Why? How is this differing from :

char[64] = "hello";

Thanks for any help in advance!

rt

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

int main(void)
{
    char array[64] = "hello"; /* in your example, you missed out the
variable name. I'm presuming that's a typo. */

    printf("array has [%s] in it right now.\n", array);

    array = "hi there";    /* WRONG */

    printf("array now has [%s] in it.\n", array);

    return EXIT_SUCCESS;

strcpy(array, "hi there");

strcpy prototype is in string.h.

That's my psychic research for the day finished. Now to get on with some
levitation.

--
Richard H

#include "sig.h"

  char carray[64];
  ...
  carray = "hello";

You can't assign to arrays, but you can assign to their constituent
elements. So

  int iarray[64];
  iarray = 3;     /* assignment to array, not allowed */

is illegal, but

  int iarray[64];
  iarray[0] = 3;  /* assignment to element, allowed */

is fine.

  char carray[64] = "hello";

the difference is that this is an initialisation, not an assignment.
As its name suggests, initialisation sets the _initial_ value for an
object.

  int i = 5; /* defines i, and initialises it to 5 */
  int j;     /* defines j, but does not initialise it */

  j = 3;     /* assigns 3 to j (not an initialisation) */

Other types of array can be initialised too, e.g.

  int iarray[64] = { 1, 2, 3, 4, 5 };

but, as with character arrays, assignment

  iarray = { 1, 2, 3, 4, 5 };

is illegal.

-- Mat.

   char a[64] = "hello";

and the identifier got lost somewhere on the way from your brain to my
newsreader. This is an initialization, and it is just a shorthand form
for writing

   char a[64] = { 'h', 'e', 'l', 'l', 'o' };

The rest of the array of 64 char will be initialized with '\0' characters.
It is _not_ an assignment. In C, assignments to arrays are not allowed.
You might want to read the section on "Arrays and Pointers" in the FAQ for
comp.lang.c

Kurt

--
| Kurt Watzka