Since we've been doing linked lists in computer science
class recently but are unfortunately using the ol' Pascal
(which I've grown to dislike) I decided to code the same
simple program we finished in class - this time in Perl to see how it works.
After reading a little on that topic I made an implementation
like this :
use strict; my $list = undef; my $lastnode = \$list; while (1) { print '<i>nput, <o>utput : '; chomp (my $input = <STDIN>); if ($input eq 'i') { ($list, $lastnode) = add ($list, $lastnode); next; } if ($input eq 'o') { show ($list); next; } } sub add { my ($list, $lastnode) = @_; chomp (my $input = <STDIN>); my $newnode = [undef, $input]; $$lastnode = $newnode; $lastnode = \$newnode -> [0]; return ($list, $lastnode); } sub show { my $list = shift; for (my $node = $list; $node; $node = $node -> [0]) { print $node -> [1] . "\n"; } }
Interestingly I found out that this wasn't working.
Debugging it with Komodo 1.1 I found that after
$$lastnode = $newnode;
$list wasn't being set, which kinda defeated it's purpose.
Then I changed the implementation to this :
use strict; my $list = undef; my $lastnode = \$list; while (1) { print '<i>nput, <o>utput : '; chomp (my $input = <STDIN>); if ($input eq 'i') { &add; next; } if ($input eq 'o') { &show; next; } } sub add { chomp (my $input = <STDIN>); my $newnode = [undef, $input]; $$lastnode = $newnode; $lastnode = \$newnode -> [0]; } sub show { for (my $node = $list; $node; $node = $node -> [0]) { print $node -> [1] . "\n"; } }
I removed the variable passing and accessed $list and $lastnode directly.
And well, now it worked. But I don't get why passing the variable
makes example 1 don't work.

I am thankful for enlightenment.

In reply to Linked List Strangeness by Mandor

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