Can References Be Reassigned in C ?

Understanding Reference Reassignment in C

In C , it's commonly stated that references require initialization upon declaration and cannot be reassigned. However, a recent experiment has raised doubts about this notion. Let's investigate the behavior of reference reassignment in the following 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; // *** Reassignment at Issue ***
    cout << "ri is : " << ri << "\n";

    getch();
    return 0;
}

The program assigns the reference ri to the integer variable i, making ri an alias for i. Initially, ri points to the value 5. When i is modified to 10, ri correctly reflects this change. However, the critical point is the next statement: ri = j.

Is this not a reassignment of the reference?

The answer, surprisingly, is no. ri is still a reference to i. To demonstrate this, you can compare the addresses of ri and i: they are identical. What you are observing is not a reassignment of the reference but rather a modification of the value stored at the memory location pointed to by ri.

In simpler terms, ri = j is equivalent to *(&ri) = j, where &ri retrieves the memory address of ri and * dereferences it to access the value.

For comparison, if you create a const int &cri = i, it will prevent any reassignment to cri, enforcing its constant nature.

In summary, while references cannot be reassigned to new variables in C , they can be used to modify the value of the object they reference, as seen in our program.

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