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Doubly linked list

Why do we refer to say
C = cur->prev->next  

instead of just saying

C = cur

for given 3 nodes A,B & C in a doubly circular sorted linked list.

If I am incorrect then please clearify what does one mean by
cur->prev->next  given that 3 nodes were given node A,B & C.
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Mallp
Asked:
Mallp
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1 Solution
 
tinchosCommented:
I would say that it's just the same.

If you post a code using that, maybe I can help you a little more.

Tincho
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MallpAuthor Commented:
Here is an example.

void SpecializedList::InsertEnd(int item)
{
      struct node *temp = get_node(item);
      temp->info = item;
      if (head==NULL)
      {       head = temp;                  // head is assigned the value of temp.
            head->prev = temp;            
            head->next = temp;
            temp->prev = head;
            temp->next = head;
            tail = temp;                  // head is also a tail if only one node exists.
      }
      else
      {      
            cur=head;
            while (cur->next != head)
                  cur=cur->next;
            temp->prev = head->prev;
            temp->next = head;
            head->prev->next = temp;
            head->prev = temp;
            tail=temp;
      }
}

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tinchosCommented:
Then Mallp

I would say that it's just the same, but I would say that it's just a matter of coding style

It may be clearer to the person who wrote it.

Tincho
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efnCommented:
Your question and your example are different.  In the form in the question, they are the same, assuming the list is properly linked.  But in the example, the code makes sense, and the substitution would not make sense.  I'll try to explain it.

You have a circular list, so at the start the tail is linked to the head:

tail -> head
tail <- head

The code is inserting the node addressed by temp between tail and head.  The first thing it does is set the pointers in temp:

          temp->prev = head->prev;
          temp->next = head;

So then we have:

tail -> head
tail <- head
tail <- temp
temp -> head

So now temp has the right pointers, but we need to change the forward pointer in tail and the backward pointer in head.  The next line changes the forward pointer in tail.

head->prev->next = temp;

So then we have

tail -> temp
tail <- head
tail <- temp
temp -> head

The old value of head->prev->next is indeed head, but we are not using the value here, we are changing it:  we are referring to head->prev->next in order to change it.  So it makes sense and definitely does not have the same effect as

head = temp;

At the risk of running the point into the ground, I'll give a non-pointer example.  If you had

a = 5;
b = a;

you could just as well code

b = 5;

but if you have

a = 5;
a = 10;

you can't just as well code

5 = 10;

--efn
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tinchosCommented:
This question has been classified as abandoned.  I will make a recommendation to the moderators on its resolution in approximately one week.  I would appreciate any comments by the experts that would help me in making a recommendation.

It is assumed that any participant not responding to this request is no longer interested in its final deposition.

If the asker does not know how to close the question, the options are here:
http://www.experts-exchange.com/help.jsp#hs5

Tinchos
EE Cleanup Volunteer
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efnCommented:
I think I answered it.

--efn
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tinchosCommented:
efn

I didn't post a recommendation for this as I'm directly involved in it and personally I would rather avoid judging the questions I participated in unless the recommendation is fairly clear to me.
I just posted the message just waiting for the asker to close his question.

Lets wait to see if he shows up and if that does not happen I'll post a recommendation for it.

Tincho
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efnCommented:
Tincho,

OK, no problem, thanks for your cleanup efforts.

--efn
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tinchosCommented:
No comment has been added lately, so it's time to clean up this TA.
I will leave the following recommendation for this question in the Cleanup topic area:

Accept: efn {http:#9705722}

Please leave any comments here within the next seven days.
PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER!

Tinchos
EE Cleanup Volunteer
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