Unsigned int weird behavior with Loops in C

In C, I encounter this weird problem:
For the following code:

#include <stdio.h>
#include <math.h>


int main(int argc, char *argv[]) {
    unsigned int i = 0;
    for (int j = i - 1; j < i + 1; j++) {
        printf("RUN1?\n");
    }
    
    int j = i - 1;
    int k = i + 1;
    for (; j < k; j++) {
        printf("RUN2?\n");
    }
    
    printf("DONE?");
}

The first loop did not loop (0 times) while the second loop can run 2 times. I don’t see what’s fundamentally different.

The below also works…

for (;j < (int)(i+1); j++) {}

I have seen some posts here, but did not find any good explanation.
Maybe someone can link one here?

>Solution :

The difference between the 2 loops is in the types used for the condition:

1. In the 1st loop the condition is: j < i + 1. Since i is unsigned (and so is i+1) j is also converted to unsigned to perform the comparison. As an unsigned it is a high number and never < i+1.
2. In the 2nd loop the condition is j < k. These are both signed values and therefore they are not converted to unsigned. j is initialized to -1 when i-1 is converted to signed int. The result is the one you expected (2 iterations of the loop).

Note:
When the unsigned i-1 is converted to signed you get an overflow. The result is implementation defined. A common result (which you also get) is -1 but it is not guaranteed.