为什么我不能投射一个指针Derived类的成员函数到相同但是类Base？

问题描述

对我来说，将 void（Derived :: *）（）转换为 void（Base :: *） ），如下面的代码所示：

To me it looks perfectly safe to cast a void(Derived::*)() to a void(Base::*)(), like in this code:

#include <iostream>
#include <typeinfo>
using namespace std;
struct Base{
    void(Base::*any_method)();
    void call_it(){
        (this->*any_method)();
    }
};
struct Derived: public Base{
    void a_method(){
        cout<<"method!"<<endl;
    }
};
int main(){
    Base& a=*new Derived;
    a.any_method=&Derived::a_method;
    a.call_it();
}

但是编译器抱怨 any_method =& Derived :: a_method; 。这是一个路障，以防止微妙的编程错误，或只是一些让生活更容易编译器作家？是否有解决方法让 Base 类具有指向 Derived 的成员函数的指针，不带类型knoweledge （即，我不能使 Base 具有模板参数 Derived 的模板）。

But the compiler complains about the cast at a.any_method=&Derived::a_method;. Is this a roadblock to prevent subtle programming errors, or just something to make life easier for compiler writers? Are there workarounds to let the Base class have a pointer to member functions of Derived without type knoweledge (that is, I cannot make Base a template with template argument Derived).

推荐答案

如果你的 Derived :: a_method（）试图使用一个数据成员在 Derived 中，而不是在 Base ，并且在 Base object（或从 Base 派生但与 Derived 无关的对象）？

What happens if your Derived::a_method() attempts to use a data member only present in Derived, not in Base, and you call it on a Base object (or an object derived from Base but not related to Derived)?

换句话说，换句话说是有意义的，这个不是。

The conversion the other way around makes sense, this one doesn't.

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