Question 36

The function convert is defined as:

void convert(int n){
    if (n < 0)
        printf("%d", n);
    else {
        convert(n / 2);
        printf("%d", n % 2);
    }
}

Let $n$ be any positive integer (e.g. $n = 5$):

1. Call convert(5): since $5 < 0$ is false, calls convert(5 / 2) = convert(2).
2. Call convert(2): since $2 < 0$ is false, calls convert(2 / 2) = convert(1).
3. Call convert(1): since $1 < 0$ is false, calls convert(1 / 2) = convert(0).
4. Call convert(0): check condition n < 0. For $n = 0$, $0 < 0$ is false!
   Therefore, execution proceeds to the else block and calls convert(0 / 2) = convert(0).

Because convert(0) repeatedly calls convert(0), the base condition n < 0 is never reached. This causes an infinite recursion.

Furthermore, all printf statements in the else block are placed after the recursive call convert(n / 2). Because the recursive calls never return, no printf statement is ever executed before the program exhausts the call stack (stack overflow crash).

Hence, it will not print anything and will not terminate (Option D).