\%foo = \%bar is fine with me, it's the is alias I was a little worried
about.
Ilya
-Original Message-
From: David L. Nicol
To: Sterin, Ilya
Cc: 'Davíð Helgason '; '[EMAIL PROTECTED] '; 'John Porter '
Sent: 07/24/2001 5:45 PM
Subject: Re: aliasing a value [...]
Sterin, Ilya wrote
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
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
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().
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; # sets
