On Jun 6, Timothy Johnson said: >"If $x = \\"bar", >then $$x returns a reference (to "bar") which is a scalar, but certainly >NOT a string." > >How is "bar" not a string?
You've misparsed my use of parentheses here. If $x = \\"bar", then $$x returns a reference which is a scalar, but certainly not a string. The '(to "bar")' part was just to inform you what the reference was to. $x is a reference to \"bar". \"bar" is a reference to "bar". -- Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/ RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/ ** Look for "Regular Expressions in Perl" published by Manning, in 2002 ** <stu> what does y/// stand for? <tenderpuss> why, yansliterate of course. [ I'm looking for programming work. If you like my work, let me know. ] -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
