my @parts = split /\./ $address;
return (@parts == 4 
   and $parts[0] == 111 
   and $parts[1] == 255 
   and (($parts[2] > 0 and $parts[2] < 56 
    and $parts[3] >= 0 and $parts[3] < 56) 
     or ($parts[2] == 0 and $parts[3] == 0)));
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First of, I assume your IP address is actually valid and in the format you mention in your post.

Now for your problem: why make life hard? The code below is quite simple:

   my @bytes = split(/\./, $ip);
   if ($bytes[2] <= 55 && $bytes[3] <= 55) {
      # ok
   }
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You may -- or even should -- want to look at Net::IP, especially at the bincomp function.

Hope this helps, -gjb-

/\A   111 
  [.] 255 
  [.] ( \d | [1234]\d | 5[012345] )
  [.] ( \d | [1234]\d | 5[012345] )
\z/x;
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...should work ( with /x modifier used to enhance readability ), but if you're not limited to a regexp, a cleaner ( and more easily reusable!) solution would be something like...

return 1 if 
  /\A 111 [.] 255 [.] (\d+) [.] (\d+) \z/x
  && $1 <= 55
  && $2 <= 55;
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    --k.

Update: Fixed typo; thanks CS!

Errrm... Concerning your RegEx: What about '111.255.11.11'? Should be a valid IP, but is not recognized by your pattern. Probably just a typo, change [234] to [1234]. ;-)
Cheers, CombatSquirrel.

---

Entropy is the tendency of everything going to hell.

from the docs:

use NetAddr::IP;
  my $ip = new NetAddr::IP 'loopback';
  if ($ip->within(new     NetAddr::IP "127.0.0.0", "255.0.0.0")) {
      print "Is a loopback address\n";
  }
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use strict;
use Test::More 'no_plan';

my $rx = qr/\A
    (?: [0-4]\d? | 5?[0-5] ) \.
    (?: [0-4]\d? | 5?[0-5] )
\z/x;

my %test = (
    '56.0' => 0,
    '0.56' => 0,
   '48.56' => 0,
   '56.10' => 0,
    '55.0' => 1,
   '55.55' => 1,
     '0.0' => 1,
   '10.10' => 1,
);

my $result;

is(/$rx/, !!$result, $_) while ($_, $result) = each %test;

__END__
ok 1 - 0.56
ok 2 - 56.10
ok 3 - 55.0
ok 4 - 55.55
ok 5 - 48.56
ok 6 - 56.0
ok 7 - 10.10
ok 8 - 0.0
1..8
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If the first digit seen is any of 0 to 4, then it is legal, and may optionally be followed by any other digit. If it is 5, it is legal, and may optionally be followed by another digit from the 0-5 range. Any other sequence is illegal.

Update: I knew I was missing something when I wrote my test cases.. sigh. And it is so obvious. So there's a third case if the first digit is 6-9: it it is legal if not followed by anything.

my $rx = qr/\A
    (?: [0-4]\d? | 5?[0-5] | [6-9] ) \.
    (?: [0-4]\d? | 5?[0-5] | [6-9]  )
\z/x;
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    '62.6' => 0,
    '9.97' => 0,
    '55.9' => 1,
    '1.49' => 1,
     '6.7' => 1,
    '06.7' => 1,
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Makeshifts last the longest.

Your regex fails on for instance "7.8".

Abigail

Anyway, here's one the treats the digits as a number. I don't think this is a good solution in this case, but it does escape the lexical prison.

my $rx = qr/\A
    (\d+) (?(?{ $1>=0 && $1<=55 }) | (?!)) \.
    (\d+) (?(?{ $2>=0 && $2<=55 }) | (?!))
\z/x;
[download]