Virtual Inheritance

When using inheritance, you can specify the virtual keyword:

struct A{};
struct B: public virtual A{};

When class B has virtual base A it means that A will reside in most derived class of inheritance tree, and thus that most derived class is also responsible for initializing that virtual base:

struct A
{
    int member;
    A(int param)
    {
        member = param;
    }
};

struct B: virtual A
{
    B(): A(5){}
};

struct C: B
{
    C(): /*A(88)*/ {}
};

void f()
{
    C object; //error since C is not initializing it's indirect virtual base `A`
}

If we un-comment /*A(88)*/ we won’t get any error since C is now initializing it’s indirect virtual base A.

Also note that when we’re creating variable object, most derived class is C, so C is responsible for creating(calling constructor of) A and thus value of A::member is 88, not 5 (as it would be if we were creating object of type B).

It is useful when solving the diamond problem.:

A                                        A   A
/ \                                       |   |
B   C                                      B   C
\ /                                        \ /
D                                          D
virtual inheritance                   normal inheritance

B and C both inherit from A, and D inherits from B and C, so there are 2 instances of A in D! This results in ambiguity when you’re accessing member of A through D, as the compiler has no way of knowing from which class do you want to access that member (the one which B inherits, or the one that is inherited byC?).

Virtual inheritance solves this problem: Since virtual base resides only in most derived object, there will be only one instance of A in D.

struct A
{
    void foo() {}
};

struct B : public /*virtual*/ A {};
struct C : public /*virtual*/ A {};

struct D : public B, public C
{
    void bar()
    {
        foo(); //Error, which foo? B::foo() or C::foo()? - Ambiguous
    }
};

Removing the comments resolves the ambiguity.