Official PYQ · 2018Q21RecursionNAT1 mark

GATE CS 2018 Q21 — Recursion

Consider the following $\text{C}$ program:

#include<stdio.h>

int counter=0;

int calc (int a, int b) {
        int c;
        counter++;
        if(b==3) return (a*a*a);
        else {
                c = calc(a, b/3);
                return (c*c*c);
        }
}

int main() {
        calc(4, 81);
        printf("%d", counter);
}

The output of this program is ______.

View worked solution

Answer: 4

To find the output of the program, we trace the execution of the recursive function calc(4, 81) and track the value of the global variable counter.

The function calc(a, b) increments counter by 1 on every entry. Let's trace the sequence of recursive calls:

First Call: calc(4, 81)
- counter is incremented to 1.
- Since $b = 81 \neq 3$ , the else branch is executed.
- It calls calc(4, 81/3), which is calc(4, 27).
Second Call: calc(4, 27)
- counter is incremented to 2.
- Since $b = 27 \neq 3$ , the else branch is executed.
- It calls calc(4, 27/3), which is calc(4, 9).
Third Call: calc(4, 9)
- counter is incremented to 3.
- Since $b = 9 \neq 3$ , the else branch is executed.
- It calls calc(4, 9/3), which is calc(4, 3).
Fourth Call: calc(4, 3)
- counter is incremented to 4.
- Since $b = 3$ , the base condition if(b == 3) is satisfied.
- It returns $4^3 = 64$ and begins unwinding the recursion.

After all recursive calls complete and return, the final value of counter printed by the main function is 4.

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Published 10 Oct 2026

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