Ok I got it working now, thanx much Jason. Only question I have is regarding
this.
>$head ||= $link;
What is this line doing ??
Mark
----- Original Message -----
From: "Jason Tiller" <[EMAIL PROTECTED]>
To: "perl" <[EMAIL PROTECTED]>
Sent: Monday, December 09, 2002 3:16 PM
Subject: Re: linked list's
> Hi, Mark, :)
>
> On Sun, 8 Dec 2002, Mark Goland wrote:
>
> > I am having problems printing this doubly list. the bug statment
> > prints the expacted so I think I have the list setup right, but I
> > cant get this list to print in reverse... any ideas ??
>
> > #!/usr/bin/perl -w
>
> # Always do this.
> use strict;
> use warnings;
>
> > $debug=1;
>
> > for($i=1;$i<10;$i++){
> > $list = {value=>$i,prev=>$list,next=>${prev} };
> > print "prev=$prev\tlist=$list\n" if( $debug);
> > $prev= $list;
> >
> > }
>
> Since $prev and $list are global variables and you always set $prev to
> list in the loop, how do you expect to go backwards ("reverse") with
> your second iterator? $prev == $list at the end of every iteration!
> The only way to go backwards would be to store the head of the list
> like so:
>
> my $head;
>
> for( my $i = 1;... ) {
> $list = ...;
> $head ||= $list;
> ...
> }
>
> That sets head equal to the first list member you create.
>
> However, you have a larger problem. You can't possibly go backwards
> because your first element has both ->{next} *and* ->{prev}
> undefined. This shouldn't be the case in a double-linked list.
>
> Also, your algorithm is incorrect. If you slightly change the debug
> output statement to look like this:
>
> printf "\$list->{prev} = %s \t \$list->{next} = %s\n",
> $list->{prev},
> $list->{next} if $debug;
>
> You'll see the problem. Here's a sample run:
>
> $ ./bob.pl
> Use of uninitialized value in printf at ./bob.pl line 12.
> Use of uninitialized value in printf at ./bob.pl line 12.
> $list->{prev} = $list->{next} =
> $list->{prev} = HASH(0xa04116c) $list->{next} = HASH(0xa04116c)
> $list->{prev} = HASH(0xa066678) $list->{next} = HASH(0xa066678)
> $list->{prev} = HASH(0xa066630) $list->{next} = HASH(0xa066630)
> $list->{prev} = HASH(0xa0665e8) $list->{next} = HASH(0xa0665e8)
> $list->{prev} = HASH(0xa0665a0) $list->{next} = HASH(0xa0665a0)
> $list->{prev} = HASH(0xa066558) $list->{next} = HASH(0xa066558)
> $list->{prev} = HASH(0xa066510) $list->{next} = HASH(0xa066510)
> $list->{prev} = HASH(0xa0664c8) $list->{next} = HASH(0xa0664c8)
>
> Umm, ->{prev} and ->{next} are the SAME FOR EVERY ELEMENT. So, you've
> in a sense created a single linked list with an additional, un-needed
> data member for each element.
>
> The problem is that when creating a linked list, you never know the
> address (er, reference in perl) of the *next* element. So, in your
> loop you have to "look back" and set the previous element's ->{next}
> "pointer" (again, "reference" in perl) to the current element. Your
> algorithm didn't do this; hence the single linked list you created.
>
> Try this on for size:
>
> my ($head, $tail);
>
> my $link;
> for( my $i = 1; $i < 10; ++$i ) {
> $link = { value => $i,
> next => undef,
> prev => $link };
> $head ||= $link;
> $link->{prev}->{next} = $link if $link->{prev};
> }
> $tail = $link;
>
> Let me know if you have any questions.
>
> ---Jason
>
>
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