22.1.4.2 Passing array arguments

The function call passes this pointer by value, like all argument values in C. However, the result is paradoxical in that the array itself is passed by reference: its contents are treated as shared memory—shared between the caller and the called function, that is. When clobber4 assigns to element 4 of array, the effect is to alter element 4 of the array specified in the call.

#include <stddef.h>  /* Defines NULL. */
#include <stdlib.h>  /* Declares malloc, */
                     /* Defines EXIT_SUCCESS. */

int
main (void)
{
  int data[] = {1, 2, 3, 4, 5, 6};
  int i;

  /* Show the initial value of element 4. */
  for (i = 0; i < 6; i++)
    printf ("data[%d] = %d\n", i, data[i]);

  printf ("\n");

  clobber4 (data);

  /* Show that element 4 has been changed. */
  for (i = 0; i < 6; i++)
    printf ("data[%d] = %d\n", i, data[i]);

  printf ("\n");

  return EXIT_SUCCESS;
}

shows that data[4] has become zero after the call to clobber4.

The array data has 6 elements, but passing it to a function whose argument type is written as int [20] is not an error, because that really stands for int *. The pointer that is the real argument carries no indication of the length of the array it points into. It is not required to point to the beginning of the array, either. For instance,

clobber4 (data+1);

passes an “array” that starts at element 1 of data, and the effect is to zero data[5] instead of data[4].

If all calls to the function will provide an array of a particular size, you can specify the size of the array to be static:

void
clobber4 (int array[static 20])
…

This is a promise to the compiler that the function will always be called with an array of 20 elements, so that the compiler can optimize code accordingly. If the code breaks this promise and calls the function with, for example, a shorter array, unpredictable things may happen.