From: Robin Stocker <robin@...>
Date: 2006-08-13T08:04:44+09:00
Subject: Re: ruby equiv of perl pos

gga wrote:
> This is simpler to write, but will be slower on longer strings.

You're right, it's much slower! I didn't think about speed while writing 
it, thanks for pointing it out.

Another question: Why do you use Regexp.last_match? Maybe to have a more 
general solution? The following seems to be simpler and faster:

def test_b
   result = ''
   pos = 0
   while pos = @string.index("\0", pos)
     result << '\%o' % pos
     pos += 1
   end
   return result
end