Consider the following ANSI C program.

The output of the program upon execution is \(\underline{\hspace{2cm}}\).
Step 1: Understand base conditions.
The function returns \( q \) when both \( x \le 0 \) and \( y \le 0 \).
Step 2: Recursive behavior.
If one variable is non-positive, recursion continues by decreasing the other by \( q \).
If both are positive, the function branches into two recursive calls.
Step 3: Evaluate \( \text{foo}(15,15,10) \).
\[
\text{foo}(15,15,10) = \text{foo}(15,5,10) + \text{foo}(5,15,10)
\]
Each of these further expands until base cases are reached.
This forms a binary recursion tree where each leaf contributes a value of \( 10 \).
Step 4: Count base case occurrences.
The recursion generates exactly 6 base case hits.
Step 5: Compute final result.
\[
\text{Result} = 6 \times 10 = 60
\]
% Final Answer
Final Answer: \[ \boxed{60} \]
Consider the following code:
main() {
int x = 126, y = 105;
{
if (x > y)
x = x - y;
else
y = y - x;
}
while (x != y)
printf("%d", x);
}
Consider the following code:
int a;
int arr[] = {30, 50, 10};
int *ptr = arr[10] + 1;
a = *ptr;
(*ptr)++;
ptr = ptr + 1;
printf("%d", a + arr[1] + *ptr);