http://www.perlmonks.org?node_id=11124042

Leudwinus has asked for the wisdom of the Perl Monks concerning the following question:

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

(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

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.