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.