Can we reassign the reference in C++?

Question

I have read everywhere that a reference has to be initialized then and there and can't be re-initialized again.

To test my understanding, I have written the following small program. It seems as if I have actually succeeded in reassigning a reference. Can someone explain to me what is actually going on in my program?

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

using namespace std;

int main()
{
    int i = 5, j = 9;

    int &ri = i;
    cout << " ri is : " << ri  <<"\n";

    i = 10;
    cout << " ri is : " << ri  << "\n";

    ri = j; // >>> Is this not reassigning the reference? <<<
    cout << " ri is : " << ri  <<"\n";

    getch();
    return 0;
}

The code compiles fine and the output is as I expect:

ri is : 5
ri is : 10
ri is : 9

Answer 1

ri = j; // >>> Is this not reassigning the reference? <<<

No, ri is still a reference to i - you can prove this by printing &ri and &i and seeing they're the same address.

What you did is modify i through the reference ri. Print i after, and you'll see this.

Also, for comparison, if you create a const int &cri = i; it won't let you assign to that.

Answer 2

It seems as if I have actually succeeded in reassigning a reference. Is that true?

No, you haven't. You are actually reassigning the value, and you are not rebinding the reference.

In your example, when you do int &ri = i;, ri is bound to i for its lifetime. When you do ri = j;, you are simply assigning the value of j to ri. ri still remains a reference to i! And it results in the same outcome as if you had instead written i = j;

If you understand pointers well, then always think of the reference as an analogical interpretation of T* const where T is any type.

Answer 3

When you assign something to a reference you actually assign the value to the object the reference is bound to. So this:

ri=j;

has the same effect as

i = j;

would have because ri is bound to i. So any action on ri is executed on i.

Answer 4

You are not reassigning the reference when executing ri = j;. You're actually assigning j to i. Try printing i after the line and you'll see that i changed value.