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GO Classes DPP | GATE CS | C Programming | Double Pointer Basics

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8 8 votes

What is the output of the following code?

#include <stdio.h>

int main() {
    int x = 10;
    int *p = &x;
    int **q = &p;

    **q = **q + 4;
    *p = *p + 1;

    printf("%d %d", x, **q);

    return 0;
}
  1. $\texttt{14 14}$
     
  2. $\texttt{15 15}$
     
  3. $\texttt{10 15}$
     
  4. $\texttt{Compilation error}$

3 Answers

0 0 votes

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

Also, $\texttt{q}$ points to $\texttt{p}$.

So,

$\texttt{*q}$ gives $\texttt{p}$

and

$\texttt{**q}$ gives the value of $\texttt{x}$.

Initially:

$\texttt{x = 10}$

First statement:

$\texttt{**q = **q + 4}$

Since $\texttt{**q}$ refers to $\texttt{x}$,

$\texttt{x = 10 + 4 = 14}$

Second statement:

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

Since $\texttt{p}$ points to $\texttt{x}$,

$\texttt{x = 14 + 1 = 15}$

Now,

$\texttt{x = 15}$

and

$\texttt{**q = 15}$
 

$\therefore$ Output : $\texttt{15 15}$
 

Answer: B

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