How to implement a linked list in C?

Question

I am creating a linked list as in the previous question I asked. I have found that the best way to develop the linked list is to have the head and tail in another structure. My products struct will be nested inside this structure. And I should be passing the list to the function for adding and deleting. I find this concept confusing.

I have implemented the initialize, add, and clean_up. However, I am not sure that I have done that correctly.

When I add a product to the list I declare some memory using calloc. But I am thinking shouldn't I be declaring the memory for the product instead. I am really confused about this adding.

Many thanks for any suggestions,

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

#define PRODUCT_NAME_LEN 128

typedef struct product_data 
{
    int product_code;
    char product_name[PRODUCT_NAME_LEN];
    int product_cost;
    struct product_data_t *next;
}product_data_t;

typedef struct list 
{
    product_data_t *head;
    product_data_t *tail;
}list_t;

void add(list_t *list, int code, char name[], int cost); 
void initialize(list_t *list);
void clean_up(list_t *list);

int main(void)
{
    list_t *list = NULL;

    initialize(list);
    add(list, 10, "Dell Inspiron", 1500);
    clean_up(list);

    getchar();

    return 0;
}

void add(list_t *list, int code, char name[], int cost)
{
    // Allocate memory for the new product
    list = calloc(1, sizeof(list_t));
    if(!list)
    {
        fprintf(stderr, "Cannot allocated memory");
        exit(1);
    }

    if(list)
    {
        // First item to add to the list
        list->head->product_code = code;
        list->head->product_cost = cost;
        strncpy(list->head->product_name, name, sizeof(list->head->product_name));
        // Terminate the string
        list->head->product_name[127] = '/0';
    } 
}

// Initialize linked list
void initialize(list_t *list)
{
    // Set list node to null
    list = NULL;
    list = NULL;
}

// Release all resources
void clean_up(list_t *list)
{    
    list_t *temp = NULL;

    while(list)
    {
        temp = list->head;
        list->head = list->head->next;
        free(temp);    
    }
    list = NULL;
    list = NULL;
    temp = NULL;
}

============================== Edited ============================

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

#define PRODUCT_NAME_LEN 64

// typedef struct product_data product_data_t;
typedef struct product_data 
{
    int product_code;
    char product_name[PRODUCT_NAME_LEN];
    int product_cost;
}product_data_t;

typedef struct list
{
    struct list *head;
    struct list *tail;
    struct list *next;
    struct list *current_node;
    product_data_t *data;

}list_t;

void add(list_t *list, int code, char name[], int cost); 

int main(void)
{
    list_t *list = NULL;
    list = initialize(list);
    add(list, 1001, "Dell Inspiron 2.66", 1299);

    add(list, 1002, "Macbook Pro 2.66", 1499);

    clean_up(list);

    getchar();

    return 0;
}

void add(list_t *list, int code, char name[], int cost)
{
    /* Allocate memory for the new product */
    product_data_t *product = (product_data_t*) calloc(1, sizeof(*product));
    if(!product)
    {
        fprintf(stderr, "Cannot allocate memory.");
        exit(1);
    }

    /* This is the first item in the list */
    product->product_code = code;
    product->product_cost = cost;
    strncpy(product->product_name, name, sizeof(product->product_name));
    product->product_name[PRODUCT_NAME_LEN - 1] = '\0';

    if(!list->head)
    {
        /* Assign the address of the product. */
        list = (list_t*) product;   
        /* Set the head and tail to this product */
        list->head = (list_t*) product;
        list->tail = (list_t*) product;
    }
    else
    {
        /* Append to the tail of the list. */
        list->tail->next = (list_t*) product;
        list->tail = (list_t*) product;
    }

    /* Assign the address of the product to the data on the list. */
    list->data = (list_t*) product;
}

Answer 1

If you are looking to better understand the basics of linked lists, take a look at the following document:

http://cslibrary.stanford.edu/103/LinkedListBasics.pdf

Answer 2

Arguably you want your list data structure to be external to the data that it stores.

Say you have:

struct Whatever
{
   int x_;
}

Then your list structure would look like this:

struct Whatever_Node
{
   Whatever_Node* next_
   Whatever* data_
}

Ryan Oberoi commented similarly, but w/o example.

