SFINAE来检查继承的成员函数

问题描述

使用SFINAE，我给定类具有某个成员函数。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中不工作（即检测到 A 有一个成员函数 foo（），但不是 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

查看此主题：

从讨论中链接到的代码派生：

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

这篇关于SFINAE来检查继承的成员函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！