Can C References Be Reassigned, or is it Just Value Modification?

Reference Reassignment in C

Declaring a reference in C requires initialization, prompting the impression that references are immutable and cannot be reassigned. However, consider the following program:

#include <iostream>

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';
    return 0;
}

The code compiles successfully and produces the expected output:

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

Contrary to popular belief, the line ri = j does not reassign the reference ri. Instead, it modifies the value of i through the reference ri, as evidenced by printing i before and after the line. This behavior is supported by the fact that &ri and &i print the same address, indicating that ri remains a reference to i.

In contrast, declaring a constant reference (const int &cri = i) prevents reassignment of the reference itself.

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