1];
    template <typename  > static char (& chk(...))[2];
    static bool const value = (sizeof(chk<T>(0)) == 1);
};

struct A {
    void foo();
};

struct B : A {};

int main()
{
    using namespace std;
    cout << boolalpha << has_foo<A, void (A::*)()>::value << endl; // true
    cout << boolalpha << has_foo<B, void (B::*)()>::value << endl; // false
}

那么，有没有办法测试继承的成员函数?

So, is there a way to test for inherited member functions?

推荐答案

看看这个线程:

http://lists.boost.org/boost-users/2009/01/44538.php

源自该讨论中链接到的代码:

Derived from the code linked to in that discussion:

#include <iostream>

template <typename Type>
class has_foo
{
   class yes { char m;};
   class no { yes m[2];};
   struct BaseMixin
   {
     void foo(){}
   };
   struct Base : public Type, public BaseMixin {};
   template <typename T, T t>  class Helper{};
   template <typename U>
   static no deduce(U*, Helper<void (BaseMixin::*)(), &U::foo>* = 0);
   static yes deduce(...);
public:
   static const bool result = sizeof(yes) == sizeof(deduce((Base*)(0)));
};

struct A {
    void foo();
};

struct B : A {};

struct C {};

int main()
{
    using namespace std;
    cout << boolalpha << has_foo<A>::result << endl;
    cout << boolalpha << has_foo<B>::result << endl;
    cout << boolalpha << has_foo<C>::result;
}

结果:

true
true
false

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