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GO Classes DPP | GATE CS | C Programming | Static Sum & Recursion

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5 5 votes

What is the output of the following code?

#include <stdio.h>

int fun(int n) {
    static int sum = 0;
    int temp;

    if (n <= 0)
        return sum;

    sum = sum + n;

    temp = fun(n - 2);

    return temp + sum;
}

int main() {
    printf("%d", fun(5));
    return 0;
}

2 Answers

2 2 votes

Here, $\texttt{sum}$ is a static variable.

So, there is only one copy of $\texttt{sum}$, and it keeps its value across recursive calls.

Initially, $\texttt{sum = 0}$.

For $\texttt{fun(5)}$:

$\texttt{sum = 0 + 5 = 5}$, then it calls $\texttt{fun(3)}$.

For $\texttt{fun(3)}$:

$\texttt{sum = 5 + 3 = 8}$, then it calls $\texttt{fun(1)}$.

For $\texttt{fun(1)}$:

$\texttt{sum = 8 + 1 = 9}$, then it calls $\texttt{fun(-1)}$.

For $\texttt{fun(-1)}$, the base condition is true, so it returns $\texttt{sum}$.

So, $\texttt{fun(-1)}$ returns $\texttt{9}$.

Now returning starts.

In $\texttt{fun(1)}$, $\texttt{temp = 9}$ and current $\texttt{sum = 9}$.

So, $\texttt{fun(1)}$ returns $\texttt{9 + 9 = 18}$.

In $\texttt{fun(3)}$, $\texttt{temp = 18}$ and current $\texttt{sum = 9}$.

So, $\texttt{fun(3)}$ returns $\texttt{18 + 9 = 27}$.

In $\texttt{fun(5)}$, $\texttt{temp = 27}$ and current $\texttt{sum = 9}$.

So, $\texttt{fun(5)}$ returns $\texttt{27 + 9 = 36}$.

$\therefore$ Output $:\texttt{36}$

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