C Programming - Pointers - Discussion

Thanks @Praful.

We use triple for a 3d array but arr[3][3] is a 2d array.

Thanks @Wikiok.

Thanks @Praful.

Now I get it. Thanks to all.

Thanks @Praful.

Thanks to all for the clear explanation.

#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.

Why use *** arr?

@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.