Perl Guru Forums: Perl Programming Help: Beginner: A simple regex question

CGI/Perl Guide | Learning Center | Forums | Advertise | Login

Site Search: in


	MAIN INDEX	SEARCH POSTS	WHO'S ONLINE	LOG IN

Home: Perl Programming Help: Beginner:
A simple regex question

lovinPerl
New User

Jan 18, 2013, 6:02 PM

Post #1 of 5 (2471 views)

A simple regex question	Can't Post

This question is going to seem super basic (because it is)

pretty much, I have these codes, and they can be of the following two forms, (where a represents any letter):

aa* or a*a

So now, I know how to do pattern matching and I could do something like (/\w{2}\*/) - this would match the first type. However, I'm confused how to get the second type. I can't do (/\w\*\w/) because this would match a*b, or b*c, or g*l, or anything like that. The issue is the first and second letter must be the same. So, b*b is all right, z*z is all right, etc.

Edit: Think I figured it out! There is something in perl called pattern memory. I have done this:

if(m/(\w)\1\*/){
print "matched case one with $1";
}
if(m/(\w)\*\1/){
print "matched case two with $1";
}

this works! Apparently putting something in parens stores it as a variable, first $1, then $2, etc.

for some reason when I try to combine the two if statements by using brackets to give the OR condition,

if(m/[(\w)\1*/|(\w)\*\1]/){
print "print $1";
}

this doesn't seem to work

(This post was edited by lovinPerl on Jan 18, 2013, 7:14 PM)

BillKSmith
Veteran

Jan 18, 2013, 8:22 PM

Post #2 of 5 (2463 views)

Re: [lovinPerl] A simple regex question [In reply to]	Can't Post

Yes, you did get it right. Note: You can combine the two forms using "non-capturing" parens.

Code

use strict; 
use warnings; 
my $string = 'xx*'; 
if ( $string =~ m/(\w)(?:\*\1|\1\*)/ ) { 
    print $1, "\n"; 
}

Good Luck,
Bill

Laurent_R
Veteran / Moderator

Jan 19, 2013, 1:08 AM

Post #3 of 5 (2450 views)

Re: [lovinPerl] A simple regex question [In reply to]	Can't Post

Bill gave you a solution, but just to point to your problem:

In Reply To

for some reason when I try to combine the two if statements by using brackets to give the OR condition,

if(m/[(\w)\1*/|(\w)\*\1]/){
print "print $1";
}

this doesn't seem to work

- The * marked in red above needs to be escaped
- the / marked in red above closes the pattern of your regex, so that the rest of your regex does not make sense.

You could have two regexes with a "or":

Code

if (/(\w)\1\*/  or  /(\w)\*\1/){

or, if you want to use an alternation, factoring the \w:

Code

if (m/(\w)((\1\*)|(\*\1))/) {

(You could modify the latest code above and add non capturing parens, but I pprefered not to do it here because it is not necessary and it would reduce overall legibility.)

lovinPerl
New User

Jan 19, 2013, 6:50 AM

Post #4 of 5 (2441 views)

Re: [lovinPerl] A simple regex question [In reply to]	Can't Post

Thank you guys for the help. :D I'm having lots of fun with regex in perl now

7stud
Enthusiast

Jan 20, 2013, 12:38 PM

Post #5 of 5 (2425 views)

Re: [lovinPerl] A simple regex question [In reply to]	Can't Post

One other thing...escaping things in regexes makes them very hard to read. One trick is to put the character inside a character class. Inside a character class, the special regex characters lose their special meaning. Also, using some whitespace to separate the parts of your regex (coupled with the x modifier) will make your regexes easier to read:

Code

if ( /(\w) \1 [*] | (\w) [*] \1/x )

(This post was edited by 7stud on Jan 20, 2013, 12:39 PM)

Search for (options)

Web Applications & Managed Hosting Powered by Gossamer Threads
Visit our Mailing List Archives