in reply to Regex question

You are being confused by list context vs. scalar context. Compare and contrast the version below.

my $test_line = 'Happy Birthday'; my ( $parse1, $parse2 ) = $test_line =~ /(Happy)(.*)/; print "parse1 = $parse1\nparse2 = $parse2\n";

The match in list context returns the matches. Excerpts from perlop:

m/PATTERN/msixpodualgc /PATTERN/msixpodualgc Searches a string for a pattern match, and in scalar co +ntext returns true if it succeeds, false if it fails. If no +string is specified via the "=~" or "!~" operator, the $_ stri +ng is searched. (The string specified with "=~" need not be +an lvalue--it may be the result of an expression evaluatio +n, but remember the "=~" binds rather tightly.) See also perl +re.
Matching in list context If the "/g" option is not used, "m//" in list context r +eturns a list consisting of the subexpressions matched by the parentheses in the pattern, that is, ($1, $2, $3...). +(Note that here $1 etc. are also set, and that this differs f +rom Perl 4's behavior.) When there are no parentheses in the pa +ttern, the return value is the list "(1)" for success. With o +r without parentheses, an empty list is returned upon fai +lure.

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Re^2: Regex question
by AnomalousMonk (Archbishop) on May 07, 2014 at 05:06 UTC

    Conversely, if the  /g modifier is used when matching in list context, regex behavior is changed yet again:

    c:\@Work\Perl>perl -wMstrict -le "my $line = 'Happy Birthday'; ;; my $p; ($p) = $line =~ /(Happy).*/; print qq{A: '$p'}; ;; ($p) = $line =~/Happy/g; print qq{B: '$p'}; " A: 'Happy' B: 'Happy'