C ++使用函数指针覆盖纯虚函数

Question

If I have a pure virtual function can it be overriden with a function pointer? Scenario below (I'm aware that it's not 100% syntactically correct):

#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)();
}

EDIT: In case anyone's interested, here's how I accomplished what I wanted to do:

#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();
}

The output is:

Hello One
Hello Two

Answer 1

No. And you use this wrong. In your code you are trying to assign member-function pointer to function-pointer - it's cannot be compiled.

C++03 standard 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.

Answer 2

As @ForEveR said, your code cannot compile. However, since what you actually need is the ability of switching B 's implementation of foo in the runtime, we do have workaround:

#include <iostream>
using namespace std;

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

class B : public A {
    private:
        void (B::*_f)();

    public:
        B() { chooseOne(); }

        void caseOne() {
            cout << "case one" << endl;
        }

        void caseTwo() {
            cout << "case two" << endl;
        }

        void chooseOne() { _f = &B::caseOne; }

        void chooseTwo() { _f = &B::caseTwo; }

        void foo() {
            (this->*_f)();
        }
};

int main(int argc, const char *argv[])
{
    A* b = new B();
    b->foo();
    ((B*)b)->chooseTwo();
    b->foo();
    return 0;
}

UPDATE :

Just found the OP added his answer in the question, which is almost the same as mine. But I think calling foo through pointer instead of instance object is better, for that can exhibit the effect of polymorphism. Besides, it's better to hide f as a private member function.

Answer 3

I think when compile time, the syntax can NOT be compiled. You should provide an override function with the certain name and same args list.