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GO Classes DPP | GATE CS | C Programming | Print Before and After Recursion

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6 6 votes

What is the output of the following code?

#include <stdio.h>

void fun(int n) {
    if (n == 0)
        return;

    printf("%d ", n);
    fun(n - 1);
    printf("%d ", n);
}

int main() {
    fun(3);
    return 0;
}
  1. $\texttt{3\ 2\ 1\ 1\ 2\ 3}$
     
  2. $\texttt{1\ 2\ 3\ 3\ 2\ 1}$
     
  3. $\texttt{3\ 2\ 1}$
     
  4. $\texttt{1\ 2\ 3}$

4 Answers

0 0 votes


The function call is $\texttt{fun(3)}$.

In this function, one $\texttt{printf}$ is before the recursive call and one $\texttt{printf}$ is after the recursive call.

The first $\texttt{printf}$ runs while going down in recursion.

So, it prints:

$\texttt{3\ 2\ 1}$

Then $\texttt{fun(0)}$ is called, and the base condition becomes true.

Now the function calls start returning.

The second $\texttt{printf}$ runs while returning from recursion.

So, it prints:

$\texttt{1\ 2\ 3}$

The complete output is $\texttt{3\ 2\ 1\ 1\ 2\ 3}$

Answer: A

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