Hi, The following line stores the first return value by the function Function1 to the variable $name:
my ($name) = @_[0] = &Function1 ($arg); but this one doesn't work: my ($name) = $_[0] = &Function1 ($arg); Eventhough I've no issues to use the first one as long as it works well for me, but I'm getting the following warning: Scalar value @_[0] better written as $_[0] I hate warnings, but how can I fix that? As I said, $_[0] doesn't work. Can someone shed some light on this? Regards, Akhthar Parvez K http://Tips.SysAdminGUIDE.COM UNIX is basically a simple operating system, but you have to be a genius to understand the simplicity - Dennie Richie -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
