Question

I’ve seen quite a few example C implementations of linked lists on this site, and most of them place the next pointer at the end of each node like so…

struct intNode1 {
   int data;
   intNode1 *next;
};

Why do they implement them like that instead of like this?

struct node {
   struct node *next;
};

struct intNode2 {
   struct node node;
   int data;
};

The latter way of implementing linked lists allows your insertion and deletion code work on any kind of node as well as allowing you to create a generic list type while the former way forces you to implement each kind of list from scratch.

For example, here is an (incomplete) implementation of a singly linked list using both kinds of nodes:

struct intList {
   struct intNode1 *head;
};

struct list {
   struct node *head;
};

Now, obviously any operation on a generic list that needs to compare it’s nodes will need a function pointer to a comparison function, but that can often be hidden away in the implementation of a less generic interface for a list. For instance:

/* Returns zero if successful or nonzero otherwise */
int list-insertInt(struct list *list, int n) {
   struct intNode2 * newNode;
   if(!(newNode = malloc(sizeof *newNode)) {
      return -1;
   }
   newNode->data = n;
   return list-insertNode(list, (struct node *)newNode);
}

/* Assumes that the list contains only int nodes. */
int list-containsInt(struct list *list, int n) {
   struct intNode2 *current = (intNode2 *)list->head;
   while (current) {
      if(current->data == n) {
         return true;
      }
      current = current->next;
   }
   return false;
}

You can of course free a list without knowing what kinds of nodes it has:

void list-free(struct list *list) {
   struct node *current = list->head;
   struct node *next;
   while(current) {
      next = current->next;
      free(current);
      current = next;
   }
}

PS. It’s a bit late (i.e. it’s early in the morning but I haven’t slept yet) as I write this. so feel free to edit this question to be more clear.

Answer 1

Because textbooks on datastructures are mostly meant to teach concepts to beginners. That kind of ‘optimization’ just adds a lot of noise to the beginner’s ear. It is what you do with your knowledge after school, that separates you from the rest…