在 C++ 中访问重写的父虚拟方法

Question

在以下代码中，如何从pBase 访问Base::g()？ （并且仍然让“pBase->g();”像下面那样工作）

#include <iostream>
using namespace std;

class Base
{
    public:
    virtual void f(){ cout << "Base::f()" << endl; }
    virtual void g(){ cout << "Base::g()" << endl; }
    void h(){ cout << "Base::h()" << endl; }
};

class Derived : public Base
{
    public:
    void f(){ cout << "Derived::f()" << endl; }
    virtual void g(){ cout << "Derived::g()" << endl; }
    void h(){ cout << "Derived::h()" << endl; }
};

int main()
{
    Base *pBase = new Derived;
    pBase->f();
    pBase->g();
    pBase->h();

    Derived *pDerived = new Derived;
    pDerived->f();
    pDerived->g();
    pDerived->h(); 
    return 0;
}

输出是：

Derived::f()
Derived::g()
Base::h()
Derived::f()
Derived::g()
Derived::h()

另外，Derived::f() 是否与Derived::g() 完全相同？ （即自动定义为virtual?）

Answer 1

1. 使用pBase->Base::g();强制在Base中调用g。

2. 是的，Derived::f 是 virtual。我个人觉得重新强调virtual 的品味很差。从 C++11 开始，您可以在重写的函数上使用 override 说明符，然后如果从基类中删除了 virtual，编译器会发出诊断信息。