How to swap two nodes in a doubly linked list

This blog post gives example C code how to swap two elements (nodes, items) of a doubly linked list. The code works for both circular and non-circular lists, even if the two arguments are the same, or if they are adjacent in the list. (It is surprisingly complicated to give a correct and elegant solution.)

#include <stdlib.h>  /* NULL */

typedef int node_data_t;  /* can be any other type as well */

struct node {
  struct node* prev;
  struct node* next;
  node_data_t data;
};

/** If node1 and node2 are non-NULL members of a doubly-linked list
 * (circular or not), swap their data fields.
 */
void swap_data(struct node* node1, struct node* node2) {
  node_data_t temp_data = node1->data;
  node1->data = node2->data;
  node2->data = temp_data;
}

/** If node1 and node2 are non-NULL members of a doubly-linked list
 * (circular or not), swap their data fields. If node1 or node2 is the head
 * of the non-circular list, return the new head.
 */
struct node* swap(struct node* node1, struct node* node2) {
  struct node* temp;
  temp = node1->next;
  node1->next = node2->next;
  node2->next = temp;
  if (node1->next != NULL)
    node1->next->prev = node1;
  if (node2->next != NULL)
    node2->next->prev = node2;
  temp = node1->prev;
  node1->prev = node2->prev;
  node2->prev = temp;
  if (node1->prev != NULL)
    node1->prev->next = node1;
  if (node2->prev == NULL)
    return node2;
  node2->prev->next = node2;
  return node1;
}

The code above is based on the discussion at http://bytes.com/topic/c/answers/219236-double-linked-list-elements-swap

swap_data is simpler, and it's also faster if the data is small (i.e. at most 2 pointers). However, swap_data cannot be used when there are external pointers inside the data.