Why the for loop doesn't stop at i == 0 in strangeForLoop?

Question

I tried the following code on https://www.onlinegdb.com/

Also tried it on Mac.

Couldn't find out why the for loop in strangeForLoop will not stop when i is equal to 0?

#include <stdio.h>
#include <stdint.h>

void strangeForLoop();
void normalForLoop();

const uint32_t COUNTDOWN = 3;
int BREAK = -2;

int main()
{
    strangeForLoop();
    printf("\n===========\n");
    normalForLoop();
    return 0;
}

void strangeForLoop() {
    for(uint32_t i = COUNTDOWN; i>=0; i--) {
        printf("strange i : %d\n", i);
        
        if (i == 0) 
            printf("--> i: %d\n", i);
        if (i == BREAK) break;
    }
}

void normalForLoop() {
    for(int i = COUNTDOWN; i>=0; i--) {
        printf("normal i : %d\n", i);
        
        if (i == 0) 
            printf("==> i: %d\n", i);
        if (i == BREAK) break;
    }
}

Any help will be greatly appreciated.

When `i == 0` and you do `i--` at the end of the loop, the value of `i` will "wrap" and become really big. — Some programmer dude, Jul 18 '22 at 18:57
As you run the code, you will notice that "--> i: 0" is printed. — Tony, Jul 18 '22 at 18:58
Also note that the `printf` format `%d` is for plain `int`, which is *signed*, while `i` is *unsigned*. Mismatching format specifier and argument value type leads to *undefined behavior*. If you use `%u` instead, you will see what really happens with `i`. — Some programmer dude, Jul 18 '22 at 18:59
The point is that (i == 0) is "true". So why doesn't the for loop stop? — Tony, Jul 18 '22 at 19:02
@David Ranieri However, when i reaches 0, the for loop should be stopped. Isn't that this should be the case? — Tony, Jul 18 '22 at 19:06
Because the `for` loop condition `i>=0;` is `true`. That is the condition for the loop to continue. It is never `false` as explained. — Weather Vane, Jul 18 '22 at 19:06
OTOH `if (i == BREAK) break;` will work when `BREAK = -2` because the integer conversion rules apply to `-2`. — Weather Vane, Jul 18 '22 at 19:09
The loop does not stop when `i` reaches zero. The loop stop when `i` is less than zero. Which is never true. — William Pursell, Jul 18 '22 at 19:09
Both David and Weather are right. Thanks for the help, everyone. — Tony, Jul 18 '22 at 19:09

score 3 · Accepted Answer · edited Jul 18 '22 at 19:08

3

i is an unsigned integer. Such a uint can represent only positive values and zero;

[0, 1, 2, 3, ... 2^number_of_bits - 1]

When an operation would decrease a uint below zero, or above its maximum, an integer overflow occurs.

In the case of this code, it wraps around to the maximum value for the integer, so the condition i >= 0 will always remain true, and the loop will never stop.

edited Jul 18 '22 at 19:08

Andreas Wenzel

22,760
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answered Jul 18 '22 at 19:06

makeVoiceBot

143
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By standard explicit definition unsigned integers do not overflow. That is because they do not represent natural numbers, but they represent equivalence classes. Standard quotes on [this answer](https://stackoverflow.com/a/54190216/2805305) and more discussion in the comments there. – bolov Jul 18 '22 at 19:24
@bolov: Yes, the definition in the ISO C standard for the word "overflow" is different from the common definition. However, in my opinion, it is still correct to use the word "overflow" in this context. That just means that the common definition is being used instead of the definition used in the ISO C standard. – Andreas Wenzel Jul 18 '22 at 19:54
@AndreasWenzel yes, that is exactly what I did in my linked answer. However I do feel it is important to at least note the standard definition. – bolov Jul 18 '22 at 20:00

Why the for loop doesn't stop at i == 0 in strangeForLoop?

1 Answers1