How does this while loop work?

[johnmerlino]

The below function confuses me. Basically, since unsigned integers use a two's complement system, we cannot take the inverse of n before processing it, should n be negative. Because the largest negative number does not have absolute value representation. Thus, we use abs here after processing n % 10 while n is still negative. But how does that while loop work? Why doesn't it loop forever? If n is -133, reducing it by a factor of 10 with each iteration, it will start spitting out -1 infinitely. Am I missing something?

void itoa(int n, char s[]) {

    int i, sign;

    sign = n;

    

    i = 0;

    do {

        s[i++] = abs(n % 10) + '0';

    } while ( n /= 10 );

    if (sign < 0)

        s[i++] = '-';

    s[i] = '\0';

    reverse(s);

}

[nonpuz]

They're using integer math, anything less then 1 is 0 , thus the loop ends... -1/10 = 0.

[johnmerlino]

But if n is a negative value, n will always be less than 1 and therefore the loop will never execute, if what you say is correct, no?

[nonpuz]

The loop ends when n /= 10 results in 0. Not less then zero, not greater then zero, exactly zero.

Besides that it's a do while loop and will always execute at least once.

EDIT: Why don't you put a printf("n = %d\n",n) right above (and/or below) line 13 and see what n is on each iteration of the loop.
Perhaps that will clear things up for you.

[grumpy]

They're using integer math, anything less then 1 is

That's not true. Integer match rounds toward zero. So -21/10 is -2, not -3.