Thread: the best way to check if command line argumets are valid

int getopt(int argc, char * const argv[], const char *optstring);

It works on Linux and gcc on Windows.

Edit: I am pasting an example from the man page itself:

Code:

#include <stdio.h>     /* for printf */
#include <stdlib.h>    /* for exit */
#include <getopt.h>

int
main (int argc, char **argv) {
	int c;
	int digit_optind = 0;

	while (1) {
		int this_option_optind = optind ? optind : 1;
		int option_index = 0;
		static struct option long_options[] = {
			{"add", 1, 0, 0},
			{"append", 0, 0, 0},
			{"delete", 1, 0, 0},
			{"verbose", 0, 0, 0},
			{"create", 1, 0, 'c'},
			{"file", 1, 0, 0},
			{0, 0, 0, 0}
		};

		c = getopt_long (argc, argv, "abc:d:012",
				 long_options, &option_index);
		if (c == -1)
			break;

		switch (c) {
		case 0:
			printf ("option &#37;s", long_options[option_index].name);
			if (optarg)
				printf (" with arg %s", optarg);
			printf ("\n");
			break;

		case '0':
		case '1':
		case '2':
			if (digit_optind != 0 && digit_optind != this_option_optind)
				printf ("digits occur in two different argv-elements.\n");
			digit_optind = this_option_optind;
			printf ("option %c\n", c);
			break;

		case 'a':
			printf ("option a\n");
			break;

		case 'b':
			printf ("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 '?':
			break;

		default:
			printf ("?? getopt returned character code 0%o ??\n", c);
		}
	}

	if (optind < argc) {
		printf ("non-option ARGV-elements: ");
		while (optind < argc)
			printf ("%s ", argv[optind++]);
		printf ("\n");
	}

	exit (0);
}