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Try to explain myself.

foo|bar is foo or bar. if it is grouped by (foo|bar), the matched $1 will be set to "foo" or "bar".

In this case ... it is not "non capturing grouping" (?foo|bar), because it is zero width look ahead assertion '(?='. Zero width look ahead assertion works like place holder and it does not eat up pos($expr) in matching.

$ is the end of line... as far as I know.

Well, it says look ahead for "end of line" or MARK and match against them as 'place holder'. I think I understand this!

#!/usr/bin/perl use strict; use warnings; my $RefLine = "(a) This is first line(once all 4 was one line). (b) Th +is is second line; ( print "original -----\n"; print "$RefLine\n"; print "original -----\n\n"; print "\n## without 'end of line or' condtion. last line fails\n"; while( $RefLine =~ /(\([a-z]\).*?)(?=\([a-z]\))/g ){ my $p=pos $RefLine; print "$-[0], $p,matched=$&\n"; print "---\n"; } print "\n## without lookahead assertion... \n"; while( $RefLine =~ /(\([a-z]\).*?)($|\([a-z]\))/g ){ my $p=pos $RefLine; print "$-[0], $p,matched=$&\n"; print "---\n"; } print "\n## with 'end of line or' condtion and zero width place holder +\n"; while( $RefLine =~ /(\([a-z]\).*?)(?=$|\([a-z]\))/g ){ my $p=pos $RefLine; print "$-[0], $p,matched=$&\n"; print "---\n"; }

Thank you very much JavaFan.


In reply to Re^4: RegEx related line split by remiah
in thread RegEx related line split by dominic01

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