Akhthar Parvez K wrote:
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
my ( $name ) = () = Function1( $arg );
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