alias(%foo, %bar) is better IMO since it conforms to other functions in
perl.
my %foo is alias = %bar; #seems a little out of scope of the language,
unless more functionality is implemented in that way.
Ilya
-Original Message-
From: Davíð Helgason
To: [EMAIL PROTECTED]; John Porter
Sent
David L. Nicol wrote:
>
>Are there really situations where
>
> $$reference = An Expression;
>
>is clearer than
>
> $reference = \(An Expression);
>
>?
Eric is confused. I don't know about in Perl 6-to-be, but in Perl 5
those two mean totally different things:
$foo = \$bar;
David L. Nicol wrote:
> Assignment to a nonexistent reference becomes an
> alias instead of a copy.
Uh, I dunno. Like Python/Ruby, but without the consistency.
I think special constructs -- defined as NOT doing assignment
-- should be allowed to set up aliases. This includes, e.g. for().
P
But how would you then copy, without having to bring the reference in
existance first. How would you copy period? Maybe I am not understanding,
hopefully someone can clear it up:)
Ilya
-Original Message-
From: David L. Nicol
To: Mark J. Reed
Cc: '[EMAIL PROTECTED] '
Sent: 07/20/2001 1:
