Challenge 2: unsigned char Overflow — Possible Solution
====================================================================

#include <stdio.h>

int main() {
    unsigned char c = 255;
    c = c + 1;
    printf("%d\n", c);
    return 0;
}

Output:

    0

Explanation: an unsigned char can represent 0 through 255 (2^8 - 1).
255 is the maximum representable value, so adding 1 has nowhere to go
within that range -- per the chapter, unsigned integer overflow in C
is well-defined: it WRAPS AROUND, computing the result modulo
2^(number of bits). 255 + 1 = 256, and 256 mod 256 = 0, so c becomes 0.

This is not a bug or an error -- it's guaranteed, specified behavior
that any standard-conforming C compiler must produce identically,
every time, on every platform. It's the exact well-defined counterpart
the chapter contrasts against signed overflow, which has no such
guarantee at all.

WHY THIS WORKS AS AN ANSWER
------------------------------
This correctly computes the wraparound result (0, via modulo
arithmetic) and explains WHY it's well-defined rather than a bug --
unsigned overflow is specified by the standard to wrap, unlike signed
overflow's undefined behavior, which is exactly the asymmetry the
chapter's warn-box highlights.