Call by reference vs call by copy restore

Question

I was looking at call by reference and call by copy/restore. And I had some confusion. Suppose I have a 2-parameter function that increments 1 to each parameter. Now what will happen if I use both the parameters as the same variable, like incr(i, i), in the case of call by reference and call by copy/restore?

Answer 1

The question What's the difference between call by reference and copy/restore covers this in part, but it isn't quite complete. The Wikipedia article on Evaluation Strategy is helpful.

Here are two implementations and calls for a function incr() that takes two arguments and increments each one.

Call by reference

#include <stdio.h>

void incr(int *pi1, int *pi2)
{
    (*pi1)++;
    (*pi2)++;
}

int main(void)
{
    int i = 57;
    printf("%d\n", i);
    incr(&i, &i);
    printf("%d\n", i);
    return 0;
}

Output:

57
59

Call by copy/restore

The incr function is actually unchanged from the call by reference version.

#include <stdio.h>

void incr(int *pi1, int *pi2)
{
    (*pi1)++;
    (*pi2)++;
}

int main(void)
{
    int i = 57;
    printf("%d\n", i);
    {
    int cr_0000 = i;    // Copy for first argument
    int cr_0001 = i;    // Copy for second argument
    incr(&cr_0000, &cr_0001);
    i = cr_0001;        // Restore for second argument
    i = cr_0000;        // Restore for first argument
    }
    printf("%d\n", i);
    return 0;
}

Output:

57
58

Note that with this incr() function, it does not matter which order the values are restored, but if the incr() function treated its arguments asymmetrically (say, added 1 to the first and 2 to the second), then the final result would depend on whether cr_0000 is restored before cr_0001 or vice versa.