Tom Arnall 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] > > The difference between the two sections of code is in Lines 6/12, i.e., > the > presence/absence of '.*?'. The regex in Line 6 seems to be using the > second ']' in the target string to satisfy its '\]'. But shouldn't the '?' > in '.*?' cause the search to terminate at the first ']' and yield the same > result as the expression in Line 12? > > also, why should removal of the 'a|' in $re1 make any difference in the > behavior of the expression? in fact, the removal causes the regex to > return '[11]', i.e., allows the '?' quantifier to work.
Hi Tom Your problem is that there are two spaces between [11] and [22] yet you have coded only a single space with ($re1\s). In the first case the first $re1 matches the string '[11] [22] ' as the .*? is forced to stretch to the single space after [22] to let the second $re1 match just the trailing 'a'. In the second case the first $re1 matches nothing, as it is optional and the earliest match for the entire regex is to leave it out and match '[11]' with the second $re1. Change lines 7 and 13 to: $re2 = qr/($re1\s+)?($re1)/; and both cases give the same result: with .*? : [11] [22] without .*? : [11] [22] You have given yourself a complicated task approaching the problem like this. What are you trying to do and what does the real data look like? It looks from this like a case for split(). HTH, Rob -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] <http://learn.perl.org/> <http://learn.perl.org/first-response>
