On 19 March 2010 11:45, <trapd...@trapd00r.se> wrote: > On 19/03/10 13:19 +0200, Chris Knipe wrote: >> >> my ($foo, $bar) = 1 >> >> I 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?
Probably not: $ perl foo.pl 11 $ cat foo.pl my ($foo,$bar) = 1 x 2; print "$foo\n"; print "$bar\n"; You've concatenated two strings containing '1' into a single scalar '11' and assigned it to $foo. You probably meant (1) x 2, and even that is not terribly readable or maintainable. Phil -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