Answer 3

In your case the head and tail could simply point to the beginning and end of a linked-list. With a singly linked-list, only the head is really needed. At it's most basic, a linked-list can be made by using just a struct like:

typedef struct listnode
{
   //some data
   struct listnode *next;
}listnodeT;

listnodeT *list;
listnodeT *current_node;
list = (listnodeT*)malloc(sizeof(listnodeT));
current_node = list;

and as long as list is always pointing to the beginning of the list and the last item has next set to NULL, you're fine and can use current_node to traverse the list. But sometimes to make traversing the list easier and to store any other data about the list, a head and tail token are used, and wrapped into their own structure, like you have done. So then your add and initialize functions would be something like (minus error detection)

    // Initialize linked list
void initialize(list_t *list)
{
    list->head = NULL;
    list->tail = NULL;
}

void add(list_t *list, int code, char name[], int cost)
{
    // set up the new node
    product_data_t *node = (product_data_t*)malloc(sizeof(product_data_t));
    node->code = code;
    node->cost = cost;
    strncpy(node->product_name, name, sizeof(node->product_name));
    node->next = NULL;

    if(list->head == NULL){ // if this is the first node, gotta point head to it
        list->head = node;
        list->tail = node;  // for the first node, head and tail point to the same node
    }else{
        tail->next = node;  // append the node
        tail = node;        // point the tail at the end
    }
}

In this case, since it's a singly linked-list, the tail is only really useful for appending items to the list. To insert an item, you'll have to traverse the list starting at the head. Where the tail really comes in handy is with a doubly-linked list, it allows you to traverse the list starting at either end. You can traverse this list like

// return a pointer to element with product code
product_data_t*  seek(list_t *list, int code){
   product_data_t* iter = list->head;
   while(iter != NULL)
       if(iter->code == code)
           return iter;
       iter = iter->next;
   }
   return NULL; // element with code doesn't exist
}

Often times, the head and tail are fully constructed nodes themselves used as a sentinel to denote the beginning and end of a list. They don't store data themselves (well rather, their data represent a sentinel token), they are just place holders for the front and back. This can make it easier to code some algorithms dealing with linked lists at the expense of having to have two extra elements. Overall, linked lists are flexible data structures with several ways to implement them.

oh yeah, and nik is right, playing with linked-lists are a great way to get good with pointers and indirection. And they are also a great way to practice recursion too! After you have gotten good with linked-list, try building a tree next and use recursion to walk the tree.

Answer 4

I am not writing the code here but you need to do the following:

• Create and object of list, this will remain global for the length of program.
• Malloc the size of product _ data _ t.
• For first element (head is NULL), point head to the malloced' address.
• To add next element, move to the end of list and then add the pointer of malloced address to next of last element. (The next of the last element will always be NULL, so thats how you traverse to end.)
• Forget tail for a while.

Answer 5

If you are learning C pointer theory this is a good exercise. Otherwise, it feels like too much indirection for code that is not generic (as in a library).

Instead of allocating a static 128 byte character string, you might want to do some more pointer practice and use an allocated exact length string that you clean up at exit.

Academically, kungfucraigs' structure looks more generic then the one you have defined.

Answer 6

I think u first need to Imagin back-end. Code are nothing to important. Go here and visualize back-end basic c code of all insertion. 1) Insertion at beginning Visit and scroll to get every instruction execution on back- end And u need front and imagin Go here Back end imagin

And All other possible insertion here.

And important thing u can use this way.

struct Node{
int data;//data field
struct Node*next;//pointer field
};

struct Node*head,*tail; // try this way

or short cut

struct Node{
int data;//data field
struct Node*next;//pointer field
}*head,*tail; //global root pointer

And

<< Click >> To visualize other linked list problem.

Thanks.

Answer 7

A demo for Singly Linked List. If you prefer, try to check Circular Linked List and Doubly Linked List.

