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";
}
[download]

Thank you very much JavaFan.