C++ virtual override functions with same name

Question

Hello there, I have something like that (simplified)

class A
{
  public:
    virtual void Function () = 0;
};

class B
{
  public:
    virtual void Function () = 0;
};

class Impl : public A , public B
{
  public:
        ????
};

How can I implement the Function () for A and the Function() for B ? Visual C++ lets you only define the specific function inline (i.e. not in the cpp file), but I suppose it's an extension. GCC complains about this. Is there a standard C++ way to tell the compiler which function I want to override?

(visual c++ 2008)

class Impl : public A , public B
{
  public:
     void A::Function () {  cout << "A::Function" << endl; }
     void B::Function () {  cout << "B::Function" << endl; }
};

Thank you!

Answer 1

You cannot use qualified names there. I you write void Function() { ... } you are overriding both functions. Herb Sutter shows how it can be solved.

Another option is to rename those functions, because apparently they do something different (otherwise i don't see the problem of overriding both with identical behavior).

Answer 2

If A and B are interfaces, then I would use virtual derivation to "join" them (make them overlap). If you need different implementations for your Function if called through a pointer to A or to B then I would strongly recommend to choose another design. That will hurt otherwise.

Impl "derives from" A and B means Impl "is a" A and B. I suppose you do not mean it.

Impl "implements interface" A and B means Impl "behaves like" A and B. then same interface should mean the same behavior.

In both cases having a different behavior according to the type of pointer used would be "schizophrenic" and is for sure a situation to avoid.

Answer 3

As a workaround, try

struct Impl_A : A
{ 
     void Function () {  cout << "A::Function" << endl; } 
}; 


struct Impl_B : B
{
    void Function () { cout << "B::function" << endl; }
};

struct Impl : Impl_A, Impl_B {};