Regex lazy behaviour

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

my $foo = "bar road";

if ($foo =~ /\b\w+\s*(road)?/) {
    print "Matched:", $1, "\n";
}
[download]

In the above code in $1 road is captured which is as expected.

Now if I change the regex to /\b\w+\s*?(road)?/, that is, make the space matching lazy then "road" is not captured.

Can someone please tell me as to why this happens as soon as I make the space matching lazy?

Comment on Regex lazy behaviour Download Code

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Re: Regex lazy behaviour by GrandFather (Saint) on Mar 03, 2009 at 10:35 UTC
You have made the road match optional. With a greedy \s* match road can be matched. If you go the minimal match route with \s? then no white space gets matched and, because the road match is optional, road doesn't get matched either. Actually a greedy \s is always ok if the next thing to be matched is \S because all the white space has to match in any case. True laziness is hard work	[reply]
Re: Regex lazy behaviour by moritz (Cardinal) on Mar 03, 2009 at 10:29 UTC
`\s?` matches zero spaces, then the regex engine tries to match `(road)?`. It succeeds with the "zero" option that the "zero-or-more" questionmark offers. No need to backtrack. If you want `road` to get captured anyway, you can make `\s` eager and `(road)?` still optional (and eager).	[reply] [d/l] [select]
Re: Regex lazy behaviour by ELISHEVA (Prior) on Mar 03, 2009 at 10:40 UTC
To clarify a bit: `\b matches before bar \w+ matches "bar" \s* matches as many spaces as it can swallow (it's greedy), i.e. " " (road)? matches "road" but \b matches before bar \w+ matches "bar" \s? matches the empty* string (it isn't greedy) so (road)? attempts to match " road" - which of course fails` [download]	[reply] [d/l]
Re: Regex lazy behaviour by abhy (Novice) on Mar 04, 2009 at 06:18 UTC
Thanks monks.	[reply]