my @data = ('a', 'b', 'c');
my ($x, $y);
$x = \@data;
@$y = @data;
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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.
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Boris

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 = [];
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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

Well the two differ semantically in the first you are taking a reference to a scalar constant and in the second you are assigning a scalar constant to a dereference scalar ref. The first will be a readonly value. In these cases Devel::Peek is your friend:

$x:

SV = RV(0x8086d20) at 0x807b2e8
  REFCNT = 1
  FLAGS = (PADBUSY,PADMY,ROK)
  RV = 0x807b30c
  SV = IV(0x807ae44) at 0x807b30c
    REFCNT = 2
    FLAGS = (IOK,READONLY,pIOK)
    IV = 12


$y:

SV = RV(0x8086d24) at 0x807b2f4
  REFCNT = 1
  FLAGS = (PADBUSY,PADMY,ROK)
  RV = 0x804c500
  SV = IV(0x807ae4c) at 0x804c500
    REFCNT = 1
    FLAGS = (IOK,pIOK)
    IV = 12
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/J\

When $x and $y aren't initialized, there is no functional difference. The difference is only noticeable when they've been previously initialized.

my ( $x, $y );

$x = \12;
$$y = 13;

my $a = $x;
my $b = $y;

$x = \14;
$$y = 15;

print("a = $$a\n");  # 12
print("b = $$b\n");  # 15
print("x = $$x\n");  # 14
print("y = $$y\n");  # 15
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$x = \14 create a reference to the constant 14. It then stores the reference in $x, erasing $x's previous content.

$$y = 15 replaces the value of the variable at which $y points with 15. In my example, $b points to that same variable. When $y isn't initialized, a new variable is created through auto-vivification. In languages without auto-vivification (C, C++ and Java, for example), you get a NULL pointer error instead. In languages with auto-vivification, you just created a new variable, possibly by accident.

In C++ parlance, $x = \14 is similar to

x = new int;
*x = 14;
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while $$y = 15 is similar to

if (y == NULL) {
   y = new int;
}
*y = 15;
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* in this case, the location is created and assigned Just In Time.

#!/usr/bin/perl

use strict;

my $x = \12;
my $$y = 12;

$$y++;
$$x++;  # <<-line 9

#################
output:
Modification of a read-only value attempted at ./tst line 9.
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