Question

Consider the following code:

struct A
{
  virtual void f() = 0;
};

struct B
{
  void f();
};

struct C : public A, public B
{};

int main()
{
  A* a = new C();
  B* b = new C();
  C* c = new C();

  // All these calls should result in B::f
  a->f();
  b->f();
  c->f();
}

The compiler states that C is abstract.
How can this situation be resolved?
The issue seems similar to diamond inheritance, but I fail to see the solution.

EDIT: Thanks, this is the working example:

#include "stdio.h"

struct A
{
  virtual void f() = 0;
};

struct B
{
  void f()
  {
      printf("B::f\n");
  }
};

struct C : public A, public B
{
  void f()
  {
      printf("C::f\n");
      B::f();
  }
};

int main()
{
  A* a = new C();
  B* b = new C();
  C* c = new C();

  printf("Calling from A\n");
  a->f();
  printf("Calling from B\n");
  b->f();
  printf("Calling from C\n");
  c->f();
}

Output:

Calling from A
C::f
B::f
Calling from B
B::f
Calling from C
C::f
B::f

Answer 1

The issue is that the two f() functions are completely unrelated, even though they happen to have the same name.

If the desired behaviour is for the C‘s virtual f() to call B::f(), you have to do it explicitly:

struct C : public A, public B
{
  void f();
};

void C::f() {
  B::f();
}