in "(options, args) = parser.parse_args" what value does args contains

Question

option_list = [
    make_option("-i", "--adapter", action="store",
        type="string", dest="dev_id"),
    make_option("-b", "--device", action="store",
        type="string", dest="address"),
            ]
parser = OptionParser(option_list=option_list)
(options, args) = parser.parse_args()

In the above code snippet, what value does args contain?

when I try to print len(args), I am getting 0 even though I am passing 2 arguments via command line
when I try to print just args it just prints [].

what actually is the use of that parameter and what does it contain?

Unless you are specifically wanting to support python <=2.6, suggest you forget optparse and learn argparse instead. See http://stackoverflow.com/questions/3217673/why-use-argparse-rather-than-optparse ;) — Antti Haapala -- Слава Україні, Apr 10 '14 at 11:00
" I am passing 2 arguments"... I don't think so. You are probably passing 2 *options*. — Bakuriu, Apr 10 '14 at 11:11

jonrsharpe · Answer 1 · 2014-04-10T13:04:22.093

1

Per the documentation:

parse_args() returns two values:

options, an object containing values for all of your options—e.g. if --file takes a single string argument, then options.file will be the filename supplied by the user, or None if the user did not supply that option

args, the list of positional arguments leftover after parsing options

In your case, len(args) == 0 because all of the arguments you pass are parsed into options.

edited Apr 10 '14 at 13:04

answered Apr 10 '14 at 10:58

jonrsharpe

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jonrsharpe, thanks for your response. when i print len(args),will it be '0' always? – Jyothi Apr 10 '14 at 11:52
@Jyothi not necessarily, it depends whether there are any arguments left over after parsing the options – jonrsharpe Apr 10 '14 at 12:07

in "(options, args) = parser.parse_args" what value does args contains

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