Fellow Monks,

I am still trying to wrap my head around pointers/references in Perl. I came up with the following program to help me better understand but was hoping you could please help me with some questions.

use warnings;
use strict;
use v5.10;

my $variable = 22;
my $pointer = \$variable;

say "The address of \$varible, which contains the value $variable,";
say "is $pointer";

$$pointer = 25;

say "Look at that!  \$variable now equals $variable";

sub sum_and_diff {
    my $a = shift @_;
    my $b = shift @_;
    my $res = \(shift @_); # why does the "\" work here?
    my $sum = $a + $b;
    $$res = $a - $b;
    return $sum;
}

my $b = 2;
my $diff;  # this is line 27
my $pointer_to_diff = \$diff;

say "the sum of 5 and $b is ", &sum_and_diff(5, $b, $pointer_to_diff);
say "and the difference is ", $pointer_to_diff;

say "the sum of 9 and $b is ", &sum_and_diff(9, $b, \$diff);
say "and the difference is ", $diff;  # this is line 34
[download]

(1) Does the backslash ("\") in the my $res line mean that $res contains the address of the third argument passed to the function? I think so but just wanted to confirm.

(2) Does the "double dollar sign" ("$$") two lines later mean to put the value of the difference of $a and $b in the memory location that is $res?

(3) Why do the lines using $pointer_to_diff work but the last two lines using \$diff and $diff not work? I thought that these lines were essentially equivalent and that $diff was defined in line 27. Instead, I get the following output:

The address of $varible, which contains the value 22,
is SCALAR(0x801e64540)
Look at that!  $variable now equals 25
the sum of 5 and 2 is 7
and the difference is 3
the sum of 9 and 2 is 11
Use of uninitialized value $diff in say at line 34.
and the difference is
[download]

Gratias tibi ago
Leudwinus

Edited to add: I thought using ${\$diff} in the last line would work but that too gave me the same error.

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