*(*(multi + 3) + 1)

With a little thought we can see that:

     *(*(multi + row) + col)             and

     multi[row][col]                     yield the same results.

The following program illustrates this using integer arrays instead of character arrays.

------------------- program 6.1 ----------------------
/* Program 6.1 from PTRTUT10.HTM              6/13/97*/
#include <stdio.h>
#define ROWS 5
#define COLS 10
int multi[ROWS][COLS];
int main(void)
{
     int row, col;
     for (row = 0; row < ROWS; row++)
     {
           for (col = 0; col < COLS; col++)
           {
                 multi[row][col] = row*col;
           }
      }
      for (row = 0; row < ROWS; row++)
      {
           for (col = 0; col < COLS; col++)
           {
                printf("\n%d ",multi[row][col]);
                printf("%d ",*(*(multi + row) + col));
           }
      }
      return 0;
}
----------------- end of program 6.1 ---------------------

Because of the double de-referencing required in the pointer version, the name of a 2 dimensional array is often said to be equivalent to a pointer to a pointer. With a three dimensional array we would be dealing with an array of arrays of arrays and some might say its name would be equivalent to a pointer to a pointer to a pointer. However, here we
have initially set aside the block of memory for the array by defining it using array notation. Hence, we are dealing with a constant, not a variable. That is we are talking about a fixed address not a variable pointer. The dereferencing function used above permits us to access any element in the array of arrays without the need of changing the value of that address (the address of multi[0][0] as given by the symbol multi).