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.

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.

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.