#include <stdio.h>
int main() {
  int abc[4][4] = {
      {0, 1, 2, 3},
      {4, 5, 6, 7},
      {8, 9, 10, 11},
      {12, 13, 14, 15},
  };
  int *base = &abc;
  int *current = NULL;
  for (int i = 0; i <= 15; i++) {
    /* The correct way of displaying an address would be
     * printf("%p ",abc[i]); but for the demonstration
     * purpose I am displaying the address in int so that
     * you can relate the output with the diagram above that
     * shows how many bytes an int element uses and how they
     * are stored in contiguous memory locations.
     *
     */
    current = base + i;
    printf("\n abc[%d], %d ", i, *current);
  }

  long int cur = (long int)(&abc);
  for (int i = 0; i <= 15; i++) {
    /* Here we cast the address to an int.
     * Manipulate it and then reinterpret as a pointer.
     */
    cur = (long int)base + i * 4;
    printf("\n abc[%d], %d ", i, *(int *)cur);
  }
  return 0;
}