Getopt- Passing string parameter for argument

Question

I have a program which takes in multiple command line arguments so I am using getopt. One of my arguments takes in a string as a parameter. Is there anyway to obtain that st

Answer 1

Read man getopt http://linux.die.net/man/3/getopt

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; if there is text in the current argv-element (i.e., in the same word as the option name itself, for example, "-oarg"), then it is returned in optarg, otherwise optarg is set to zero.

A sample code:

#include 
#include 

int main (int argc, char *argv[])
{
  int opt;
  while ((opt = getopt (argc, argv, "i:o:")) != -1)
  {
    switch (opt)
    {
      case 'i':
                printf ("Input file: \"%s\"\n", optarg);
                break;
      case 'o':
                printf ("Output file: \"%s\"\n", optarg);
                break;
    }
  }
  return 0;
}

Here in the optstring is "i:o:" the colon ':' character after each character in the string tells that those options will require an argument. You can find argument as a string in the optarg global var. See manual for detail and more examples.

For more than one character option switches, see the long options getopt_long. Check the manual for examples.

EDIT in response to the single '-' long options:

From the man pages

getopt_long_only() is like getopt_long(), but '-' as well as "--" can indicate a long option. If an option that starts with '-' (not "--") doesn't match a long option, but does match a short option, it is parsed as a short option instead.

Check the manual and try it.