#include <stdio.h>
#include <stdlib.h>
#include <conio.h>

int main(void)
{
  int **rptr;
  int *aptr;
  int *testptr;
  int k;
  int nrows = 5;     /* Both nrows and ncols could be evaluated */
  int ncols = 10;    /* or read in at run time */
  int row, col;
     /* we now allocate the memory for the array */
  aptr = malloc(nrows * ncols * sizeof(int *));
  if(aptr == NULL)
  {
    puts("\nFailure to allocate room for the array");
    exit(0);
  }
     /* next we allocate room for the pointers to the rows */
  rptr = malloc(nrows * sizeof(int *));
  if(rptr == NULL)
  {
    puts("\nFailure to allocate room for pointers");
    exit(0);
  }
     /* and now we 'point' the pointers */
  clrscr();
  for(k = 0; k < nrows; k++)
  {
     rptr[k] = aptr + (k * ncols);
  }
  printf("\n\n\nIndex   Pointer(hex)   Pointer(dec)   Diff.(dec)");

  for(row = 0; row < nrows; row++)
  {
    printf("\n%d         %p         %d", row, rptr[row], rptr[row]);
    if(row > 0)
      printf("              %d",(int)(rptr[row] - rptr[row-1]));
  }
  for(row = 0; row < nrows; row++)
  {
     for(col = 0; col < ncols; col++)
     {
       rptr[row][col] = row + col;
       printf("%d ", rptr[row][col]);
     }
     putchar('\n');
  }
  puts("\n\n\n");

     /* and here we illustrate that we are, in fact, dealing with
        a 2 dimensional array in a _contiguous_ block of memory. */

  testptr = aptr;
  for(row = 0; row < nrows; row++)
  {
    for(col = 0; col < ncols; col++)
    {
      printf("%d ", *(testptr++));
    }
    putchar('\n');
  }
  return 0;
}
.