How to find all the elements in a circular linked list

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

Hello , I wanted to know if there is a way in perl to iterate a circular linked list and capture all the elements in it .
so if I have a list like 2->4->5->4->6->2->7->8->2 , where after 8 it points back to the head 2 . Is thee a way in which I can iterate through this list and push the elements in an array ?
I tried something like this below , but this does not seem to work. Any advise on how this can be solved ?
Consider that the each node in the linked list is having the following data structure

$head ={
       data=>2
       next=>undef;      ## or the next node to which it is linked to
}
[download]



my @elements;

my $curr=$head;    ## I am trying to make my curr pointer start from t
+he head

while (1){
    last unless ($curr->{next} == $head);       # here I am trying to 
+check that the next element is not head. And if is then break out of 
+the loop
    print "data : $curr->{data} \n";
    push(@elements,$curr->{data});
    $curr = $curr->{next};        
}
[download]

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Re: How to find all the elements in a circular linked list
by Athanasius (Archbishop) on Feb 10, 2016 at 03:32 UTC

Hello punitpawar,

The problem is with this line:

last unless ($curr->{next} == $head);
[download]

First, the logic is back-to-front: you want to break out of the loop when $curr->{data} is equal to $head, so you need if, not unless.

Second, that test comes too soon. If you break out of the loop when the next node is $head, you will fail to push the last value onto @elements. So you need to move the test to follow the push statement: