swap in doubly linked list - c

Question

I am trying to swap two nodes in a doubly linked list. Below is the part of program having swap function.
int swap (int x, int y)
{
struct node *temp = NULL ;
struct node *ptr1, *ptr2;
temp = (struct node *)malloc(sizeof(struct node));
if (head == NULL )
{
printf("Null Nodes");
}
else
{
ptr1 = ptr2 = head;
int count = 1;
while (count != x)
{
ptr1 = ptr1->next;
count++;
}
int count2 = 1;
while (count2 != y)
{
ptr2 = ptr2->next;
count2++;
}
ptr1->next->prev = ptr2;
ptr1->prev->next = ptr2;
ptr2->next->prev = ptr1;
ptr2->prev->next = ptr1;
temp->prev = ptr1->prev;
ptr1->prev = ptr2->prev;
ptr2->prev = temp->prev;
temp->next = ptr1->next;
ptr1->next = ptr2->next;
ptr2->next = temp->next;
}
return 0;
}
When I run this program, in case of 1st and 2nd node, it crashes. while in case of any other nodes, it gives infinite loop output. (eg:- 2->4 2->4 2->4....so on)`.
I know there are some more questions about node swappings, but I didn't find any one similar to my problem. Please help me out..!!
Thanks in Advance.

Answer 1

The code will fail if ptr1 == head (ptr1->prev == NULL) or ptr2 == head (ptr2->prev == NULL), because it ends up trying to use head->next, which doesn't exist. There also needs to be a check for the end of a list, if ptr1->next == NULL or ptr2->next == NULL, which can be handled using a local tail pointer. Using pointers to pointer to node can simplify the code. For example the pointer to next pointer to ptr1 could be &ptr1->prev->next or &head. The pointer to prev pointer to ptr2 could be &ptr2->next->prev or &tail (and set tail = ptr2).
Using pointers to pointer to node fixes the issue with swapping adjacent nodes. Also temp can be a pointer to node.
Example code using pointers to nodes (instead of counts) to swap:
typedef struct node NODE;
/* ... */
NODE * SwapNodes(NODE *head, NODE *ptr1, NODE *ptr2)
{
NODE **p1pn; /* & ptr1->prev->next */
NODE **p1np; /* & ptr1->next->prev */
NODE **p2pn; /* & b->prev->next */
NODE **p2np; /* & b->next->prev */
NODE *tail; /* only used when x->next == NULL */
NODE *temp; /* temp */
if(head == NULL || ptr1 == NULL || ptr2 == NULL || ptr1 == ptr2)
return head;
if(head == ptr1)
p1pn = &head;
else
p1pn = &ptr1->prev->next;
if(head == ptr2)
p2pn = &head;
else
p2pn = &ptr2->prev->next;
if(ptr1->next == NULL){
p1np = &tail;
tail = ptr1;
} else
p1np = &ptr1->next->prev;
if(ptr2->next == NULL){
p2np = &tail;
tail = ptr2;
}else
p2np = &ptr2->next->prev;
*p1pn = ptr2;
*p1np = ptr2;
*p2pn = ptr1;
*p2np = ptr1;
temp = ptr1->prev;
ptr1->prev = ptr2->prev;
ptr2->prev = temp;
temp = ptr1->next;
ptr1->next = ptr2->next;
ptr2->next = temp;
return head;
}

Answer 2

This can be compacted, but if you are having problems, it can help to spell it out in detail.
typedef struct node Node;
void link( Node* a, Node* b )
{
a->next = b;
b->prev = a;
}
void swap_nodes( Node* a, Node* b )
{
if(a==b) return; // don't swap with yourself
// handle adjacent nodes separately
if( a->next == b )
{
Node* bef = a->prev;
Node* aft = b->next;
link( bef, b); // link bef, b, a, aft
link( b, a );
link( a, aft );
}
else if( b->next == a )
{
Node* bef = b->prev;
Node* aft = a->next;
link( bef, a); // link bef, a, b, aft
link( a, b );
link( b, aft );
}
else
{
Node* a_prv = a->prev;
Node* a_nxt = a->next;
Node* b_prv = b->prev;
Node* b_nxt = b->next;
link( a_prv, b ); link( b, a_nxt ); // links b in a's old position
link( b_prv, a ); link( a, b_nxt ); // links a in b's old position
}
}
Also note that your head node should never be null, it should be a sentry node that links to itself if your list is empty. This means that there are never a first node, nor a last, nor is the list ever empty. This removes a ton of special cases. See here