reference polymorphism

[Fred Ikk]

#include<iostream>
using namespace std;

class Base{
	public:
		int i=0;
		Base(int i) {this->i=i;}
		virtual ~Base() {}
		virtual void print(){
			cout<<"Base, i="<<i<<endl;
		}
};

class Derived:public Base{
	public:
		Derived(int i):Base(i){}		
		virtual ~Derived(){}
		virtual void print(){
			cout<<"Derived, i="<<i<<endl;
		}

};

int main(){
        Base& rb1=*new Base(1);
	rb1.print();
	rb1=*new Derived(2);
	rb1.print();// why Base::print() is called instead of Derived::print()?
	Base& rb2=*new Derived(3);
	rb2.print();
        return 0;
}

Base, i=1
Base, i=2
Derived, i=3

Shouldn't line 28 call the print() method of a Derived object? Could someone explain to me what is happening and why the same example written with pointers instead of references works as I would expect, meaning something like:

        Base* rb1=new Base(1);
	rb1->print();
	rb1=new Derived(2);
	rb1->print();

gives output:

Base, i=1
Derived, i=2

[Peter87]

No the equivalent code written with pointer would look like:

Base* rb1 = new Base(1);
rb1->print();
*rb1 = *new Derived(2); // Note that this doesn't change the pointer
rb1->print();

It is not possible to change the object that a reference is referring to. Line 27 is using the copy assignment operator (Base::operator=) to assign to the object being referred to by rb1. rb1 will still be referring to the same object as it was initialized to on line 25.

[Fred Ikk]

Thanks for the reply, so it copies from the Derived object only the part that "is related" to Base object?

[Peter87]

Yes that's right. It will only copy the Base::i in this case.