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From: worley@compass.com (Dale Worley)
Newsgroups: comp.lang.perl
Subject: Puzzle
Message-ID: <1991Apr1.224210.28384@uvaarpa.Virginia.EDU>
Date: 1 Apr 91 22:42:10 GMT
Sender: mmdf@uvaarpa.Virginia.EDU (Uvaarpa Mail System)
Reply-To: worley@compass.com
Organization: The Internet
Lines: 24

The following program:

    $f = '/usr';
    $a = ($f =~ m#/usr($|/)#);
    $b = ($f =~ m#${f}($|/)#);
    print "\$a = '$a'\n";
    print "\$b = '$b'\n";

produces the output:

    $a = '1'
    $b = ''

Why are the two apparently similar pattern matches doing different
things?  Is ${f} absorbing the following parenthesized pattern as a
subscript?  If so, how is it succeeding, since "$|/" is not a valid
expression?  Am I going crazy?

Dale

Dale Worley		Compass, Inc.			worley@compass.com
--
If you live to a ripe old age and get what you want, you probably
aren't a fool.	-- Walter M. Miller, Jr.