conditional expression testing

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Why doesn't below work? (I am following perl rexp tutorial from perldoc.perl.org)
Using perl version 5.8.4


use warnings;
use strict;

my @dic = qw#beriberi hanabbs#;

for (@dic) {
   print "trying to match $_\n";
   print $_ if $_ =~ m#^(hana)(bbs)?(?(1)\2\1|\1)$#;
}
[download]

yahoo@myserver ~/.yahoo/perl> !$
simple_grep.pl
trying to match beriberi
trying to match hanabbs
yahoo@myserver ~/.yahoo/perl>
[download]

Comment on conditional expression testing Select or Download Code

Replies are listed 'Best First'.
Re: conditional expression testing by ikegami (Patriarch) on Jun 01, 2010 at 16:26 UTC
Checks if the numbered capturing buffer has matched something. `beri beri ^, then '(hana)' fails. Fail.` [download] `hana bbs ^(hana) (bbs), then '\2' fails. Backtrack; ^(hana)\2, then '\1' fails. Fail.` [download] `(?(1)...)` doesn't make any sense since it's impossible to match the regex without the first capture matching something. Did you mean `(?(2)...)`? `'beriberi' =~ m#^(beri)(bbs)?(?(2)\2\1\|\1)$#;` [download] `beri beri ^(beri)(bbs)?\1 $ Match.` [download] `'beribbsbbsberi' =~ m#^(beri)(bbs)?(?(2)\2\1\|\1)$#;` [download] `beri bbs bbsberi ^(beri)(bbs)?\2 \1 $ Match.` [download] Update: Added working examples	[reply] [d/l] [select]
Re^2: conditional expression testing by Anonymous Monk on Jun 01, 2010 at 16:35 UTC
thank you guys. total misunderstanding on my part works perfectly now `use warnings; use strict; my @dic = qw#beriberi hanabbsbbshana#; for (@dic) { my $pat = $_; print "trying to match $pat\n"; print "matched $pat\n" if $_ =~ m#^(hana)(bbs)?(?(1)\2\1\|\1)$#; } my @pat = qw#AAAAG AC GC AZG ZC AAAG#; for ( @pat ) { my $pat2 = $_; print "matching $pat2\n"; print "matched $pat2\n" if $_ =~ m#[ATGC]+(?(?<=AA)G\|C)$#; }` [download]	[reply] [d/l]
Re^3: conditional expression testing by ikegami (Patriarch) on Jun 01, 2010 at 16:51 UTC
`m#^(hana)(bbs)?(?(1)\2\1\|\1)$#;` [download] is the same as `m#^(hana)(bbs)?\2\1$#;` [download] since `(?(1)...)` is always true. And since you don't use the captures, that can be be simplified to `m#^hana(?:bbsbbs)?hana$#;` [download]	[reply] [d/l] [select]
Re: conditional expression testing by choroba (Cardinal) on Jun 01, 2010 at 16:24 UTC
`hanabbsbbshana` matches your RE, though. Are you sure you wanted to use `(?(1)`? 1 is always true, so the no-pattern after \| is never considered, so your RE is equivalent to `/(hana)(bbs)?\2\1$/`	[reply] [d/l] [select]