CODEWITHSUNDEEP
cpointers
ikm104
ikm104

Reputation: 43

In chaos regarding function of pointers in C Language

I wrote a program to test pointer actions but I didn't get the logic behind output.

As if I use arr+1 then instead of 2nd location I get 7th and similar is the case with *arr+1 and I feel weird for * expression is showing address not value.

CAN ANYONE EXPLAIN OUTPUT, PLEASE !

Program is :

#include<stdio.h>
int main()
{
    static int arr[2][3][2] = {1,2,3,4,5,6,7,8,9,10,11,12};
        int i,j,k;
        printf("%d\t",arr);
        printf("%d\t",*arr);
        printf("%d\n",**arr);
        printf("%d\n",***arr);
        printf("%d\t",arr+1);
        printf("%d\t",*arr+1);
        printf("%d\n",**arr+1);
        printf("%d\n",***arr+1);
        for(i=0;i<2;i++)
        {
            for(j=0;j<3;j++)
            {
            for(k=0;k<2;k++)
                printf("%d      %u \n",*(*(*(arr+i)+j)+k),&arr[i][j][k]);
                printf("\n");
            }
        printf("\n");
        }
return 0;
}

And OUTPUT is:

4202512 4202512 4202512
1
4202536 4202520 4202516
2
1      4202512
2      4202516

3      4202520
4      4202524

5      4202528
6      4202532

7      4202536
8      4202540

9      4202544
10      4202548

11      4202552
12      4202556

Upvotes: 3

Views: 109

Answers (2)

oblitum
oblitum

Reputation: 12016

you declare a int [2][3][2] array, which means that:

  • the type of the expression arr is an int [2][3][2] array, so you can only get a pointer from that type.
  • the type of the expression arr[0], or *arr is an int [3][2] array, so you can only get a pointer from that type.
  • the type of the expression arr[0][0], or **arr is an int [2] array, so you can only get a pointer from that type.
  • the type of the expression arr[0][0][0], or ***arr is an int, so you can only get a int value from that type.

variations of this process will give similar outcome.

you always get a same base address because multidimensional arrays are flat in memory and dimension sizes are used for partitioning that flat memory.

int [2][2] is like:

0, 1
2, 3

Where the numbers resemble the order of elements in memory, and the bidimensional display resemble the array's line/column access.

Upvotes: 2

Leeor
Leeor

Reputation: 19736

Your first 3 prints are showing you the same address but have different meanings (and types)-

  1. arr is the address of your entire 3d array, of type int*** (you cast the pointer to int, not a good practice since on some systems it may be bigger)
  2. *arr is the address of the "flattened" 2d array of type int**, effectively located at arr[0]
  3. **arr is the address of the twice "flattened" 1d array of type int*, effectively located at arr[0][0]
  4. Only ***arr finally brings you to an actual int value of arr[0][0][0]

Now you start pointer arithmetics, that's determined by the type of the pointer as mentioned abode and by the size of each dimension (which is known to the compiler). So:

  1. arr+1 gives you the address of arr[1], which is removed from arr[0] by the length of 2*3 int elements (6*4=24 here)
  2. *arr+1 gives you the address arr[0][1], which is removed from arr[0][0] by the length of 2 int elements (2*4=8 here)
  3. **arr+1 gives you the address of arr[0][0][1], which is removed from arr[0][0][0] by the length of one int element (4 here)
  4. ***arr+1 gives you arr[0][0][1], which is simply 2

See also this answer - How are 3D arrays stored in C?

Upvotes: 1

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