Why is this regex greedy? (pbeckingham)

Why is this regex greedy?

pbeckingham

created: 2004-07-02 16:16:16

perlmonks

node
author

prlmnks

node
author

Given a string that looks suspiciously like a path name:

  '/a/b/c/d/e/abc00000'

yet is not a path name, I am trying to extract the abc000000 identifier from the end. My regex doesn't work, and while I know how to write a better one that works, I don't know specifically why this one does not:

  my $s = '/a/b/c/d/e/abc00000';
  my ($id) = $s =~ m{/(.+?\d+)$};
  print $id, "\n";

  __OUTPUT__
  a/b/c/d/e/abc00000

I can fix it easily by using:

  my ($id) = $s =~ m{/([^/]+\d+)$};

Which I understand. It's just that I don't understand why the first version is greedy.

Re: Why is this regex greedy?

ercparker

created: 2004-07-02 16:28:02

perlmonks

prlmnks

your regex is matching:
/ followed by one or more any character up to the first digit til end of line

updated: removed regex example sice you just want it explained

Re^2: Why is this regex greedy?

pbeckingham

created: 2004-07-02 16:38:46

perlmonks

prlmnks

Thanks, but I have no shortage of correctly functioning regexes - I want to know precisely why the one listed doesn't work. chromatic knows.

Re^3: Why is this regex greedy?

mcogan1966

created: 2004-07-06 09:21:28

perlmonks

prlmnks

It's simple, the .+ is capturing everything.

Re: Why is this regex greedy?

chromatic

created: 2004-07-02 16:32:19

perlmonks

prlmnks

The regex engine prefers leftmost, longest matches. Nothing in your regex prevents it from matching everything between the first slash and the end of line.

Re^2: Why is this regex greedy?

pbeckingham

created: 2004-07-02 16:34:29

perlmonks

prlmnks

So it is not treating that .+? the way I expected? Even though there are more-minimal matches?

I guess leftmost-longest trumps non-greedy.

Re^3: Why is this regex greedy?

duff

created: 2004-07-02 16:55:20

perlmonks

prlmnks

Right and right.

I don't understand why people expect non-greedy matching to actually mean "globally shortest match". Perhaps it's in the language we use. Just keep the left-to-rightness as the most prominent feature of your mental model of how perl's RE engine works and you shouldn't go wrong though.

duff

Re: Why is this regex greedy?

dpavlin

created: 2004-07-02 19:32:13

perlmonks

prlmnks

If you are running perl 5.6 or newer (and you are, right?) you might be able to insert use re 'debug'; which will give you detailed output from regex engine. That's a good way to debug your regular expressions.

2share!2flame...

Re: Why is this regex greedy?

kscaldef

created: 2004-07-02 20:15:40

perlmonks

prlmnks

Short answer, the regex engine works left to right.

Long answer, go read Friedl's articles in The Perl Journal or his book on Mastering Regular Expressions.

Re: Why is this regex greedy?

heroin_bob

created: 2004-07-02 21:51:48

perlmonks

prlmnks

Here's a good tutorial by chromatic that might be of help, if you haven't already checked it out.

~hb

Re: Why is this regex greedy?

japhy

created: 2004-07-02 22:50:11

perlmonks

prlmnks

I'd suggest you use m{.*/(.*)} to get whatever is after the last '/' (assuming there are no newlines in your data). Or perhaps a module like File::Basename

_____________________________________________________
Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;

Re: Why is this regex greedy?

Stevie-O

created: 2004-07-04 13:04:27

perlmonks

prlmnks

Using a regular expression along isn't necessarily what you want.

Remember: Perl is more than just regular expressions :)

$str = '/a/b/c/d/e/abc00000';
$end = (split '/', $str)[-1]; # pull the last element split() returns
print "we wanted '$end'";

--Stevie-O

$"=$,,$_=q>|\p4<6 8p
.q>.<4-KI;$,
.=pack'N*',"@{[unpack'C*',$_]
}"for split/