Re^2: how do $x = \12 differ from $$y = 12 ?

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

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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]

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Re^3: how do $x = \12 differ from $$y = 12 ?
by japhy (Canon) on May 10, 2005 at 13:15 UTC

$x = \@array;
$y = [];
[download]

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]