getopt_long problem

Question

i have a problem with parsing arguments from a program that i'm writing, the code is below:

void parse_args(int argc, char** argv)
{
    char ch;
    int index = 0;

    struct option options[] = {       
        { "help", no_argument, NULL, 'h'   },      
        { "port", required_argument, NULL, 'p'  },      
        { "stop", no_argument, NULL, 's' },         
        { 0,    0,    0,    0   }       
    };

    while ((ch = getopt_long(argc, argv, "hp:s", options, &index)) != -1) {
        switch (ch) {
            case 'h':   
                printf("Option h, or --help.\n");
                break;
            case 's':
                printf("Option s, or --stop.\n");

                break;
            case 'p':
                printf("Option p, or --port.\n");
                if (optarg != NULL)
                    printf("the port is %s\n", optarg);
                break;
            case '?':
                printf("I don't understand this option!!!\n");

            case -1:  
                break;
            default:
                printf("Help will be printed very soon -:)\n");
        }
    }
}

When i run my program i got some strange output :

./Server -p 80
Option p, or --port.
the port is 80

./Server -po 80
Option p, or --port.
the port is o

./Server -por 80
Option p, or --port.
the port is or

./Server -hoho
Option h, or --help.
Server: invalid option -- o
I don't understand this option!!!

Answer 1

I think the confusion stems from a misunderstanding of long get opt. Essentially it will do partial string match only when you use the -- form. When you only use - it will fallback to standard parsing so -por 80 is matched as -p or 80 (as in, the option is -p and the argument is or). Try the same thing with --po and --por. As for help, try --he or--hel