C getopt - <integer>

Question

How to get the -10 parameter from command line arguments - "tail -10". The getopt function finds the character "1". But how do I access the string "10"?

If this can be done using getopt_long, an example will help. Thank.

Answer 1

If you do not intend to use the -1parameter 0as an argument, there is no answer to it. getoptonly for processing parameters that correspond to the standard syntax settings of POSIX utilities. Perhaps you can use GNU for this purpose getopt_long, or you can just write your own parser argv(this is easy).

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Answer 2

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Answer 3

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Answer 4

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int ch;
long num;
char *numarg;

while ((ch = getopt(argc, argv, "0123456789")) != -1) {
    switch (ch) {
    case '0': case '1': case '2': case '3': case '4':
    case '5': case '6': case '7': case '8': case '9':
        numarg = argv[optind - 1];
        if (numarg[0] == '-' && numarg[1] == ch && numarg[2] == '\0') {
            num = '0' - ch;
        } else if ((numarg = argv[optind]) != NULL && numarg[1] == ch) {
            char *ep;
            int numoptind = optind;

            num = strtol(numarg, &ep, 10);
            if (*ep != '\0') {
                fprintf(stderr, "illegal number -- %s\n", numarg);
                return EXIT_FAILURE;
            }

            /* Advance getopt internal state to next argument. */
            while (optind == numoptind) {
                ch = getopt(argc, argv, "0123456789");
                assert(ch >= '0' && ch <= '9');
            }
        } else {
            fprintf(stderr, "number after other options -- %s\n", numarg);
            return EXIT_FAILURE;
        }
        break;

    default:
        /* Unrecognized option character.  Error printed by getopt. */
        return EXIT_FAILURE;
    }
}

getopt getopt_long .