This document discusses the diamond problem that can occur with multiple inheritance in C++. Specifically, it shows an example where a class "four" inherits from classes "two" and "three", which both inherit from class "one". This results in two copies of the base class "one" being present in objects of class "four", leading to ambiguity when trying to access attributes from the base class. The document presents two ways to resolve this issue: 1) manual selection using scope resolution to specify which attribute to access, and 2) making the inheritance of the base class "one" virtual in classes "two" and "three", which ensures only one copy of the base class exists in class "four" objects. The virtual

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