是否可以检查是否为类定义了成员函数，即使该成员是从未知基类继承的 [英] Is it possible to check if a member function is defined for a class even if the member is inherited from an unknown base class

问题描述

我发现了类似的问题和解答，例如这一个。然而，正如我试验，这个SFINAE测试只成功，如果被测试的成员直接在被测试类中定义。例如下面的类 B ， D1 print HAS 而另外两个打印 NOT HAS 。是否有一种方法来确定一个类是否有一个成员，无论它是由自身定义还是一个基类，并且基类的名称在这种情况下是未知的。动机是，我想写一个通用函数，如果它存在（从基础或非，参数的类型是通用的，沿着其可能的基础的类型），将调用某个方法。

I found similar questions and answers like this one. However, as I tried out, this SFINAE tests only succeeded if the tested member is directly defined in the class being tested. For example the following, class B, D1 print HAS while the other two print NOT HAS. Is there a way to determine that if a class has a member, whether it is defined by itself, or a base class, and the name of the base class is not known in this case. The motivation is that I want to write a generic function that will call a certain method if it exists (from base or not, the type of the parameter is generic, leave along the type of its possible base).

#include <iostream>

class HasFoo
{
    public :

    typedef char Small;
    typedef struct {char; char;} Large;

    template <typename C, void (C::*) ()> class SFINAE {};

    template <typename C> static Small test (SFINAE<C, &C::foo> *)
    {
        std::cout << "HAS" << std::endl;
    }

    template <typename C> static Large test (...)
    {
        std::cout << "NOT HAS" << std::endl;
    }
};

class B
{
    public :

    void foo () {}
};

class D1 : public B
{
    public :

    void foo () {} // overide
};

class D2 : public B
{
    public :

    using B::foo;
};

class D3 : public B {};

int main ()
{
    HasFoo::test<B>(0);
    HasFoo::test<D1>(0);
    HasFoo::test<D2>(0);
    HasFoo::test<D3>(0);
}

推荐答案

可能至少在C ++ 03中，我也怀疑在C ++ 11中。

Unfortunately it wouldn't be possible at least in C++03 and I doubt in C++11 also.

几个重要的点：

1. 建议的SFINAE只有在方法 public


2. 才有效。即使SFINAE可以工作对于基本方法，点（1）
   适用;因为 private 和 protected 继承SFINAE
   可能最终无用


3. 假设您只想处理 public 方法/继承，
   代码 HasFoo :: test<> code>可以增强多个
参数，其中基类也可以传递;
std :: is_base_of<> 可用于进一步验证
基本/派生关系;然后对基类
和



1. The proposed SFINAE works only if the method is public
2. Even if the SFINAE would have worked for base methods, the point (1) applies; because for private and protected inheritance the SFINAE may end up useless
3. Assuming you may want to deal only with public method/inheritance, the code HasFoo::test<> can be enhanced for taking multiple parameters where a base class also can be passed; std::is_base_of<> can be used for further validation of the base/derived relationship; then apply the same logic for base class also

这篇关于是否可以检查是否为类定义了成员函数，即使该成员是从未知基类继承的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IT屋！