We know that for int array[5]; &array is int (*)[5]
and we can assign and use them as
int array[5] = {1,2,3,4,5};
int (*p)[5] = &array;
// to access first element
printf("%d", p[0][0]);
This above code is for 1D array.
How to use this pointer to array approach to access 2D or n-D arrays?
What will be &array for int array[n][n]?
Generally, if p is pointer name,i row number and j column number,
(*(p+i)+j) would give a memory address of a element in 2D array. i is row no. and j is col no.,*(*(p+i)+j) would give the value of that element.*(p+i) would access the ith row
to access columns, add column number to *(p+i). You may have to declare the pointer as (*p)[columns] instead of just *p. Doing so, you are declaring pointer to an 2D array.Using pointer arithmetic is treating 2d array like 1D array. Initialize pointer *Ptr to first element (int *Ptr = *data) and then add an no. (Ptr + n) to access the columns. Adding a number higher than column number would simply continue counting the elements from first column of next row, if that exists. Source
The & operator simply returns a pointer to the whole array, so for example you can assign it to be the first part of a 1-level higher dimension array.
To understand this, we can show the difference by this code snippet:
int array[5] = {1, 2, 3, 4, 5};
printf("array address: %p\n&array address: %p\n", array, &array);
/* now test incrementing */
printf("array+1 address: %p\n&array+1 address: %p\n", array+1, &array+1);
A sample output of the above code is:
array address: 0x7fff4a53c310
&array address: 0x7fff4a53c310
array+1 address: 0x7fff4a53c314
&array+1 address: 0x7fff4a53c324
As you can see, if we increment the array pointer, it increments address by four (As Integer takes 4-bytes in my compiler). And if we increment the &array pointer, it increments by 20 that is 0x14.
&arrayforint array[n][n]" -int (*p)[n][n]