Why is this regex greedy?

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

Given a string that looks suspiciously like a path name:

  '/a/b/c/d/e/abc00000'
[download]

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
[download]

I can fix it easily by using:

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

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

Comment on Why is this regex greedy? Select or Download Code

Replies are listed 'Best First'.
Re: Why is this regex greedy? by chromatic (Archbishop) on Jul 02, 2004 at 20:32 UTC
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.	[reply]
Re^2: Why is this regex greedy? by pbeckingham (Parson) on Jul 02, 2004 at 20:34 UTC
So it is not treating that `.+?` the way I expected? Even though there are more-minimal matches? I guess leftmost-longest trumps non-greedy.	[reply] [d/l]
Re^3: Why is this regex greedy? by duff (Parson) on Jul 02, 2004 at 20:55 UTC
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	[reply]
Re: Why is this regex greedy? by ercparker (Hermit) on Jul 02, 2004 at 20:28 UTC
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	[reply]
Re^2: Why is this regex greedy? by pbeckingham (Parson) on Jul 02, 2004 at 20:38 UTC
Thanks, but I have no shortage of correctly functioning regexes - I want to know precisely why the one listed doesn't work. chromatic knows.	[reply]
Re^3: Why is this regex greedy? by mcogan1966 (Monk) on Jul 06, 2004 at 13:21 UTC
It's simple, the `.+` is capturing everything.	[reply] [d/l]
Re: Why is this regex greedy? by dpavlin (Friar) on Jul 02, 2004 at 23:32 UTC
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...	[reply] [d/l]
Re: Why is this regex greedy? by kscaldef (Pilgrim) on Jul 03, 2004 at 00:15 UTC
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.	[reply]
Re: Why is this regex greedy? by heroin_bob (Sexton) on Jul 03, 2004 at 01:51 UTC
Here's a good tutorial by chromatic that might be of help, if you haven't already checked it out. ~hb	[reply]
Re: Why is this regex greedy? by japhy (Canon) on Jul 03, 2004 at 02:50 UTC
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:??;`	[reply] [d/l]
Re: Why is this regex greedy? by Stevie-O (Friar) on Jul 04, 2004 at 17:04 UTC
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'";` [download] --Stevie-O `$"=$,,$_=q>\|\p4<6 8p<M/_\|<('=> .q>.<4-KI<l\|2$<6%s!<qn#F<>;$, .=pack'N',"@{[unpack'C',$_] }"for split/</;$_=$,,y[A-Z a-z] {}cd;print lc` [download]	[reply] [d/l] [select]