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GO Classes DPP | GATE CS | C Programming | Static Variable in Recursion

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6 6 votes

What is the output of the following code?

#include <stdio.h>

int fun(int n) {
    static int x = 0;
    int y;

    if (n == 0)
        return 0;

    x++;
    y = fun(n - 1);

    return y + x;
}

int main() {
    printf("%d", fun(3));
    return 0;
}

3 Answers

0 0 votes

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

A static variable is initialized only once and keeps its value across recursive calls.

 


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

For $\texttt{fun(3)}$, $\texttt{x++}$ makes $\texttt{x = 1}$, then it calls $\texttt{fun(2)}$.

For $\texttt{fun(2)}$, $\texttt{x++}$ makes $\texttt{x = 2}$, then it calls $\texttt{fun(1)}$.

For $\texttt{fun(1)}$, $\texttt{x++}$ makes $\texttt{x = 3}$, then it calls $\texttt{fun(0)}$.

For $\texttt{fun(0)}$, the base condition becomes true and it returns $\texttt{0}$.

Now returning starts.

In $\texttt{fun(1)}$, $\texttt{y = 0}$ and current $\texttt{x = 3}$.

So, it returns $\texttt{0 + 3 = 3}$.

In $\texttt{fun(2)}$, $\texttt{y = 3}$ and current $\texttt{x = 3}$.

So, it returns $\texttt{3 + 3 = 6}$.

In $\texttt{fun(3)}$, $\texttt{y = 6}$ and current $\texttt{x = 3}$.

So, it returns $\texttt{6 + 3 = 9}$.

$\therefore$ Output $:\texttt{9}$

• edited by
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