How does virtual inheritance solve the diamond multiple inheritance ambiguity

Question

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(); 
    a->eat(); 
} 

I understand the diamond problem, and the code above does not have it.

How can virtual inheritance address the issue?

What I mean is that when I say A *a = new D();, the compiler wants to know if an object of type D may be allocated to a reference of type A, but it has two options that it cannot choose between.

So, how can virtual inheritance address the problem (assist the compiler in making a decision)?

Answer 1

You desire: (Achievable with virtual inheritance)

  A  
 / \  
B   C  
 \ /  
  D      

And not: (What occurs in the absence of virtual inheritance)

A   A  
|   |
B   C  
 \ /  
  D    

Because of virtual inheritance, there will only be one instance of the base A class, rather than two.

Your type D would contain two vtable references (shown in the first figure), one for B and one for C, both of which essentially inherit A. D's object size has risen since it now holds two pointers, although there is only one A.

As a result, B::A and C::A are the same, and D cannot make any confusing calls. If virtual inheritance is not used, the second diagram above is displayed. Any call to a member of A becomes ambiguous, and you must indicate which path to take.