call base class function or derived class function

Hello everyone,


I am just a little confusing and not 100% confident about the following case. In the following sample, function call foo (called in function goo in class Foo) will call foo in class Goo, other than foo in class Foo, because we are using an instance of Goo g and no matter in which class, right?

(I have tested the result is correct, and I am asking the reason here.)

Code:

---

#include <iostream>

using namespace std;

class Foo {
public:
        void goo()
        {
                foo(); // will call foo in class Goo
        }
private:
        virtual void foo() = 0;
};

void Foo::foo()
{
        cout << "I am here. " << endl;
}

class Goo : public Foo {
public:
        void foo()
        {
                Foo::foo();
        }
};

int main()
{
        Goo g;
        g.foo();
        return 0;
}

---

thanks in advance,
George

The reason is that foo is a virtual function. When a virtual function is called, the dynamic type of the object is used to determine which version of that function is executed. In this case, the type of the object is a Goo, so the Goo's version of the function is called.

Thanks Daved,


Question answered. What makes me confused is the "pure" virtual. :-)

Quote:

---

Originally Posted by Daved

The reason is that foo is a virtual function. When a virtual function is called, the dynamic type of the object is used to determine which version of that function is executed. In this case, the type of the object is a Goo, so the Goo's version of the function is called.

---

regards,
George

Pure virtual is just another virtual function which makes the class abstract, thus requiring the derived to override it or it will become abstract too. Aside from that, it behaves just like any other virtual function.
But you should know this, shouldn't you?

I understand it now, thanks Elysia!

Quote:

---

Originally Posted by Elysia

Pure virtual is just another virtual function which makes the class abstract, thus requiring the derived to override it or it will become abstract too. Aside from that, it behaves just like any other virtual function.
But you should know this, shouldn't you?

---

regards,
George