Need help with sorting a linked list?

Quote:

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Originally Posted by mike_g

Ok, you have problems with you add function as well.

If this is what you have:

Code:

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NODE new_list_add(NODE *sent, NODE *high)
{
    NODE *new_node;
    new_node = (NODE *)malloc(sizeof(NODE));
    sent->prev->next = new_node;
    new_node->prev = sent->prev;
    new_node->next = sent;
    sent->prev = new_node;
    new_node->total = high->total;
}

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Firstly, you want to be returning a pointer to your new_node, not a NODE struct.

Next, you need to figure out whats linking to where.
"sent" would be the head of your sorted list.
"high" would be the node to add.
So, "sent->next" should point to new_node and "new_node->prev" should point to "sent"

since you have a doubly linked list "new_node->next" should be set to NULL to terminate the list.

Once you have created your new node, you should return the pointer to it as it will be the new head of your list.

I think you want something like:

Code:

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NODE* new_list_add(NODE *sent, NODE *high)
{
    NODE *new_node;
    new_node = (NODE *)malloc(sizeof(NODE));
    sent->next = new_node;
    new_node->prev = sent;
    new_node->next = NULL;
    new_node->total = high->total;
    return new_node;
}

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On a side note, in relation to delete_current, it should probably be declared void as it does not return anything. IE:

Code:

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void delete_current(NODE *current)

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I dont know why.. im still getting a seg fault in the same functions..
it all makes sense and it looks like it should work..

Quote:

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mike_g: You need to understand one thing about what he is coding. In this list, the last node points back to the first node, not to NULL. In other words it is a ring-list. You can confirm this by looking at his print_list function.

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Yeah, i only just realised that.

However, its probaly easiest to do this as a standard doubly linked list; then make it loop round once its working. the OP seems to have attempted a lot of stuff at once w/o testing it.

Quote:

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I dont know why.. im still getting a seg fault in the same functions

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Hmm... As a random guess maybe I was doing:

Code:

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    sent->next = new_node;
    new_node->prev = sent;
    new_node->next = NULL;

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when it should have been:

Code:

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    sent->prev = new_node;
    new_node->next = sent;
    new_node->prev = NULL;

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I think you need to focus on getting things working one bit at a time. First sort out your insertion, test it, then focus on your sort function.

Quote:

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Originally Posted by iMalc

Okay, I avoided posting my definition of Remove and Put, so that you could finish it on your own (hopefully). I'll describe what they need to do though.
Firstly though, You need to remove the node from the unsorted list before you add it to the sorted list. .

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I dont understand.. that frees the memory of the old node and then i have nothing to put into the new list...

Quote:

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Originally Posted by mike_g

Yeah, i only just realised that.

However, its probaly easiest to do this as a standard doubly linked list; then make it loop round once its working. the OP seems to have attempted a lot of stuff at once w/o testing it.


Hmm... As a random guess maybe I was doing:

Code:

---

    sent->next = new_node;
    new_node->prev = sent;
    new_node->next = NULL;

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when it should have been:

Code:

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    sent->prev = new_node;
    new_node->next = sent;
    new_node->prev = NULL;

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I think you need to focus on getting things working one bit at a time. First sort out your insertion, test it, then focus on your sort function.

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Ive been working on this for days nonstop.. I had no programming knowledge before the begining of september.. so i dont know how to 'focus' on different sections that work off each other... if i was to comment out the 'sort' function.. how would i know that the 'insertion' portion works... i need the sort function to test it. same with the other way around... im so lost! :( :( :(