c++ - 调用在基类中私有(private)的派生类的虚函数时出现访问说明符错误

Question

为什么我会收到以下代码的访问说明符错误(私有(private)成员)？

#include<iostream>
using namespace std;

class Derived;

class Base {
private:
    virtual void fun() { cout << "Base Fun"; }
};

class Derived: public Base {
public:
    void fun() { cout << "Derived Fun"; } //this should be called
};

int main()
{
   Base *ptr = new Derived;
   ptr->fun();
   return 0;
}

这里应该调用派生类的fun()，因为是public的，应该不会出错。

Answer 1

根据标准(N4140):

11.5 Access to virtual functions

¹ 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. [ Example:

class B {
public:
    virtual int f();
};
class D : public B {
private:
    int f();
};
void f() {
    D d;
    B* pb = &d;
    D* pd = &d;
    pb->f(); // OK: B::f() is public,
             // D::f() is invoked
    pd->f(); // error: D::f() is private
}

—end example ]

顺便说一句，在一般情况下，在编译期间不可能知道在运行时将调用哪个类。

考虑

Base *ptr = GetBaseOrDerivedObject();
ptr->fun();

哪里GetBaseOrDerivedObject , 根据具体的运行时情况，可以返回一个指向 Base 类型对象的指针或 Derived .