C++ 使用函数指针重写纯虚函数

Question

如果我有一个纯虚函数，可以用函数指针覆盖它吗？下面的场景(我知道它在语法上不是 100% 正确):

#include<iostream>
using namespace std;

class A {
    public:
        virtual void foo() = 0;
};

class B : public A {
    public:
        B() { foo = &B::caseOne; }
        void caseOne() { cout << "Hello One" << endl; }
        void caseTwo() { cout << "Hello Two" << endl; }
        void (B::*foo)();
        void chooseOne() { foo = &B::caseOne; }
        void chooseTwo() { foo = &B::caseTwo; }
};

int main() {
    B b;
    b.(*foo)();
}

编辑:如果有人感兴趣，以下是我完成我想做的事情的方法:

#include<iostream>
using namespace std;

class A {
    public:
        virtual void foo() = 0;
};

class B : public A {
    public:
        B() { f = &B::caseOne; }
        void caseOne() { cout << "Hello One" << endl; }
        void caseTwo() { cout << "Hello Two" << endl; }
        void (B::*f)();
        void chooseOne() { f = &B::caseOne; }
        void chooseTwo() { f = &B::caseTwo; }
        void foo() { (this->*f)(); }
};

int main() {
    B b;
    b.foo();
    b.chooseTwo();
    b.foo();
}

输出为:

Hello One
Hello Two

Answer 1

没有。而你用错了这个。在您的代码中，您尝试将成员函数指针分配给函数指针 - 它无法编译。

C++03 标准 10.3/2

If a virtual member function vf is declared in a class Base and in a class Derived, derived directly or indirectly from Base, a member function vf with the same name and same parameter list as Base::vf is declared, then Derived::vf is also virtual (whether or not it is so declared) and it overrides Base::vf.