Elegant implementation of circular singly-linked list in C?

Question

Going through classic data structures and have stopped on linked lists.Just implemented a circular singly-linked list, but I'm under overwhelming impression that this list could be expressed in a more elegant manner, remove_node function in particular. Keeping in mind efficiency and code readability, could anybody present a more concise and efficient solution for singly-linked circular list?

#include <stdio.h>
#include <stdlib.h>


struct node{
    struct node* next;
    int value;
};


struct list{
    struct node* head;
};


struct node* init_node(int value){
    struct node* pnode;
    if (!(pnode = (struct node*)malloc(sizeof(struct node)))){
        return NULL;
    }
    else{
        pnode->value = value;   
    }
    return pnode;
}

struct list* init_list(){
    struct list* plist;
    if (!(plist = (struct list*)malloc(sizeof(struct list)))){
        return NULL;        
    }
    plist->head = NULL;
    return plist;
}


void remove_node(struct list*a plist, int value){

    struct node* current, *temp;
    current = plist->head;
    if (!(current)) return; 
    if ( current->value == value ){
        if (current==current->next){
            plist->head = NULL; 
            free(current);
        }
        else {
            temp = current;
            do {    
                current = current->next;    
            } while (current->next != plist->head);

            current->next = plist->head->next;
            plist->head = current->next;
            free(temp);
        }
    }
    else {
        do {
            if (current->next->value == value){
                temp = current->next;
                current->next = current->next->next;
                free(temp);
            }
            current = current->next;
        } while (current != plist->head);
    }
}

void print_node(struct node* pnode){
    printf("%d %p %p\n", pnode->value, pnode, pnode->next); 
}
void print_list(struct list* plist){

    struct node * current = plist->head;

    if (!(current)) return;
    if (current == plist->head->next){
        print_node(current);
    }
    else{
        do {
            print_node(current);
            current = current->next;

        } while (current != plist->head);
    }

}

void add_node(struct node* pnode,struct list* plist){

    struct node* current;
    struct node* temp;
    if (plist->head == NULL){
        plist->head = pnode;
        plist->head->next = pnode;
    }
    else {
        current = plist->head;
        if (current == plist->head->next){
            plist->head->next = pnode;
            pnode->next = plist->head;      
        }
        else {
            while(current->next!=plist->head)
                current = current->next;

            current->next = pnode;
            pnode->next = plist->head;
        }

    }
}

Answer 1

Take a look at the circular linked list in the Linux kernel source: http://lxr.linux.no/linux+v2.6.36/include/linux/list.h

Its beauty derives from the fact that you don't have a special struct for your data to fit in the list, you only have to include the struct list_head * in the struct you want to have as a list. The macros for accessing items in the list will handle the offset calculation to get from the struct list_head pointer to your data.

A more verbose explanation of the linked list used in the kernel can be found at kernelnewbies.org/FAQ/LinkedLists (Sorry, I dont have enough karma to post two hyperlinks).

Edit: Well, the list is a double-linked list and not a single-linked one like you have, but you could adopt the concept and create your own single-linked list.

Answer 2

List processing (particularly of circular lists) gets way easier when you treat the list head like an element of the list (a so-called "sentinel"). A lot of special cases just disappear. You can use a dummy node for the sentinel, but if the next pointer is first in the struct, you don't need to do even that. The other big trick is to keep a pointer to the next pointer of the previous node (so you can modify it later) whenever you modify the list. Putting it all together, you get this:

struct node* get_sentinel(struct list* plist)
{
    // use &plist->head itself as sentinel!
    // (works because struct node starts with the next pointer)
    return (struct node*) &plist->head;
}

struct list* init_list(){
    struct list* plist;
    if (!(plist = (struct list*)malloc(sizeof(struct list)))){
        return NULL;        
    }
    plist->head = get_sentinel(plist);
    return plist;
}

void add_node_at_front(struct node* pnode,struct list* plist){
    pnode->next = plist->head;
    plist->head = pnode;
}

void add_node_at_back(struct node* pnode,struct list* plist){
    struct node *current, *sentinel = get_sentinel(plist);

    // search for last element
    current = plist->head;
    while (current->next != sentinel)
        current = current->next;

    // insert node
    pnode->next = sentinel;
    current->next = pnode;
}

void remove_node(struct list* plist, int value){
    struct node **prevnext, *sentinel = get_sentinel(plist);
    prevnext = &plist->head; // ptr to next pointer of previous node
    while (*prevnext != sentinel) {
        struct node *current = *prevnext;
        if (current->value == value) {
            *prevnext = current->next; // remove current from list
            free(current); // and free it
            break; // we're done!
        }
        prevnext = &current->next;
    }
}

void print_list(struct list* plist){
    struct node *current, *sentinel = get_sentinel(plist);
    for (current = plist->head; current != sentinel; current = current->next)
        print_node(current);
}

Answer 3

A few comments:

• I think the remove function doesn't correctly adjust the circular list pointers when you delete the head node and the list is larger than 3 elements. Since the list is circular you have to point the last node in the list to the new head.
• You might be able to shorten the remove function slightly by creating a "find_node" function. Since the list is circular, however, there will still be the edge case of deleting the head node which will be more complex than in a non-circular list.
• Code "beauty" is in the eye of the beholder. As code goes yours is easy to read and understand which beats a lot of code in the wild.

Answer 4

I use the following to create a dynamic circular singly linked list. All it requires is the size.

Node* createCircularLList(int size)
{
    Node *it; // to iterate through the LList
    Node *head;

    // Create the head /1st Node of the list
    head = it = (Node*)malloc(sizeof(Node));
    head->id = 1;

    // Create the remaining Nodes (from 2 to size)
    int i;
    for (i = 2; i <= size; ++i) {
        it->next = (Node*)malloc(sizeof(Node)); // create next Node
        it = it->next;                          // point to it
        it->id = i;                             // assign its value / id
        if (i == 2)
            head->next = it; // head Node points to the 2nd Node
    }
    // close the llist by having the last Node point to the head Node
    it->next = head;

    return head;    // return pointer to start of the list
}

And i define Node ADT like so:

typedef struct Node {
    int id;
    struct Node *next;
} Node;