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


typedef struct node {
    int val;
    struct node * next;
} node_t;


// Iterating over a list
void
print_list(node_t *head)
{
    node_t *current = head;

    while(current != NULL)
    {
        printf("%d\n", current->val);
        current = current->next;
    }
}


// Adding an item to the end of the list
void
push_end(node_t *head, int val)
{
    node_t *current = head;
    while (current->next != NULL)
    {
        current = current->next;
    }

    current->next = malloc(sizeof(node_t));
    current->next->val = val;
    current->next->next = NULL;
}

// Adding an item to the head of the list
void
push_head(node_t **head, int val)
{
    node_t *new_node = NULL;

    new_node = malloc(sizeof(node_t));
    new_node->val = val;
    new_node->next = *head;

    *head = new_node;
}

// Removing the head item of the list
int
pop_head(node_t **head)
{
    int retval = -1;
    node_t *next_node = NULL;

    if (*head == NULL) {
        return -1;
    }

    next_node = (*head)->next;
    retval = (*head)->val;
    free(*head);
    *head = next_node;

    return retval;
}

// Removing the last item of the list
int
pop_last(node_t *head)
{
    int retval = 0;
    node_t *current = NULL;

    if (head->next == NULL) {
        retval = head->val;
        free(head);
        return retval;
    }

    /* get to the second to last node in the list */
    current = head;
    while (current->next->next != NULL) {
        current = current->next;
    }

    /* now current points to the second to last item of the list.
       so let's remove current->next */
    retval = current->next->val;
    free(current->next);
    current->next = NULL;

    return retval;
}

// Removing a specific item
int
remove_by_index(node_t **head, int n)
{
    int i = 0;
    int retval = -1;
    node_t *current = *head;
    node_t *temp_node = NULL;

    if (n == 0) {
        return pop_head(head);
    }

    for (i = 0; i < n - 1; i++) {
        if (current->next == NULL) {
            return -1;
        }
        current = current->next;
    }

    temp_node = current->next;
    retval = temp_node->val;
    current->next = temp_node->next;
    free(temp_node);

    return retval;
}


int
main(int argc, const char *argv[])
{
    int i;
    node_t * testnode;

    for (i = 0; i < argc; i++)
    {
        push_head(&testnode, atoi(argv[i]));
    }

    print_list(testnode);

    return 0;
}

// http://www.learn-c.org/en/Linked_lists
// https://www.geeksforgeeks.org/data-structures/linked-list/

Answer 8

The linked list implementation inspired by the implementation used in the Linux kernel:

// for 'offsetof', see: https://stackoverflow.com/q/6433339/5447906.
#include <stddef.h>

// See: https://stackoverflow.com/q/10269685/5447906.
#define CONTAINER_OF(ptr, type, member) \
    ( (type *) ((char *)(ptr) - offsetof(type, member)) )

// The macro can't be used for list head.
#define LIST_DATA(ptr, type, member) \
    CONTAINER_OF(ptr, type, member);

// The struct is used for both: list head and list nodes.
typedef struct list_node
{
    struct list_node *prev, *next;
}
list_node;

// List heads must be initialized by this function.
// Using the function for list nodes is not required.
static inline void list_head_init(list_node *node)
{
    node->prev = node->next = node;
}

// The helper function, mustn't be used directly.
static inline void list_add_helper(list_node *prev, list_node *next, list_node *nnew)
{
    next->prev = nnew;
    nnew->next = next;
    nnew->prev = prev;
    prev->next = nnew;
}

// 'node' must be a list head or a part of a list.
// 'nnew' must not be a list head or a part of a list. It may
//   be uninitialized or contain any data (even garbage).
static inline void list_add_after(list_node *node, list_node *nnew)
{
    list_add_helper(node, node->next, nnew);
}

// 'node' must be a list head or a part of a list.
// 'nnew' must not be a list head or a part of a list. It may
//   be uninitialized or contain any data (even garbage).
static inline void list_add_before(list_node *node, list_node *nnew)
{
    list_add_helper(node->prev, node, nnew);
}

// 'node' must be part of a list.
static inline list_node *list_del(list_node *node)
{
    node->prev->next = node->next;
    node->next->prev = node->prev;
    return node->prev;
}

Example of usage:

#include <stdio.h>

// The struct must contain 'list_node' to be able to be inserted to a list
typedef struct
{
    int       data;
    list_node node;
}
my_struct;

// Convert 'list_node *' to 'my_struct*' that contains this 'list_node'
static inline my_struct* get_my_struct(list_node *node_ptr)
{
    return LIST_DATA(node_ptr, my_struct, node);
}

void print_my_list(list_node *head)
{
    printf("list: {");
    for (list_node *cur = head->next; cur != head; cur = cur->next)
    {
        my_struct *my = get_my_struct(cur);
        printf(" %d", my->data);
    }
    printf(" }\n");
}

