On 19/03/10 13:19 +0200, Chris Knipe wrote:
my ($foo, $bar) = 1I am getting more and more occurances where when I use the later as above, $bar would not have a defined value... I'm not quite sure I understand why.
Does; my ($foo,$bar) = 1 x 2; do what you want? -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
