Is there a case where bitwise swapping won't work?-CodePudding

At school someday several years ago I had to do a swap function that swaps two integers, I wanted to do this using bitwise operations without using a third variable, so I came up with this:

void swap( int * a, int * b ) {
    *a = *a ^ *b;
    *b = *a ^ *b;
    *a = *a ^ *b;
}

I thought it was good but when my function was tested by the school's correction program it found an error (of course when I asked they didn't want to tell me), and still today I don't know what didn't work, so I wonder in which case this method wouldn't work.

CodePudding user response：

I wanted to do this using bitwise operations without using a third variable

Do you mind if I ask why? Was there a practical reason for this limitation, or was it just an intellectual puzzle?

when my function was tested by the school's correction program it found an error

I can't be sure what the correction program was complaining about, but one class of inputs this sort of solution is known to fail on is exemplified by

int x = 5;
swap(&x, &x);
printf("%d\n", x);

This prints 0, not 5.

You might say, "Why would anyone swap something with itself?" They probably wouldn't, as I've shown it, but perhaps you can imagine that, in a mediocrely-written sort algorithm, it might end up doing the equivalent of

if(a[i] < a[j]) {
    /* they are in order */
} else {
    swap(&a[i], &a[j]);
}

Now, if it ever happens that i and j are the same, the swap function will wrongly zero out a[i].