On 6/26/06, tom arnall <[EMAIL PROTECTED]> wrote:
> > 1 #!/usr/bin/perl -w > > 2 > > 3 $_ = " [11] [22] a "; > > 4 > > 5 #with .*? > > 6 $re1 = qr/a|\[.*?\d\d\]/; > > 7 $re2 = qr/($re1\s)?$re1/; > > 8 ($f) = /($re2)/; > > 9 print "with .*? : $f\n"; > > 10 > > 11 #without .*? > > 12 $re1 = qr/a|\[\d\d\]/; > > 13 $re2 = qr/($re1\s)?$re1/; > > 14 ($f) = /($re2)/; > > 15 print "without .*? : $f\n"; > > > > gets this result: > > > > with .*? : [11] [22] a > > without .*? : [11]
but it gobbles to the 2nd ']' even if there is only one space between the ']' and '['.
When you have two spaces, the optional clause in $re2 has to continue until the space before the 'a' in order to make a match. It can't stop after '[11]'. When there's only one space between the brackets, the optional clause can succeed after "[11] ", allowing "[22]" to match: with .*? : [11] [22] without .*? : [11] [22] Perhaps it's confusing because sometimes you've got two matches of $re1 and sometimes just one? From your earlier code, when you had two spaces:
> > with .*? : [11] [22] a
First $re1 matched '[11] [22]', second matched 'a'.
> > without .*? : [11]
First $re1 matched empty string, second matched '[11]'. Is it getting any clearer? Cheers! --Tom Phoenix Stonehenge Perl Training -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] <http://learn.perl.org/> <http://learn.perl.org/first-response>
