Skip to content
Related Articles

Related Articles

Improve Article

C++ | Inheritance | Question 12

  • Difficulty Level : Easy
  • Last Updated : 28 Jun, 2021




#include<iostream>
using namespace std;
  
class Base
{
protected:
    int a;
public:
    Base() {a = 0;}
};
  
class Derived1:  public Base
{
public:
    int c;
};
  
  
class Derived2:  public Base
{
public:
    int c;
};
  
class DerivedDerived: public Derived1, public Derived2
{
public:
    void show()  {   cout << a;  }
};
  
int main(void)
{
    DerivedDerived d;
    d.show();
    return 0;
}

(A) Compiler Error in Line “cout << a;"
(B) 0
(C) Compiler Error in Line “class DerivedDerived: public Derived1, public Derived2”


Answer: (A)

Explanation: This is a typical example of diamond problem of multiple inheritance. Here the base class member ‘a’ is inherited through both Derived1 and Derived2. So there are two copies of ‘a’ in DerivedDerived which makes the statement “cout << a;" ambiguous.

The solution in C++ is to use virtual base classes. For example, the following program works fine and prints

#include<iostream>
using namespace std;

class Base
{
protected:
int a;
public:
Base() {a = 0;}
};

class Derived1: virtual public Base
{
public:
int c;
};



class Derived2: virtual public Base
{
public:
int c;
};

class DerivedDerived: public Derived1, public Derived2
{
public:
void show() { cout << a; }
};

int main(void)
{
DerivedDerived d;
d.show();
return 0;
}


Quiz of this Question

Want to learn from the best curated videos and practice problems, check out the C Foundation Course for Basic to Advanced C.
My Personal Notes arrow_drop_up
Recommended Articles
Page :