// Print 'cmd' and run it. Note: newline is not printed.
#define TRACE(cmd) \
    (printf("%s -> ", #cmd), (cmd))

int main()
{
    // The head of the list and the list itself. It doesn't contain any data.
    list_node head;
    list_head_init(&head);

    // The list's nodes, contain 'int' data in 'data' member of 'my_struct'
    my_struct el1 = {1};
    my_struct el2 = {2};
    my_struct el3 = {3};

    print_my_list(&head); // print initial state of the list (that is an empty list)

    // Run commands and print their result.
    TRACE( list_add_after (&head    , &el1.node) ); print_my_list(&head);
    TRACE( list_add_after (&head    , &el2.node) ); print_my_list(&head);
    TRACE( list_add_before(&el1.node, &el3.node) ); print_my_list(&head);
    TRACE( list_del       (head.prev)            ); print_my_list(&head);
    TRACE( list_add_before(&head    , &el1.node) ); print_my_list(&head);
    TRACE( list_del       (&el3.node)            ); print_my_list(&head);

    return 0;
}

The result of execution of the code above:

list: { }
list_add_after (&head , &el1.node) -> list: { 1 }
list_add_after (&head , &el2.node) -> list: { 2 1 }
list_add_before(&el1.node, &el3.node) -> list: { 2 3 1 }
list_del (head.prev) -> list: { 2 3 }
list_add_before(&head , &el1.node) -> list: { 2 3 1 }
list_del (&el3.node) -> list: { 2 1 }

http://coliru.stacked-crooked.com/a/6e852a996fb42dc2

Of course in real life you will most probably use malloc for list elements.

Answer 9

In memory your items are linked by pointers in the list structure

item1 -> item2

Why not make the list structure part of your item?

Then you allocate a product item, and the list structure is within it.

typedef struct product_data 
{
    int product_code;
    char product_name[PRODUCT_NAME_LEN];
    int product_cost;
    struct list_t list; // contains the pointers to other product data in the list
}product_data_t;

Answer 10

You're calloc'ing space for your list_t struct, just pointers to list head and tail, which isn't what you want to do.

When you add to a linked list, allocate space for an actual node in the list, which is your product_data_t struct.

Answer 11

You're allocating the wrong chunk of memory. Instead of allocating memory for each list element, you're allocating for the list head and tail.

For simplicity, get rid of the separate structure for the head and tail. Make them global variables (the same scope they're in now) and change them to be listhead and listtail. This will make the code much more readable (you won't be needlessly going through a separate structure) and you won't make the mistake of allocating for the wrong struct.

You don't need a tail pointer unless you're going to make a doubly linked list. Its not a major element to add once you create a linked list, but not necessary either.

Answer 12

In C language, we need to define a Node to store an integer data and a pointer to the next value.

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

To add a new node, we have a function add which has data as an int parameter. At first we create a new Node n. If the program does not create n then we print an error message and return with value -1. If we create the n then we set the data of n to have the data of the parameter and the next will contain the root as it has the top of the stack. After that, we set the root to reference the new node n.

Answer 13

#include <stdio.h>
struct node
 {
  int data;
  struct node* next;
 };

int main()
{

//create pointer node for every new element

struct node* head = NULL;
struct node* second = NULL;
struct node* third = NULL;

//initialize every new pointer with same structure memory

head = malloc(sizeof(struct node));
second = malloc(sizeof(struct node));
third = malloc(sizeof(struct node));

head->data = 18;
head->next = second;

second->data = 20;
second->next = third;


third->data = 31;
third->next = NULL;

//print the linked list just increment by address 

  for (int i = 0; i < 3; ++i)
    {
      printf("%d\n",head->data++);
      return 0;
    }
 }

This is a simple way to understand how does pointer work with the pointer. Here you need to just create pointer increment with new node so we can make it as an automatic.

Answer 14

Go STL route. Declaring linked lists should be agnostic of the data. If you really have to write it yourself, take a look at how it is implemented in STL or Boost.

You shouldn't even keep the *next pointer with your data structure. This allows you to use your product data structure in a various number of data structures - trees, arrays and queues.

Hope this info helps in your design decision.

Edit:

Since the post is tagged C, you have equivalent implementations using void* pointers that follow the basic design principle. For an example, check out:

Documentation | list.c | list.h