I've been looking into the int rand() function from <stdlib.h> in C11 when I stumbled over the following cppreference-example:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
 
int main(void)
{
    srand(time(NULL)); // use current time as seed for random generator
    int random_variable = rand();
    printf("Random value on [0,%d]: %d\n", RAND_MAX, random_variable);
 
    // roll a 6-sided die 20 times
    for (int n=0; n != 20; ++n) {
        int x = 7;
        while(x > 6) 
            x = 1 + rand()/((RAND_MAX + 1u)/6); // Note: 1+rand()%6 is biased
        printf("%d ",  x); 
    }
}

Specifically this part:

[...]
        while(x > 6) 
            x = 1 + rand()/((RAND_MAX + 1u)/6); // Note: 1+rand()%6 is biased
[...]

Questions:

1. Why the addition of + 1u? Since rand() is [0,RAND_MAX] I'm guessing that doing rand()/(RAND_MAX/6) -> [0,RAND_MAX/(RAND_MAX/6)] -> [0,6]? And since it's integer division (LARGE/(LARGE+small)) < 1 -> 0, adding 1u gives it the required range of [0,5]?

2. Building on the previous question, assuming [0,5], 1 + (rand()/((RAND_MAX+1u)/6)) should only go through [1,6] and never trigger a second loop?

Been poking around to see if rand() has returned float at some point, but that seems like a pretty huge breakage towards old code? I guess the check makes sense if you add 1.0f instead of 1u making it a floating point division?

Trying to wrap my head around this, have a feeling that I might be missing something..

(P.s. This is not a basis for anything security critical, I'm just exploring the standard library. D.s)