Challenge 2: calloc's Zero-Initialization, Observed Directly — Possible Solution
====================================================================

#include <stdio.h>
#include <stdlib.h>

int main() {
    int *nums = calloc(3, sizeof(int));
    if (nums == NULL) {
        return 1;
    }

    printf("%d %d %d\n", nums[0], nums[1], nums[2]);

    free(nums);
    return 0;
}

Output:

    0 0 0

Explanation: per the chapter, calloc zero-initializes the memory it
allocates -- every byte of the requested block is explicitly set to 0
before calloc returns, which for an int array means every element
reads as the integer 0, even though nothing was ever written to it
directly in this program.

Contrast with malloc: malloc(3 * sizeof(int)) would allocate the same
AMOUNT of memory, but would NOT zero it -- the bytes would contain
whatever leftover values happened to already be sitting in that memory
from whatever previously used it. Printing those three ints without
writing to them first would show unpredictable, essentially garbage
values -- not necessarily 0, and not consistent from one run to the
next, since it depends entirely on the memory's prior contents.

WHY THIS WORKS AS AN ANSWER
------------------------------
This confirms calloc's output is reliably 0 0 0 -- a guaranteed,
deterministic result -- and explicitly contrasts it with malloc's
genuinely undefined/garbage contents for the identical allocation
size, which is precisely the distinction the chapter calls "genuinely
important safety-wise."