#Note that even with single quotes, '\\' is 
#interpolated to '\'; this doesn't break the regex, 
#as both sides of =~ will see only '\'.

$regex = '^[24\\\\\[\]\{\}\^\$]{24}$';

$text = '2' x 24; # will also match, for example.
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That's the type of approach I took. I try to beef it up, then, by using (?<=...) to require that the text used in the regex so far MUST be there. Oh, and if were to find a regex, it can't use $ -- it must use \z, since $ can match before a newline at the end of a string.

japhy -- Perl and Regex Hacker

Replacing $ with \z is trival:

$regex = '^[z26\\\\\[\]\{\}\^\$]{26}\z';
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I've also worked out what the 'meta match' is for this expression (that is, $re2 below will only match $regex):

$re2 = '^\^\\[z26\\\\\\\\\\\\\[\\\\\]\\\\\{\\\\\}\\\\\^\\\\\$\\]\\{26\
+\}\\\\z\z';
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So, the idea is your want, in the regex, to have $re2 attempt to match $regex, then have $regex left around that basically is a large character class with a {n} modifier (Thus, if we need to add any new characters like '(', ')', or '?', we work with $regex to add them, modify $re2 above, and we're set).

However, I can't figure out how to do this part. I'm still thinking about it, but this is a rather interesting problem.

---

Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain

        123456789_123456789_123456789_123456789_123456789_123456
$re = q{^.*(??{$&ne'^.*(??{'.substr($re,7,-4).'})\z'&&'(?!)'})\z};
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Am I missing something?

  print 
   q|^.*(??{                                    })\z| =~
   m|^.*(??{$&ne'^.*(??{'.substr($&,7,36).'})\z'})\z|

  # outputs 1
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Update to japhy's update: In that case, why not simply:

  $re = q{^.*(??{$&ne$re&&'(?!)'})\z};
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   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

D'oh. Silly japhy.

japhy -- Perl and Regex Hacker

Close, but not quite yet:

First, here's a 5.005 version that's the same thing:

$regex = '^(.*)(?(?{my$x=substr$1,9,-4;$1ne"^(.*)(?(?{$x}))\\z"}))\z';
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And here's the stinger:

$test  = '^(.*)(?(?{my$japhy;my$x=substr$1,9,-4;$1ne"^(.*)(??{$x}))\\z
+"}))\z';
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(Mind you, I'm playing with this on 5.005, and things that I don't expect to match are matching - I wonder if there's a bug in the RE engine at this point. However, I'm pretty sure that adding any text prior to "my$x..." will still match...)

I think that you need to specify the exact length of the substr that you're looking at, which means to just simply do some character counting.

---

Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain

use Magic;
tilt() if ($str =~ m/$Magic::selfish/ 
    && ${str}abc !~ m/$Magic::selfish/);
tilt() if ($Magic::selfish =~ m/$str/ 
    && ${Magic::selfish}abc !~ m/$str/);
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die "42" if ${Magic::selfish}z !~ m/$Magic::selfish/;
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After Compline,
Zaxo

my $a = 'a';
print $a =~ $a; # prints 1
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No, because the regex 'a' also matches 'this is a line'. The idea was to only match itself, and no other string.

The 15 year old, freshman programmer,
Stephen Rawls