GATE CS 2016 Set 1 Q.45 (CS Q.35): Functions, Recursion, Scope and Storage Classes

What will be the output of the following C program?

#include <stdio.h>
void count(int n) {
    static int d = 1;
    printf("%d ", n);
    printf("%d ", d);
    d++;
    if (n > 1) count(n - 1);
    printf("%d ", d);
}
void main() {
    count(3);
}

3 1 2 2 1 3 4 4 4

3 1 2 1 1 1 2 2 2

3 1 2 2 1 3 4

3 1 2 1 1 1 2

Official answer and solution

Official answer: A
From: Official GATE 2016 answer key

Solution

Trace recursive activation frames and track the persistent state of the static variable.

In C, a static local variable inside a function is initialized only once and retains its value across all subsequent invocations.

Trace execution starting from count(3) with d = 1 initially:

Call count(3):
- Prints n (3): output 3
- Prints d (1): output 1
- d++ $\implies d$ becomes 2
- Since $n > 1$ ( $3 > 1$ ), calls count(2)

Call count(2):
- Prints n (2): output 2
- Prints d (2): output 2
- d++ $\implies d$ becomes 3
- Since $n > 1$ ( $2 > 1$ ), calls count(1)

Call count(1):
- Prints n (1): output 1
- Prints d (3): output 3
- d++ $\implies d$ becomes 4
- Since $n > 1$ ( $1 > 1$ ) is false, does not recurse
- Prints d (4): output 4

Unwinding recursive calls:
- Control returns to count(2): prints d (4): output 4
- Control returns to count(3): prints d (4): output 4

Concatenating all printed values gives: 3 1 2 2 1 3 4 4 4.

Final answer

Common trap

Assuming d is re-initialized to 1 on each recursive call, or forgetting the post-recursive print statement.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2016 Computer Science and Information Technology (CS) Session 1
Question: 45 of 65, printed as CS Q.35
Section: Computer Science and Information Technology

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