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Virtual inheritance has quite a few common pitfalls, one of which is related to the construction of the virtual base.

Consider the following code – what will eventually be printed?

#include <iostream>
using std::cout;

struct A {
    A (int x=0) { cout << x; }
};

struct B : virtual A {
    B () : A(1) {}
};

struct C : virtual A {
    C () : A(2) {}
};

struct D : B, C {

};

int main () {
    D d;
}

Hint: what class actually invokes A’s constructor?

§ 3 Responses to Constructor selection with virtual inheritance

• Sumant says:

  19/08/2009 at 23:13

  D invokes A’s constructor directly when diamond-shaped hierarchy is present with a virtual base class. Constructor invocations of A from B and C are ignored when D is being instantiated. Try this.

  struct D : B, C {
    D() : A(10) {}
  };
  
  int main () {
      D d;
      B b;
      C c;
  }

  Reply
  • rmn says:

    20/08/2009 at 19:45

    That is correct, thanks for posting!

    To add to that, when virtual inheritance is involved, the compiler actually creates two constructors for the directly inheriting class: one that includes the call to A’s constructor (to be used when we create a regular object of class B) and one that doesn’t (used here).

    Reply
• rmn says:

  29/08/2009 at 01:12

  Another very nice usage of the fact that the most derived class is the invoker of the virtual class’ constructor can be found here: http://www.research.att.com/~bs/bs_faq2.html#no-derivation

  Reply