C Programming - Pointers - Discussion

I feel the best way to think of 3D arrays is to imagine a cube and put the numbers in their corresponding vertices.

For 2x2x2 array, we can imagine a cube with a total of 8 vertices, 4 in the front and 4 at the back.

10, 2, 3, 4, 5, 6, 7, 8 can be written here as

10(5) 2(6)
3(7) 4(8)

/*
Where the numbers without brackets are the vertices of the nearer face of the cube and those in brackets are the vertices of the farther face.
*/

Now, a[1][1][1] is the bottom right vertex of the farther face of the cube -- which is 8.

And "arr" is the address of the first element of the array -- here it is 10.

So, we get 10 and 8 as the answers.

arr[2][2][2], this array representation says that there are 2Blocks each with 2Rows and 2Columns

i,e we can judge that every Block will have 4 Elements each
so Block1-------[10 2 3 4]
i,e {10 2} in Row1
{3 4} in Row2
Again, Block2----[5 6 7 8]
i,e {5 6} in Row1
{7 8} in Row2

Now, let me explain how elments are described in 3D array representaion:- [Block][Row][Column]
so, the member of Block1 are described as below,
10 is [0][0][0]
2 is [0][0][1]
3 is [0][1][0]
4 is [0][1][1]
now, the members of Block2,
5 is [1][0][0]
6 is [1][0][1]
and so on
Thus, arr[1][1][1] is 8

Here the 3-D array the value can be store as:

a[2][2][2]={
             {
               10,2
               3,4   // First Block
             }
             {
                5,6
                7,8  //Second Block
             }
           }
a[ 0 ] [ 0 ] [ 0 ] =10;
   |     |     |
Block  Rows  Columns

a [ 1 ] [ 1 ] [ 1 ] = 8;      
    |     |     |
 2'Block 2'Row 2'column 

and the *arr represent the first number of a array is 10.


Then hence the number is (8,10).

This is 3D array you can look at this diagram so that easly understand it

1stblock R1| 10 . 2|     2nd block R1| 5 . 6 |
           |_______|                 |_______|
        R2 | 3 ,  4|               R2| 7 . 8 |
           |_______|                 |_______| 
            C1    C2                  C1   C2

arr[1][1][1]==>block 2,row 2,colum 2; ==>[8].
arr mean arr[0][0][0]==>block 1 row 1 column; ==>[10]

a[0][0][0]=10;
a[0][1][1] =2;

0 1 ->blocks
0 1 0 1 ->sub blocks
10 2 3 4 5 6 7 8 ->values
0 1 0 1 0 1 0 1 ->sub blocks of sub blocks

3-D REPRESENTATION OF ARRY IS

blocks-subblocks-subblocks of subblocks-values
then a [0] [1] [1] = 4
a [1] [1] [0] = 7 .so i think it is correct.

1. a[1][1][1] is the last element. So a[1][1][1]=8.
2. The pointer *q always points to the first element of the array arr[][][], because of the initialization of q = (int*) arr. So *q represents the first element of the array.

Therefore the answer is 8,10.

Elements stored as:

arr[0][0][0] = 10
arr[0][0][1] = 2
arr[0][1][0] = 3
arr[0][1][1] = 4
arr[1][0][0] = 5
arr[1][0][1] = 6
arr[1][1][0] = 7
arr[1][1][1] = 8

So *p=arr[1][1][1]=8.

And *q=10(first element).

@haneef:

a[2][1][1] cannot be accessed, coz array is initialized only for the 2nd dimension(i.e 0,1) a[2]... will be like accessing the thrid element which is not initialized.

Hope you got it.

It is three dimensional array,
so,arr[1][1][1]=8,
*p,
p=&arr[1][1][1],
*p means value at address i.e=8.
(int*)arr always consist first value as,
starting value at arr address,so
o/p= 8,10

@shubham

No need of casting here as q in an integer point here.So q = (int*) arr; and q = arr; would result in same output.