Question

In the Code Below there are Two Classes. One Object of type two is created and then it is assigned to pointer of class one.

On Calling the out function, the out function of the class one is called.

#include<iostream>
using namespace std;

class one
{
    public :
        void  out()
        {
            cout<<"one ";
        }
};

class two
{
    public : 
        void out()
        {
            cout<<"two ";
        }
};

int main()
{ 
    two dp[3];
    one *bp = (one *)dp;
    for (int i=0; i<3;i++)
    (bp++)->out();
}    

OUTPUT

one one one

Output according to me should be two instead of one.
When we created the object of type two, the memory location of that object contained the address of the function of out of class two, then why on assignment, out of class one is called ?

EDIT –
moreover even if we change the name of function in class two, the output is not changed.

Answer 1

It’s not uncommon for a novice to assume that all C++ member functions “belong” to an object.
As you noticed, they don’t.

Conceptually – the precise procedure is a compiler implementation detail – your out member functions are transformed into “free” functions that look like these:

void one_out(one* this) { cout << "one"; }
void two_out(two* this) { cout << "two"; }

For non-member functions, that’s all that’s needed.

When the compiler sees

  (bp++)->out();

it knows that bp is a pointer to one (it doesn’t know that you lied), and so it calls

one_out(bp++);

because that’s what compilers do.