Method Override

is this method override? ,

Code:

#include<iostream>
using namespace std;

class A
{
   protected:
           void show()
           {
                cout<<"A";     
           }
      
};

class B:public A
{
   public:
          void show()
          {
               cout<<"B"; 
               A::show();    
          }   
};

int main() 
{ 
    A *a = new B();
    a->show();                          // tries to call A::show() 
    getchar();
    return 0;
}

a->show() is calling A::show() rather than B::show() , so i suspect this isnt method override

however if it isnt, this means B has not overriden A's show(), and hence B has 2 show() methods with same signature name and return type, but different access specifiers

so i tried this:-

Code:

#include<iostream>
using namespace std;

class A
{           
   protected:
           void show()
           {
                cout<<"A";     
           }
           
   public:
           void show()
           {
                cout<<"B";     
           }      
};


int main() 
{ 
    A *a = new A();
    a->show(); 
    getchar();
    return 0;
}

now compiler reports error in overloading show() because both have same signature, that means 2 same functions cannot even exist in different access specifiers

both conclusions are contradicting each other

which is correct?

Originally Posted by vb.bajpai

however if it isnt, this means B has not overriden A's show(), and hence B has 2 show() methods with same signature name and return type, but different access specifiers

Untrue. The names are in different namespaces. One function is A::show, the other is B::show. There's no reason these two distinct names can't have different accessibilities.

both conclusions are contradicting each other

No, because A::show and B::show are different things. So both observed behaviors are correct.

And you aren't overriding the method, because it isn't virtual. You're just masking the parent's definition when in the scope of class B.

in example 2 you are overloading the function show() within the same namespace (a class in this case). The problem is that you have the same function signature for both functions (specifically void A::*() )

In example one, you are shadowing the function show(). if you add the keyword "virtual" to the base function it will work as you expect.

you're calling A::show() in B::show(). this has nothing to do with overriding a method.

the sequence is

Code:

    A *a = new B(); // creates a B object with an A interface
    a->show(); // lookup virtual function table and call B::show()
        == in B::show()========
        cout<<"B";  // prints B
        A::show(); // directly calls A::show(). no vtable lookup
            == in A::show() ========
            cout << "A"; // prints A