Question

#include "stdafx.h"
#include <iostream>
using namespace std;

class A
{
public:
  int a,b,c;
  A()
  {
      a=0;
      b=0;
      c=0;
  }
};

class B:public A
{
public:
    void get()
    {
        A *a2 =new A;
        a2->a=10;
        a2->b=20;
        a2->c=30;
        cout<<a2->a<<""<<a2->b<<""<<a2->c<<""<<endl;
        cout<<"Checking!"<<endl;
    }
};

int main()
{
    A *a1 = new A;
    B *b1 = new B;
    cout<<a1->a<<""<<a1->b<<""<<a1->c<<""<<endl;
    b1->a=10;
    b1->b=20;
    b1->c=30;
    cout<<b1->a<<""<<b1->b<<""<<b1->c<<""<<endl;

    b1->get();//cant able to change the variables of the base class object with the derived class object
    cout<<a1->a<<""<<a1->b<<""<<a1->c<<""<<endl;//will print the same values..

    //b1->get();

    return 0;
}

output:

000
102030
102030
Checking!
000
Press any key to continue . . .

//The address of the variables that is holded by the derived class object is different from the address of the variables that is holded by the base class object.Isnt..
//But is there any possiblity to change the variables of the base class through the derived class object..

Answer 1

In your get() method, you are actually creating a temporary instance of Base class A, which has nothing to do with the class getting derived. Simply remove a2 from everywhere in your get() and try it. You will be able to see the changes happening.

[Note: Base class object is automatically allocated when derived object is allocated. That means when you do new B, it will allocate memory for B and A]

Edit:

For your question, your get() should be like this:

void get()
{
    a=10;
    b=20;
    c=30;
    cout<<a<<b<<c<<endl;
    cout<<"Checking!"<<endl;
}