Can C References Be Reassigned?

Reassigning References in C

In C , references are often described as immutable bindings to memory addresses. However, a recent question has raised doubts about this understanding. To explore this topic, let's examine the following C program:

#include <iostream>

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

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

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

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

    return 0;
}

This code compiles successfully, and its output is:

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

The question arises: doesn't the assignment ri = j; contradict the notion of immutable references?

No, ri is still a reference to i. This can be verified by printing the addresses of ri and i and observing that they are the same.

What has actually occurred is that the ri reference is used to modify the value of i. Assigning a new value to ri would indeed be forbidden, because the reference is immutable and must always point to the same memory location.

For comparison, consider the following code:

const int &cri = i;

This code will not allow an assignment to cri, because it is a reference to a constant. This demonstrates that while references cannot be reassigned to a new memory location, they can still be used to modify the value at the address they refer to, provided that value is mutable.

In conclusion, the ri = j; assignment in the original program is not a reassignment of the reference itself, but rather a modification of the value it references.

The above is the detailed content of Can C References Be Reassigned?. For more information, please follow other related articles on the PHP Chinese website!