On Wed, 2004-01-14 at 16:17, Tim Johnson wrote:
As far as I can see...
The split() function returns a list, not a scalar. When you tried to assign it to
a scalar, it tried to assign the result to @_ and then assign the number of items in
@_ to $type. Maybe I'm wrong, someone else will probably correct me i so. In any
case, I'm almost 100% sure that isn't what you want.
-----Original Message-----
From: Eric Walker [mailto:[EMAIL PROTECTED]
Sent: Wednesday, January 14, 2004 3:03 PM
To: perlgroup
Subject: error.
Does anyone know what this means...
code..
for ($i = 0;$i <= $size; $i+=$temp){
$type = split(::,shift (@hold));
}
Warning:
Use of implicit split to @_ is deprecated at .//test.pl line 21
help, thanks
perlknucklehead
ok thanks...
