运行时检查实例（Base *）是否覆盖父函数（Base :: f（）） [英] runtime-check whether an instance (Base*) override a parent function (Base::f())

问题描述

如何确定基类（ B ）的指针是否（多态）覆盖基类的某个虚函数？

How to determine whether a pointer of base (B) class is (polymorphism-ly) override a certain virtual function of the base class?

class B{
    public: int aField=0;
    virtual void f(){};
};
class C : public B{
    public: virtual void f(){aField=5;};
};
class D: public B{};

int main() {
    B b;
    C c;
    D d;
    std::vector<B*> bs;
    bs.push_back(&b);
    bs.push_back(&c);
    bs.push_back(&d);
    for(int n=0;n<3;n++){
        //std::cout<<(bs[n]->f)==B::f<<std::endl;
        //should print "true false true"
    }
}

我试图将函数指针 bs [n]-> f 的地址与 B :: f 进行比较，但无法编译。

I tried to compare the address of function pointer bs[n]->f against B::f, but it is uncompilable.

演示

我觉得这个问题可能重复了，但我找不到（抱歉）。

I feel that this question might be duplicated, but I can't find (sorry).

推荐答案

GCC具有扩展名，该扩展名允许您获取虚拟成员函数的地址。

GCC has an extension that allows you to get the address of a virtual member function.

可以这样使用：

#include <vector>
#include <iostream>

class B{
    public: int aField=0;
    virtual void f(){};
};
class C : public B{
    public: virtual void f(){aField=5;};
};
class D: public B{};

int main() {
    B b;
    C c;
    D d;
    std::vector<B*> bs;
    bs.push_back(&b);
    bs.push_back(&c);
    bs.push_back(&d);
    for(int n=0;n<3;n++){
        // This might need to be: (void*) B{}.*&B::f == (void*) (...)
        std::cout << ((void*) &B::f == (void*)(bs[n]->*&B::f)) << '\n';
    }
}

^演示

您可能会发现此质量检查很有趣。

自然地，这是非标准行为。如果您希望在标准C ++中具有类似的行为，则实际上可能正在寻找纯虚拟函数：

Naturally, this is nonstandard behavior. If you wanted similar behavior in standard C++, you might actually be looking for pure virtual functions:

class B{
    public: int aField=0;
    virtual void f() = 0;
};

否则，您必须具有一些其他通信机制，例如 bool 返回类型为 f（）。

Otherwise, you'd have to have some other mechanism to communicate, such as a bool return type on f().

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