Thread: pointers and arrays

Code:

int a[2][3] = { { 1, 2, 3}, {3, 6, 9 } };
int i, j;
for( i = 0; i < 2; ++i )
    for( j = 0; j < 3; ++j )
        printf("%d\n",*(*(a+i)+j));

The identifier "a" can be looked at as a pointer-to-pointer-to-int (int **a). "a+i" is a pointer-to-pointer-to-int of row "i" of the 2d array which when dereferenced as in "*(a+i)" yields a pointer-to-int indicating the particular row. Adding "j" to this as in "*(a+i)+j" indicates a pointer-to-int to the particular column "j" of the particular row "i" we are looking at. Dereferencing that as in "*(*(a+i)+j)" yields the value at row "i" column "j".

Consider this

Code:

 int arr[4][2]={
                          {1,2},
                          {3,4},
                          {5,6},
                          {7,8}
                      };

Now , the compiler knows how many columns are there in the
array arr ,since we specified this in the array declaration .
So it interprets the expression arr[0] as (arr+0), and arr[1] as
(arr+1).
This way we are able to reach each individual row . Now
suppose I want to refer to the element arr[2][1] using pointers.
arr[2][1] can also be interpreted as (arr[2]+1) . value at this
address can be obtained by using the value at address operator
* i.e *(arr[2]+1) . arr[2][1] can be interpreted as *(*(arr+2)+1)

hence all these expressions are same.

arr[2][1]
*(arr[2]+1)
*(*(arr+2)+1)

> The identifier "a" can be looked at as a pointer-to-pointer-to-int (int **a)
Erm, no it isn't, and no it can't.

> int a[2][3] = { { 1, 2, 3}, {3, 6, 9 } };
When you say 'a', you get a pointer to the first element of a (the type being int (*)[3])
This is NOT the same as int**
http://www.eskimo.com/~scs/C-faq/q6.3.html

So when you do (a+i), you're really getting a pointer to a[i] (with the aforementioned type). In memory terms, you now have a pointer to the start of an array containing say 3,6,9 (when i=1).

Now when you do *(a+1), this is when you get an int* pointer.
(*(a+1)+j), you have a pointer to a specific element of 3,6,9 (say a pointer to a memory location containing 9, when j=2 say).

*(*(a+1)+j) dereferences that pointer to an int to get the actual integer (9 say).

Code:

#include <stdio.h>

int main ( void ) {
  int a[2][3] = { { 1, 2, 3}, {3, 6, 9 } };
  int i, j;
  for( i = 0; i < 2; ++i )
      for( j = 0; j < 3; ++j ) {
        int (*p1)[3] = a;
        int *p2;
        int val;
        p1 += i;
        p2 = *p1 + j;
        val = *p2;
        printf("%p %p %d\n",(void*)p1,(void*)p2,val);
      }
  return 0;
}

Simply mushing p1 to be an int** for example will get you

Code:

$ gcc -Wall bar.c
bar.c: In function `main':
bar.c:8: warning: initialization from incompatible pointer type
$ ./a.exe
Segmentation fault (core dumped)

>to access arrays, you need to type array[which_one_it_is] = " whatever you want its value to be ";
You don't need to, though that's by far the easiest way to get an an element in the array. However, your post really has no point since the question was concerning how to access arrays using pointer arithmetic, not the subscript operator.