Moeb Moeb - 2 months ago 11
C++ Question

How does virtual inheritance solve the "diamond" (multiple inheritance) ambiguity?

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

int main(){
A *a = new D();

I understand the diamond problem, and above piece of code does not have that problem.

How exactly does virtual inheritance solve the problem?

What I understand:
When I say
A *a = new D();
, the compiler wants to know if an object of type
can be assigned to a pointer of type
, but it has two paths that it can follow, but cannot decide by itself.

So, how does virtual inheritance resolve the issue (help compiler take the decision)?


You want: (Achievable with virtual inheritance)

 / \  
B   C  
 \ /  

And not: (What happens without virtual inheritance)

 / \  
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