From: eden li <eden.li@...>
Date: 2007-03-09T10:24:15+09:00
Subject: Re: generate UUID

He probably meant to_s(16), not to_a(16).  Anyway, rand_hex_2 won't
zero-pad, so you might want to replace the to_s with a format
operator:

def rand_hex_3(l)
  "%0#{n}x" % rand(1 << n*4)
end

def rand_uuid
  [8,4,4,4,12].map {|n| rand_hex_3(n)}.join('-')
end

puts rand_uuid

On Mar 9, 6:17 am, Aaron Smith <beingthexempl...@gmail.com> wrote:
> Robert Klemme wrote:
> > On 08.03.2007 22:44, Aaron Smith wrote:
> >>     hexCode = ""
> >>     1.upto(l) do |i|
> >>       hexCode << validChars[rand(length-1)]
> >>     end
> >>     hexCode
> >> end
>
> >> r =
> >> "#{rand_hex(8)}-#{rand_hex(4)}-#{rand_hex(4)}-#{rand_hex(4)}-#{rand_hex(12)}"
> >> puts r
>
> > Make validChars and length constants declared outside the method.  And
> > also, use rand(length) otherwise you never get a "9".  You could also
> > use ?A..?F and ?0..?9 - no idea whether that makes a performance diff.
>
> > But I guess this one is even faster:
>
> > def rand_hex_2(l)
> >    rand(1 << (l*4)).to_a(16)
> > end
>
> > Kind regards
>
> >   robert
>
> def rand_hex_2(l)
>    rand (1 << (l*4) ).to_a(16)
> end
> puts rand_hex_2(9)
>
> ArgumentError: wrong number of arguments (1 for 0)
>
> --
> Posted viahttp://www.ruby-forum.com/.