OptionParser returning bool instead of argument?

Question

When I run this sample from the OptionParser documentation:

require 'optparse'
options = {}
OptionParser.new do |opts|
    opts.banner = "Usage: example.rb [options]"
    opts.on("-v", "--[no-]verbose", "Run verbosely") do |v|
        options[:verbose] = v
    end
end.parse!
p options
p ARGV

and type: ruby test.rb -v 100, it returns:

{:verbose=>true}
["100"]

Shouldn't verbose be 100, not a boolean?

I have no idea about this, does anyone have any advice?

When I ran the minimal section from the link provided it produced `{:verbose=>true}` as well — ScottJShea, Mar 06 '12 at 03:53
`--[no]verbose` specifies that `verbose` is a boolean switch. — the Tin Man, Nov 28 '12 at 02:56

mu is too short · Accepted Answer · 2012-08-14T05:19:46.987

52

You've specified that the -v option does not have an argument:

opts.on("-v", ...

If you want it to take an argument then you have to say so:

opts.on("-v n", "--verbose=n", ...
#-----------^

And if you want to force n to be an integer, then:

opts.on('-v n', '--verbose=n', OptionParser::DecimalInteger, ...

You want to start reading at the make_switch docs (such as it is) and then reverse engineer the examples.

Don't feel bad about being confused, the OptionParser documentation isn't quite the best thing ever.

edited Aug 14 '12 at 05:19

answered Mar 06 '12 at 03:52

mu is too short

426,620
70
833
800

i got it.... i chane my program. and it works. Thanks a lot . i will go to read the make_switch docs. Thanks ! – CrazyLion Mar 06 '12 at 04:17

OptionParser returning bool instead of argument?

1 Answers1

Linked