Question

What is the best way to pass three dimensional arrays into functions in C?

Answer 1

typedef is your friend.

#include <stdio.h>
typedef int dimension1[20]; /* define dimension1 as array of 20
                               elements of type int */
typedef dimension1 dimension2[10]; /* define dimension2 as array of 10
                                      elements of type dimension1 */

int foo(dimension2 arr[], size_t siz);

int main(void) {
  dimension2 dimension3[7] = {0}; /* declare dimension3 as an array of 7
                                     elements of type dimension2 */
  dimension3[4][3][2] = 9999;
  dimension3[4][0][12] = 1;
  dimension3[3][8][18] = 42;

  printf("%d\n", foo(dimension3, 7));

  return 0;
}

int foo(dimension2 arr[], size_t siz) {
  int d1, d2, d3;
  int retval = 0;
  for (d3=0; d3<siz; d3++) {
    for (d2=0; d2<sizeof *arr / sizeof **arr; d2++) {
      for (d1=0; d1<sizeof **arr / sizeof ***arr; d1++) {
        retval += arr[d3][d2][d1];
      }
    }
    /* edit: previous answer used definite types for the sizeof argument */
    //for (d2=0; d2<sizeof (dimension2) / sizeof (dimension1); d2++) {
    //  for (d1=0; d1<sizeof (dimension1) / sizeof (int); d1++) {
    //    retval += arr[d3][d2][d1];
    //  }
    //}
  }
  return retval;
}

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Edit

I don’t like the use of definite types as the argument to sizeof.
I added the way to get the sizes of the (sub-)arrays without directly specifying their types, but rather let the compiler infer the right type from the object definitions.

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2nd Edit

As Per Eckman notes typedef-ing “bare” arrays can be dangerous. Note that in the code above, I’m not passing arrays themselves to the function foo. I am passing a pointer to a “lower level” array.

foo(), in the code above, accepts a pointer to an object of type dimension2. The dimension3 object is an array of elements of dimension2 type, not an object of dimension3 type (which isn’t even defined).

But remember Per Eckman’s note.