Question

I’m trying to use multiple inheritance to solve a complex hierarchy that I’m developing. The situation is the following:

class A {
  virtual void foo();
}

class B {
  virtual void foo();
}

class C : public B {

}

class D : public A, public C {
  void foo() { ... }
}

class ClientClass {
  void method() {
    A *a = new D();
    a->foo();
  }

What I wonder is: will D have just one function foo() in the end? I’m thinking about it because the method is virtual in both parents so they should like collimate onto the same one, but I thin this just because I come from Java and I feel that it may be different in C++. I have to declare the virtual function foo() twice because ClientClass doesn’t know B or C but just A. This is a requirement I would like to keep.

EDIT: does the same answer apply even if both foo() in A and B are pure virtual? (eg = 0)

Answer 1

Actually, the implementation of D::foo will override both A::foo and B::foo at the same time.

Both inherited functions will still be available, of course, to be called by their full name. But that’s not different than in the single inheritance case.

About the ClientClass, you create a D object and then call foo through a pointer to A. So the D::foo override will be called.

If you want different versions of foo for the override of A::foo and C::foo (for example, because they are unrelated but happen to be called the same way) then you will need a bit of work:

class A2 : public A
{
public:
    virtual void foo()
    {
        A_foo();
    }
    virtual void A_foo()
    {
        A::foo();
    }
};
class C2 : public C
{
public:
    virtual void foo()
    {
        C_foo();
    }
    virtual void C_foo()
    {
        C::foo();
    }
};
class  D: public A2, C2
{
public:
    virtual void A_foo()
    { /* ... */ }
    virtual void C_foo()
    { /* ... */ }
};

Now the use:

{
    A *a = new D;
    a->foo(); //will call A2::foo -> D::A_foo
    B *b = new D;
    b->foo(); //will call C2::foo -> D::C_foo
    D *d = new D;
    d->foo(); //error: ambiguous call!
}