opterr declaration in getopt

Question

The following is example code from http://www.gnu.org. As surely most of you will see, it's getopt and I am having a question about the variable declarations. Why is there no type or anything written in front of

opterr = 0;

I have never seen that before.

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

int
main (int argc, char **argv)
{
  int aflag = 0;
  int bflag = 0;
  char *cvalue = NULL;
  int index;
  int c;

  opterr = 0;


  while ((c = getopt (argc, argv, "abc:")) != -1)
    switch (c)
      {
      case 'a':
        aflag = 1;
        break;
      case 'b':
        bflag = 1;
        break;
      case 'c':
        cvalue = optarg;
        break;
      case '?':
        if (optopt == 'c')
          fprintf (stderr, "Option -%c requires an argument.\n", optopt);
        else if (isprint (optopt))
          fprintf (stderr, "Unknown option `-%c'.\n", optopt);
        else
          fprintf (stderr,
                   "Unknown option character `\\x%x'.\n",
                   optopt);
        return 1;
      default:
        abort ();
      }
  printf ("aflag = %d, bflag = %d, cvalue = %s\n",
          aflag, bflag, cvalue);

  for (index = optind; index < argc; index++)
    printf ("Non-option argument %s\n", argv[index]);
  return 0;
}

Answer 1

opterr(3) is declared as an extern variable in unistd.h:

extern int optind, opterr, optopt;

So it's a global variable defined in a different translation unit, in this case your standard C library.

The reason for setting it to 0 is also explained in the manpage:

If getopt() does not recognize an option character, it prints an error message to stderr, stores the character in optopt, and returns '?'. The calling program may prevent the error message by setting opterr to 0.