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

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.

CodePudding user response：

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.