Append to end of doubly linked list

How would I change this function so that it appends new nodes to the end of the list instead of at the beginning?


struct node *append(struct node *list, int data)
{
struct node *new;
new = malloc(sizeof(struct node));
if(new == NULL) {
printf("malloc failed to allocate storage - exit\n");
exit(0);
}

new->value = data;
new->nextPTR = list;
/* headPTR is a global pointer of type struct node */
new->prevPTR = headPTR;
return new;
}

Append to the end of 'tailPtr'.

tailPtr->next = thisNode;
thisNode->prev = tailPtr;
tailPtr = thisNode;

Quzah.

I hope that's what you needed

struct node *append(struct node *list, int data)
{
struct node *new;
new =(struct node*) malloc(sizeof(struct node));
if(new == NULL)
{
printf("malloc failed to allocate storage - exit\n");
exit(0);
}

/*a loop to get the last node*/
for( ; list->nextPTR != NULL ; list = list->nextPTR)
/*you can also use a global pointer of type struct node not to find the last node every time
and it makes your code faster
last_node=new;*/

new->prevPTR=list;
new->value = data;
new->nextPTR = NULL;
return new;
}

Quote:

---

/*a loop to get the last node*/
for( ; list->nextPTR != NULL ; list = list->nextPTR)
/*you can also use a global pointer of type struct node not to find the last node every time
and it makes your code faster
last_node=new;*/

new->prevPTR=list;
new->value = data;
new->nextPTR = NULL;
return new;
}

---

If you already have a 'last node' pointer, then the fastest way to do it as I've shown. If the pointer exists, simply make it's next point to the new node, and the new node's prev point back to it. Then change it so 'last node' now points to the new one. There is no faster way.

Quzah.

yes,i know there is no faster way
and it's the basic way
but we have a 'list' parameter and we don't know what is shown by it.

kdt, maybe your code is just lacking one more statement, namely:

Code:

---


list->nextPTR = new;

---

puto amo, why do you think so?

After loop,

Code:

---


for( ; list->nextPTR != NULL ; list = list->nextPTR)

---

list->nextPTR == NULL

And that means that list is now pointing to last node.

After you add the new node (*new), you want the former last node (*list) to point to *new.

Quote:

---

yes,i know there is no faster way
and it's the basic way
but we have a 'list' parameter and we don't know what is shown by it.

---

But the original question was: How do I change this function to append to (the end of) a list.

As such, all you have to do is:

1) use a global tail and head pointer.
2) use this function:

Code:

---

void append( Node *n )
{
    if( n )
    {
        if( listTail )
        {
            listTail->next = n;
            n->prev = listTail;
        }
        listTail = n;
    }
}

---

After all, if we're changing the function, why bother with that prototype? Actually, you could also do:

Code:

---

void append( int data )
{
    Node *n = malloc( sizeof( Node ) );
    n->data = data;
    if( listTail )
    {
            listTail->next = n;
            n->prev = listTail;
    }
    listTail = n;
}

---

Quzah.

Why checking for listTail to be true?

Code:

---


if ( listTail )

---

maybe empty list?

> Why checking for listTail to be true?


Because if it is null, and I do this:

listTail->next = n;

Bad Things(TM) happen.

Quzah.

Ahaa. Sounds reasonable.