From: Austin Ziegler <austin@...>
Date: 2003-04-30T12:00:01+09:00
Subject: Ruby regexp backreferences

I'm doing something that required RLE, and the code that I
translated from Perl to do this included the following regexp:

  /^(.*?)((.)\2{2,127})(.*?)$/ois
    # Later code using $1 (.*?), $2 (...), $3 (.*?)

Now, you'd think that this simply translates as:

  /^(.*?)((.)\2{2,127})(.*?)$/m
    # Later code using $1 (.*?), $2 (...), $3 (.*?)

However, it doesn't work because of the way that Ruby builds the
regexp backreferences. The backreferences built are $1, $2, $3, and
$4 -- the (.) in the ((.)\2{2,127}) is treated as $3 -- whereas in
Perl, it's simply consumed and ignored. Obviously, I can't simply
replace (.) with (?:.), because I need the backreference within the
regexp itself. Thus, the translated version is:

  /^(.*?)((.)\3{2,127})(.*?)$/m

While what's happening makes sense, I'm wondering if it's correct --
how deep should backreferences be nested and considered part of the
process?

-austin
-- Austin Ziegler, austin@halostatue.ca on 2003.04.29 at 22:59:16