Reputation: 348
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;
}
Upvotes: 3
Views: 3170
Reputation: 181845
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 inoptopt
, and returns'?'
. The calling program may prevent the error message by settingopterr
to 0.
Upvotes: 7