C Programming - Pointers - Discussion

#include<stdio.h>

int main()
{
int arr[3][3] = {1, 2, 3, 4};
printf("%d\n", *(*(arr+1)));
return 0;
}

Why is answer 4 here?

Ans: 2-D array is nothing but the collection 1-D arrays that are placed after one another.

Here int arr[3][3] = {1, 2, 3, 4};

This array is initialized with only 4 elements, remaining all elements will be initialized to zero.

i.e int arr[3][3] = {1, 2, 3, 4,0,0,0,0,0};

Internally this 2-D array will be treated as:

int arr[3][3] = {{1, 2, 3}, {4,0,0},{0,0,0}}; //three 1-D arrays(0th,1st and 2nd 1-D array).

//3 rows and 3 columns:

1 2 3.
4 0 0.
0 0 0.

*(*(arr+0))points to 1st elements of 0th 1-D array ->1 2 3.

*(*(arr+1))points to 1st elements of 1st 1-D array ->4 0 0.

*(*(arr+2))points to 1st elements of 2nd 1-D array ->0 0 0.

@Niki, arr,*arr, arr[0] give same result and arr[0][0] and *(*(arr+i) +j) are same representation in 2d array.

arr[0][0][0] and *(*(*(arr+i)+j)+k) are same representation in 3d array.

So @Niki your representation is invalid in 2d array.

@Lavaanya.

*(arr+1) is one unit here it shifts 3(cause 3 row).

So. it will point to next column.

You can access any element via this = *(*(arr+column)+row).

#include<stdio.h>

int main()
{
int arr[3][3] = {1, 2, 3, 4};
printf("%d\n", *(*(arr+1)));
return 0;
}

Why is answer 4 here.

Array declared are two dimensional but in the we want to print three dimensional value due to this reason type is not match so it thrown error.

Array declared as two dimensional but in the printf statement there is three dimensional pointer notation so complier thrown error.

The given array is a 2-D array, so we can access all the elements using double pointer. Triple pointer is not needed.

We use triple pointer for 3-Dimensional array only.

But in above program 2-dimensional array is given.

As we have given two dimensional error so it can have only pointer to pointer means only two *.

Can we take it like this?

*(*(*(arr))=*(*(*arr[0][0]))=*(*(1))

Some one please explain me.