Implies that you have I believe an anonymous array element being passed and
you are shifting off of the front of the anonymous array into $type.
Wags ;)
-----Original Message-----
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
Sent: Sunday, November 04, 2001 02:33
To: [EMAIL PROTECTED]
Subject: what is @$data
i have come a cross a array-scalar combo that i cant figure out, maybe
someone can help. the line off code is:
my ($type) = shift @$data;
does the $ in @$data mean to place $type after @data as opposed to before
in the new @data?
--
- josh
N8MSO
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