Challenge 2: A Function That Tries to Double a Parameter — Possible Solution
====================================================================

#include <stdio.h>

void double_it(int x) {
    x = x * 2;  // only changes the local copy
}

int main() {
    int n = 10;
    double_it(n);
    printf("%d\n", n);
    return 0;
}

Output:

    10

Explanation: n is still 10, not 20, because C passes arguments by
value -- calling double_it(n) copies n's current value (10) into the
function's own local parameter x. Reassigning x inside double_it only
changes that local copy; it has no connection back to main's n at all
once the function returns. This is exactly the chapter's own
try_to_change example, demonstrating the same behavior with a
different operation.

WHY THIS WORKS AS AN ANSWER
------------------------------
This confirms n is unchanged (10, not 20) and explains WHY using the
chapter's own pass-by-value/copy terminology, rather than treating the
output as surprising -- it's the expected, guaranteed behavior of
every C function call, not an edge case.