c++ - 友情与传承

Question

假设我有这段代码:

#include <iostream>
using namespace std;

class A
{
protected:
    virtual ~A() { cout << "A destructor reached." << endl;}

    friend class Z; 
};

class B : public A
{
protected:
    virtual ~B() { cout << "B destructor reached." << endl; }
};

class Z
{
public:
    void Test();

    friend class A;
};

void Z::Test()
{
    A* derived = (A*) new B();

    delete derived;
}

int main()
{
    Z test;
    test.Test();
}

会发生什么，B的析构函数会被调用吗？合法吗？如果不是，是否有任何方法可以调用派生的构造函数而不使每个类都派生自 A friend of Z？

Answer 1

标准，§11.5/1“访问虚函数”说

The access rules (Clause 11) for a virtual function are determined by its declaration and are not affected by the rules for a function that later overrides it.

因此只要您有权访问A::~A，就可以调用B::~B。但是您必须通过 A 调用它，因为 Z 无法访问 B。

顺便说一句，Z 中的友元声明是无用的，因为其中没有任何内容是私有(private)的或 protected 。