Question

I was just doing some stuff and wrote this program. I got the following output. I know that function resolution happens at run time while binding of the variable takes place during compile time, but what I could not understand was the first output I got (this->n). Can anyone explain this to me?

    # include <iostream>
    # include <stdio.h>
    # include <conio.h>

    using namespace std;

    class A
    {
      int n;      
      public:
      virtual void Fun1(int no=10)
       {
         int n=no;        
         cout<<"A::Fun1() "<<n <<"\n";
       }
    };

    class B :public A
    {
     int n;      
     public:
     virtual void Fun1(int no=20)
     {
      int n=no;        
      cout<<"B::Fun1() " << this->n << "\n"; // WHY SO ? gives B::Fun1() 40
      cout<<"B::Fun1() " << n << "\n";
     }
    };

    int main()
    {
     B b;
     A &a =b;
     a.Fun1();
     A a1=b;
     a1.Fun1();
     getch();
     return 0;
    }

The output I got was

    B::Fun1() 40
    B::Fun1() 10
    A::Fun1() 10    

Answer 1

For the first output, this->n refers to the member variable B::n, which you never initialise – this could print anything.

For the second, n refers to the local variable n, which you initialise with the function argument. Since default arguments are provided by the caller, and you’re calling the function via a reference to A, you get the default value specified in A::Fun1, not the one in B::Fun1. If you were to call b.Fun1(), then it would print B::Fun1() 20.

For the third, you’ve “sliced” the B object, giving an object of type A, hence the call to A::Fun1.