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GO Classes DPP | GATE CS | C Programming | Mutual Recursion

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7 7 votes

What is the output of the following code?

#include <stdio.h>

void g(int n);

void f(int n) {
    if (n <= 0)
        return;

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

void g(int n) {
    if (n <= 0)
        return;

    printf("%d ", n);
    f(n - 2);
}

int main() {
    f(5);
    return 0;
}
  1. $\texttt{5\ 4\ 3\ 2\ 1}$
     
  2. $\texttt{5\ 4\ 2\ 1}$
     
  3. $\texttt{5\ 3\ 1}$
     
  4. $\texttt{4\ 2\ 1}$

3 Answers

0 0 votes

 

The function call is $\texttt{f(5)}$.

In $\texttt{f(5)}$, $\texttt{5}$ is printed and then $\texttt{g(4)}$ is called.

In $\texttt{g(4)}$, $\texttt{4}$ is printed and then $\texttt{f(2)}$ is called.

In $\texttt{f(2)}$, $\texttt{2}$ is printed and then $\texttt{g(1)}$ is called.

In $\texttt{g(1)}$, $\texttt{1}$ is printed and then $\texttt{f(-1)}$ is called.

In $\texttt{f(-1)}$, the condition $\texttt{n <= 0}$ is true, so it returns.

Therefore, the output is $\texttt{5\ 4\ 2\ 1}$

Answer: B

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