1. VIRTUAL BASE CLASS

2. DIAMOND PROBLEM
Base Class
Derived 1 Derived 2
Derived 3

3. EXAMPLE
class one
{
public:
int a;
public:
one() { cout<<"onen"; }
};
class two : public one
{
public:
int a2;
public:
two(){ cout<<"twon"; }
};

4. class three : public one
{
public:
int a3;
public:
three(){ cout<<"threen"; }
};
class four: public two , public three
{
private:
int a4;
public:
four(){ a4=6;cout<<"fourn"; }
};

5. int main()
{
four f;
return 0;
}
OUTPUT:
One
Two copies copies of one present
Two in an object of type four.
One
Three
Four

6. int main()
{
four f;
cout<<f.a; //error : request for member ‘a’ is ambiguous
return 0;
}
 Ambiguous because variable „a‟ is present twice in the object
of four, one in two class and one in three class.
 Because there are two copies of „a‟ present in object „f‟, the
compiler doesn‟t not know which one is being referred.

7. HOW TO RESOLVE
Two ways:
 Manual Selection.
 Using virtual base class.

8. MANUAL SELECTION
 Manually select the required variable ‘a’ by the using
scope resolution operator.
int main()
{
four f;
f.two::a=40; OUTPUT:
f.three::a=50; 40
50
cout<<f.two::a<<“n”;
cout<<f.three::a<<“n”;
return 0;
}

9.  f.two::a=40; and f.three::a=50;
 In the above statements, use of scope resolution operators
resolves the problem of ambiguity.
 But this is not an efficient way because still two copies of
„a‟ is available in object „f‟.
 How to make only one copy of „a‟ available in the object
„f‟ which consumes less memory and easy to access without
using scope resolution operator.
 This can be done by using virtual base class.

10. VIRTUAL BASE CLASS
class one class two : virtual public one
{ {
public: public:
int a; int a2;
public: public:
one() {cout<<"onen"; } two(){ cout<<"twon"; }
}; };
class three:virtual public one class four: public two , public three
{ {
public: private:
int a3; int a4;
public: public:
three(){cout<<"threen"; } four(){ a4=6;cout<<"fourn"; }
}; };

11. int main()
{
four f;
return 0;
}
OUTPUT:
ONE Only one copy is maintained.
TWO No ambiguity.
THREE
FOUR

12. int main()
{
four f;
f.a=40; // umambigious since only one copy of ‘a’ is present in
object ‘f’
cout<<f.a;
return 0;
}
 Now that both two and three have inherited base as virtual, any
multiple inheritance involving them will cause only one copy of base to
be present.
 Therefore, in FOUR, there is only one copy of base.

13. THANK YOU