Hide virtual function with non-virtual override

Question

Having

#include <iostream>

using namespace std;

class A {
public:
    virtual void foo() {
        cout << "A" << endl;
    }
};

class B : public A {
public:
    void foo() {
        cout << "B" << endl;
    }
};

class C : public B {
public:
    void foo() {
        cout << "C" << endl;
    }
};

int main() {
    C c;
    B* b = &c;
    b->foo();

    return 0;
}

The output is C , but I expected B .

I didn't declare B::foo() with the virtual modifier, so I expect the function call to be determined by the static type (no polymorphism).

Why is C::foo() being called?

Is it possible to provide a non-virtual function in a derived class, that hides the virtual function in the base? What signature should the derived member function have so that b->foo() calls it, and not (b->*&A::foo)()

Answer 1

The principle of virtual inheritance of a member function is is a direct consequence of the C++ 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, parameter-type-list , cv-qualification, and refqualifier (or absence of same) as Base::vf is declared, then Derived::vf is also virtual .

So regardless of the level of inheritance, the function will be virtual in all the classes derived somehow from A. There is no need to put the keyword virtual .

The goal of this polymorphic approach is to ensure that you always call the appropriate function corresponding to the real idendity of your object, regardless the fact that you use a pointer to a base or a pointer to the real class of the object. This is why you obtain "C" !

In this related SO question I explain a trick to give the impression of removing virtuality at one single level, using multiple inheritance. However it can be done only for a single level (you should do it for the class of your base pointer).

*By the way, you could write pb->B::foo(); no need of *& .