C++ multiple inheritance - same method names - can I somehow remove one of them?

Question

I have this structures in C++11

struct A {
   int idA;   
   void setId(int i) { idA = i;}
   int getId() { return idA;}

   virtual void foo() = 0;
};

struct B {
   int idB;   
   void setId(int i) { idB = i;}
   int getId() { return idB;}

   virtual void foo2() = 0;
};

struct AB : public A, public B
{
   void foo() override {}
   void foo2() override {}
};

Now in main I can call it like this:

AB * ab = new AB();
ab->B::setId(10);

but I dont really like it.

Is there any other solution? Methods setId/getId from struct A are not needed in AB. I just did the inheritance, because I need foo() and other stuff from A, but all other methods are unique.

Answer 1

Since you said that you don't need A's version of those methods, you could write

struct AB : public A, public B
{
    void foo() override {}
    void foo2() override {}

    using B::setId;
    using B::getId;
};

This will put B's implementation of those methods into AB's namespace and make calling them unambiguous.

Answer 2

I believe that both mine and the accepted answers introduces rather annoying problem, consider:

AB ab;
A *a = &ab;
a->setId(10); //problem
a->foo(); //calls AB::foo() correctly

When the name hiding is used (accepted answer) the AB object never gets called while my answer (wrapping the call) does not account for this either. The correct approach in this case in my opinion is to use private inehritance of A and exposing only foo(), so only one thing would change from original OP's code:

struct AB : private A, public B