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GO Classes DPP | GATE CS | C Programming | Pointer to Array

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9 9 votes

What is the output of the following code?

#include <stdio.h>

int main() {
    int a[2][3] = {
        {1, 2, 3},
        {4, 5, 6}
    };

    int (*p)[3] = a;

    printf("%d %d %d", **p, *(*(p + 1) + 2), *(*p + 1));

    return 0;
}
  1. $\texttt{1 5 2}$
     
  2. $\texttt{1 6 2}$
     
  3. $\texttt{4 6 2}$
     
  4. $\texttt{1 6 3}$

2 Answers

2 2 votes


Here,

$\texttt{int (*p)[3] = a;}$

means $\texttt{p}$ is a pointer to an array of $\texttt{3}$ integers.

So, $\texttt{p}$ points to the first row of the 2D array.

The array is:

$\texttt{a[0][0] = 1}$, $\texttt{a[0][1] = 2}$, $\texttt{a[0][2] = 3}$

$\texttt{a[1][0] = 4}$, $\texttt{a[1][1] = 5}$, $\texttt{a[1][2] = 6}$

Now,

$\texttt{**p}$

means first row, first element.

So,

$\texttt{**p = a[0][0] = 1}$

Next,

$\texttt{((p + 1) + 2)}$

Here, $\texttt{p + 1}$ moves to the second row.

So,

$\texttt{*(p + 1)}$ represents row $\texttt{1}$.

Then,

$\texttt{*(p + 1) + 2}$ points to $\texttt{a[1][2]}$.

So,

$\texttt{((p + 1) + 2) = a[1][2] = 6}$

Next,

$\texttt{*(*p + 1)}$

Here, $\texttt{*p}$ represents the first row.

So,

$\texttt{*p + 1}$ points to $\texttt{a[0][1]}$.

Therefore,

$\texttt{*(*p + 1) = a[0][1] = 2}$
 

$\therefore$ Output : $\texttt{1 6 2}$
 

Answer: B

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