访问在C++中覆盖父虚方法

Question

在下面的代码,我怎么能访问Base::g()的pBase？(仍然可以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()

还有,Derived::f()完全一样Derived::g()吗？(即自动定义为virtual？)

Answer 1

1. 用pBase->Base::g();强制的号召g在Base.

2. 是的,Derived::f是virtual.我个人觉得重新强调的virtual是品味不佳.从C++ 11开始,您可以override在重写函数上使用说明符,然后编译器virtual在从基类中删除时发出诊断信息.