CODE EXAMPLE FOR C

passing two dimensional array to function in c

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

#define ROWS 3
#define COLS 2

void fun1(int (*)[COLS], int);

int main()
{
  int array_2D[ROWS][COLS] = { {1, 2}, {3, 4}, {5, 6} };
  int rows = ROWS;

  /* works here because array_2d is still in scope and still an array */
  printf("MAIN: %zu
",sizeof(array_2D)/sizeof(array_2D[0]));

  fun1(array_2D, rows);

  return EXIT_SUCCESS;
}

void fun1(int (*a)[COLS], int rows)
{
  int i, j;
  int n, m;

  n = rows;
  /* Works, because that information is passed (as "COLS").
     It is also redundant because that value is known at compile time (in "COLS"). */
  m = (int) (sizeof(a[0])/sizeof(a[0][0]));
 
  /* Does not work here because the "decay" in "pointer decay" is meant
     literally--information is lost. */
  printf("FUN1: %zu
",sizeof(a)/sizeof(a[0]));

  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("array[%d][%d]=%d
", i, j, a[i][j]);
    }
  }
}