C - getopt incorrectly interprets next option as argument for previous option

Question

I'm using getopt the following way:

     while ((c = getopt(argc, argv, ":a:b")) != -1) {
        switch (c) {
        case 'a':
          printf("option: a argument: %s\n", optarg);
          break; 
        case 'b':
          printf("option: b\n");
          break;
        case ':':
          if(optopt == 'a') {
            printf("option: a argument: default\n");
          else {
            fprintf(stderr, "argument required!\n");
          }
    }

The idea is to have option a take an optional argument, and use a default if none is provided.

However, if I run my program like this: ./main -a -b , what ends up getting printed is:

option a: argument -b

What I want is:

option: a argument: default
option: b

How can I get getopt to realize what's following -a is in fact another option and not its argument?

Answer 1

getopt incorrectly interprets next option as argument for previous option

No, getopt correctly interprets the next argument as an argument, as it should.

How can I get getopt to realize what's following -a is in fact another option and not its argument?

You cannot - getopt does not support optional arguments. You may: accept the limitations, roll your own option parsing method or use a different utility for option parsing.

Answer 2

from the MAN page:

optstring is a string containing the legitimate option characters.   If
   such  a  character is followed by a colon, the option requires an argument, so getopt() places a pointer to the following text  in  the  same
   argv-element,  or  the  text  of the following argv-element, in optarg.
   Two colons mean an option takes an optional arg;

therefore, this: ":a:b" would be better written as: `"a::b" which allows '-a' to have an optional argument and 'b' to have no argument.

read the details for getopt and how to use getopt