Can i use a method overriding a non-virtual method?

Question

I am trying to understand a point here in C++. If class A has a non-virtual method, and class B, which extends A, overrides that method, can i create an instance of B and somehow use the method defined in B? Is there a point to override a non-virtual method?

Answer 1

Is there a point to override a non-virtual method?

You are not actually overriding, but this is the behavior, i.e

B* b = new B();
A* a = new B();
b->method(); //Calls B's method
a->method(); // Calls A's method

So, the pointer/reference type determines the method called.

can i create an instance of B and somehow use the method defined in B?

Yes. The pointer/reference type has to be of type B. (see previous example).

If you don't declare method to be virtual, you cannot override it, but you can hide it.

Answer 2

If B inherits from A, and redefines a method defined in A, then new instances of B will call B's version. However, if the method is not virtual, then there is no polymorphic behavior, so if an instance of B is referenced as an A, then the method will be A's. For example:

struct A {
    void foo () { std::cout << "A::foo" << std::endl; }
};

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

B b;
b.foo();
A *a = &b;
a->foo();

The output of the code above would be:

B::foo
A::foo

However, if the foo method had been virtual, then B::foo would have been printed twice.

Answer 3

If a function is not virtual then the type of the variable determines which implementation is dispatched too:

#include <iostream>

using namespace std;

struct A {
    void f() { cout << "A" << endl; }
};

struct B : public A {
    void f() { cout << "B" << endl; }
};

int main(int args, char** argv) {

    B b;
    A& a = b;

    b.f();
    a.f();

    return 0;
}