Why am I able to rebind a reference after its initialization?

Question

According to C++ Primer, by Stanley B. Lippman, Josée Lajoie, and Barbara E. Moo:

Once initialized, a reference remains bound to its initial object. There is no way to rebind a reference to refer to a different object.

How, then, am I seemingly able to rebind the reference I initialized to another object in the following code?

#include <iostream>

int main()
{
     int num1 = 10;
     int num2 = 20;

     int &rnum1 = num1;

     std::cout << rnum1 << std::endl;   // output: 10

     rnum1 = num2;

     std::cout << rnum1 << std::endl;   // output: 20

     return 0;
}

From my understanding num1 and num2 are two different objects. The same type, yes, but two completely different objects.

It's only confusing if you think of references as hidden pointers — M.M, Jul 17 '17 at 10:13
This is the reason I dislike references from C++ and prefer plain old pointers in C. It is more explicit. — Jonathon Reinhart, Jul 17 '17 at 10:14
@JonathonReinhart: For me they are most useful in their `const` guise as function parameters. — Bathsheba, Jul 17 '17 at 10:16
You didn't collect enough evidence to suggest a rebound reference (`rnum1 == num2` says nothing about identity). Look at the value of `num1` as well, and examine `rnum1` after, say, `num2 = 30`. — molbdnilo, Jul 17 '17 at 10:21
Add some code to show the values of `num1` and `num2` before and after the assignment to `rnum1`. Then add more code to change the value of `num2` and look at those values again. — Pete Becker, Jul 17 '17 at 11:34

Bathsheba · Answer 1 · 2022-12-02T18:03:01.433

7

rnum1 = num2; is not rebinding the reference.

It is setting rnum1 (and therefore num1) to the value of num2.

edited Dec 02 '22 at 18:03

answered Jul 17 '17 at 10:12

Bathsheba

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David Haim · Answer 2 · 2017-07-17T10:17:36.650

rnum1 is another name for num1

when you write rnum1 = ... it is if you were to write num1 = ..., because rnum1 is another name for num1.

So the reference itself is not rebound to another variable, what you have here is simple assignment of num1.

To really prove it , you can print the memory addresses of num1, num2 and rnum1:

#include <iostream>

int main()
{
     int num1 = 10;
     int num2 = 20;

     int &rnum1 = num1;

     std::cout << rnum1 << std::endl;   // output: 10

     rnum1 = num2;

     std::cout << rnum1 << std::endl;   // output: 20
     std::cout << &num1 << " " << &rnum1 <<" "<< &num2 << std::endl;
     return 0;
}

Why am I able to rebind a reference after its initialization?

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