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

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8 8 votes

What is the output of the following code?

#include <stdio.h>

void update(int n, int *p) {
    if (n <= 0)
        return;

    *p = *p + n;

    update(n - 2, p);

    *p = *p + n;
}

int main() {
    int x = 1;

    update(5, &x);

    printf("%d", x);

    return 0;
}

3 Answers

1 1 vote

Here, pointer $\texttt{p}$ points to $\texttt{x}$.

So, every change made using $\texttt{*p}$ directly changes the original variable $\texttt{x}$.

Initially, $\texttt{x = 1}$.

The call is $\texttt{update(5, \&x)}$.

In $\texttt{update(5)}$:

$\texttt{*p = *p + 5}$

So, $\texttt{x = 1 + 5 = 6}$

Then it calls $\texttt{update(3, p)}$.

In $\texttt{update(3)}$:

$\texttt{*p = *p + 3}$

So, $\texttt{x = 6 + 3 = 9}$

Then it calls $\texttt{update(1, p)}$.

In $\texttt{update(1)}$:

$\texttt{*p = *p + 1}$

So, $\texttt{x = 9 + 1 = 10}$

Then it calls $\texttt{update(-1, p)}$.

Since $\texttt{n <= 0}$, it returns immediately.

Now returning starts.

Back in $\texttt{update(1)}$:

$\texttt{*p = *p + 1}$

So, $\texttt{x = 10 + 1 = 11}$

Back in $\texttt{update(3)}$:

$\texttt{*p = *p + 3}$

So, $\texttt{x = 11 + 3 = 14}$

Back in $\texttt{update(5)}$:

$\texttt{*p = *p + 5}$

So, $\texttt{x = 14 + 5 = 19}$
 

$\therefore$ Output : $\boxed{19}$

• edited by
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