Re: how do $x = \12 differ from $$y = 12 ?

There is a world of difference. Forget that you're using a constant (12) for now. Imagine this:

my @data = ('a', 'b', 'c');
my ($x, $y);
$x = \@data;
@$y = @data;
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$x refers to @data (that is, it holds a reference to @data). But $y doesn't; rather, the array referenced in $y holds a shallow copy of the elements in @data.

Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

Comment on Re: how do $x = \12 differ from $$y = 12 ? Download Code

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Re^2: how do $x = \12 differ from $$y = 12 ? by borisz (Canon) on May 10, 2005 at 13:12 UTC
Thanks, that takes me closer to the answer. I wonder why `$y` is undefined now. `use Scalar::Util qw/weaken/; my ( $x, $y ); $x = \12; $$y = 12; weaken($x); weaken($y); print $x; print $y; __OUTPUT__ SCALAR(0x1807f24) Use of uninitialized value in print at 5432.pl line 11.` [download] Boris	[reply] [d/l]
Re^3: how do $x = \12 differ from $$y = 12 ? by japhy (Canon) on May 10, 2005 at 13:15 UTC
Someone else showed you Devel::Peek code that explains that. $x has a refcount of 2, and $y only has a refcount of 1. It's the same as doing: `$x = \@array; $y = [];` [download] There, $x has a refcount of 2 (from $x and @array), whereas $y has a refcount of only 1. Jeff `japhy` Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and `perl` hacker How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart	[reply] [d/l]