Why is there ambiguity in this diamond pattern?

Question

#include <iostream>
using namespace std;

class A             { public: void eat(){ cout<<"A";} };
class B: public A   { public: void eat(){ cout<<"B";} };
class C: public A   { public: void eat(){ cout<<"C";} };
class D: public B,C { public: void eat(){ cout<<"D";} };

int main(){
    A *a = new D();
    a->eat();
}

I am not sure this is called diamond problem or not, but why doesn't this work?

I have given the defination for eat() for D. So, it doesn't need to use either B's or C's copy (so, there should be no problem).

When I said, a->eat() (remember eat() is not virtual), there is only one possible eat() to call, that of A.

Why then, do I get this error:

'A' is an ambiguous base of 'D'

---

What exactly does A *a = new D(); mean to the compiler??

and

Why does the same problem not occur when I use D *d = new D();?

Answer 1

The diamond results in TWO instances of A in the D object, and it is ambiguous which one you are referring to - you need to use virtual inheritance to solve this:

class B: virtual public A   { public: void eat(){ cout<<"B";} };
class C: virtual public A   { public: void eat(){ cout<<"C";} };

assuming that you actually only wanted one instance. I also assume you really meant:

class D: public B, public C { public: void eat(){ cout<<"D";} };

Answer 2

The error you're getting isn't coming from calling eat() - it's coming from the line before. It's the upcast itself that creates the ambiguity. As Neil Butterworth points out, there are two copies of A in your D, and the compiler doesn't know which one you want a to point at.

Answer 3

Note that the compile error is on the "A *a = new D();" line, not on the call to "eat".

The problem is that because you used non-virtual inheritance, you end up with class A twice: once through B, and once through C. If for example you add a member m to A, then D has two of them: B::m, and C::m.

Sometimes, you really want to have A twice in the derivation graph, in which case you always need to indicate which A you are talking about. In D, you would be able to reference B::m and C::m separately.

Sometimes, though, you really want only one A, in which case you need to use virtual inheritance.

Answer 4

Imagine a slightly different scenario

class A             { protected: int a; public: void eat(){ a++; cout<<a;} };
class B: public A   { public: void eat(){ cout<<a;} };
class C: public A   { public: void eat(){ cout<<a;} };
class D: public B,C { public: void eat(){ cout<<"D";} };

int main(){
    A *a = new D();
    a->eat();
}

If this would work, would it increment the a in B or the a in C? That's why it's ambiguous. The this pointer and any non-static data member is distinct for the two A subobjects (one of which is contained by the B subobject, and the other by the C subobject). Try changing your code like this and it will work (in that it compiles and prints "A")

class A             { public: void eat(){ cout<<"A";} };
class B: public A   { public: void eat(){ cout<<"B";} };
class C: public A   { public: void eat(){ cout<<"C";} };
class D: public B, public C { public: void eat(){ cout<<"D";} };

int main(){
    A *a = static_cast<B*>(new D());
      // A *a = static_cast<C*>(new D());
    a->eat();
}

That will call eat on the A subobject of B and C respectively.

Answer 5

For a truly unusual situation, Neil's answer is actually wrong (at least partly).

With out virtual inheritance, you get two separate copies of A in the final object.

"The diamond" results in a single copy of A in the final object, and is produced by using virtual inheritance:

Since "the diamond" means there's only one copy of A in the final object, a reference to A produces no ambiguity. Without virtual inheritance, a reference to A could refer to either of two different objects (the one on the left or the one on the right in the diagram).

Answer 6

You want: (Achievable with virtual inheritance)

  D
  / \
B   C
  \ /
  A

And not: (What happens without virtual inheritance)

    D
   /   \
  B   C
  |     |
  A   A

Virtual inheritance means that there will be only 1 instance of the base A class not 2.

Your type D would have 2 vtable pointers (you can see them in the first diagram), one for B and one for C who virtually inherit A. D's object size is increased because it stores 2 pointers now; however there is only one A now.

So B::A and C::A are the same and so there can be no ambiguous calls from D. If you don't use virtual inheritance you have the second diagram above. And any call to a member of A then becomes ambiguous and you need to specify which path you want to take.

Wikipedia has another good rundown and example here