In the getopt_long
example here, why are the short options separated by a colon at abc:d:f:
and why are abc
are grouped together like that?
#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
/* Flag set by ‘--verbose’. */
static int verbose_flag;
int
main (int argc, char **argv)
{
int c;
while (1)
{
static struct option long_options[] =
{
/* These options set a flag. */
{"verbose", no_argument, &verbose_flag, 1},
{"brief", no_argument, &verbose_flag, 0},
/* These options don’t set a flag.
We distinguish them by their indices. */
{"add", no_argument, 0, 'a'},
{"append", no_argument, 0, 'b'},
{"delete", required_argument, 0, 'd'},
{"create", required_argument, 0, 'c'},
{"file", required_argument, 0, 'f'},
{0, 0, 0, 0}
};
/* getopt_long stores the option index here. */
int option_index = 0;
c = getopt_long (argc, argv, "abc:d:f:",
long_options, &option_index);
/* Detect the end of the options. */
if (c == -1)
break;
switch (c)
{
case 0:
/* If this option set a flag, do nothing else now. */
if (long_options[option_index].flag != 0)
break;
printf ("option %s", long_options[option_index].name);
if (optarg)
printf (" with arg %s", optarg);
printf ("\n");
break;
case 'a':
puts ("option -a\n");
break;
case 'b':
puts ("option -b\n");
break;
case 'c':
printf ("option -c with value `%s'\n", optarg);
break;
case 'd':
printf ("option -d with value `%s'\n", optarg);
break;
case 'f':
printf ("option -f with value `%s'\n", optarg);
break;
case '?':
/* getopt_long already printed an error message. */
break;
default:
abort ();
}
}
CodePudding user response:
On GNU's page, it is said that
An option character in this string can be followed by a colon (
:
) to indicate that it takes a required argument.
For example, xy:zq
accepts the options x
, y
, z
, q
; y
requires an additional argument.