命令行参数不起作用。某种逻辑错误

Question

我在使命令行参数正常工作时遇到问题。当 argc 高于 3 时，参数似乎会失败，之后只有写 -fwhatever 而不是 -fwhatever< 才能让它们起作用。/em> 应该如此。 -t 还复制 -o 的值。

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

    /* Are we in root? */

    if(geteuid() !=0)
        {
            printf("Root access is required to run this program. \n");

            exit(0);        
        }   

    /* How do we use the program */

    if ((argc < 6) || (argc > 8))
        {
            usage(argv[0]);

            exit (0);
        } 

    while (1)
        {
            static struct option long_options[] =
                {
                    /* Options */
                {"send",       no_argument,       0, 's'},
                {"recieve",    no_argument,       0, 'r'},
                {"file",       required_argument, 0, 'f'},
                {"data",       required_argument, 0, 'd'},
                {"destip",     required_argument, 0, 'i'},
                {"destport",   required_argument, 0, 'p'},
                {"sourceip",   required_argument, 0, 'o'},
                {"sourceport", required_argument, 0, 't'},
                {0, 0, 0, 0}
                };

               int option_index = 0;

               c = getopt_long (argc, argv, "srf:d:i:p:o:t:",
                            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 's':
                   puts ("option -s\n");
                   break;

                 case 'r':
                   puts ("option -r\n");
                   break;

                 case 'f':
                   printf ("option -f with value `%s'\n", optarg);
                   break;

                 case 'd':
                   printf ("option -d with value `%s'\n", optarg);
                   break;

                 case 'i':
                   printf ("option -i with value `%s'\n", optarg);
                   break;

                 case 'p':
                   printf ("option -p with value `%s'\n", optarg);
                   break;

                 case 'o': 
                   printf ("option -o with value `%s'\n", optarg);

                 case 't': 
                   printf ("option -t with value `%s'\n", optarg);


                 case '?':
                   /* Error message printed */
                   break;

                 default:
                   abort ();
                 }
             }

               /* Print any remaining command line arguments (not options). */
            if (optind < argc)
        {
            printf ("non-option ARGV-elements: ");
            while (optind < argc)
            printf ("%s ", argv[optind++]);
            putchar ('\n');
        }

        getchar ();
        exit (0);

}

这里显然有一些可怕的错误，但它似乎找不到它。

Answer 1

关于复制到 -t 的 -o，您忘记在 case 末尾添加 break;。

另外，我会删除 argc 检查。让 getopt 发挥它的魔力。您可以在解析结束时检查是否设置了所有强制选项。您还可以检查未知参数。