I'm trying to pop the last node from a doubly and circular linked list but I'm getting a headache with pointers. I've got a function that traverses the entire list til the end of it but somehow, the tmp pointer doesn't move or update. I'm traversing the entire list with a for loop instead of a while one as its more comfortable to me (and yes, ive tried with a while loop)
Here is my code:
typedef struct List {
unsigned int size;
Node *p_head;
Node *p_tail;
} List;
typedef struct Node {
void *p_value;
struct Node *p_next;
struct Node *p_previous;
} Node;
Bool RemoveAtEnd(List *list) {
Node *p_node = list->p_head; /* tmp ptr */
for (int i = 0; i < GetSize(list); ++i) {
p_node = (p_node)->p_next;
}
/*
* p_node var should be the last node or tail of the list shouldn't it?
*/
printf("Tail here is %d\n", *(int *) p_node->p_value);
list->p_tail = p_node->p_previous;
list->p_tail->p_next = list->p_head;
list->p_head->p_previous = list->p_tail;
list->size--;
DestroyNode(p_node);
return TRUE;
}
When trying to free the node within the function valgrind says "Invalid read" when trying to traverse the list outside the function because as far i seen, the tail is pointing back to the head instead of the tail
List *p_intList = CreateList();
for (int i = 0; i < 5; ++i) {
int a = i + 1;
InsertAtEnd(p_intList, &a, sizeof(int));
}
RemoveAtEnd(p_intList);
for (int i = 0; i < GetSize(p_intList); ++i) {
printf("%d\n", *(int *) GetValueAt(p_intList, i)); // Invalid read
}
DestroyList(p_intList);```
in pseudocode i would do like this
you know that last-1 -> last -> head
and what you need is the list
to become like last-1 > head
save the last node
last = list.getHead()->prev;
link last-1 with node head
last-1 = list.getHead()->prev->prev;
last-1.next = list.getHead();
list.getHead().prev = last-1
free removed node
free(last);
I divided the code in two files, .h and .c
The definition of function names is in .h, the implementation of the function is in .c
in my main file:
struct no
{
tipo info;
struct no *ant;
struct no *nxt;
};
struct list
{
no_t *head;
no_t *tail;
int size;
};
this is in my .h file:
typedef struct no no_t;
typedef struct list list_t;
typedef int tipo;
...again in main
void list_destroy(list_t **l)
{
if ((*l) == NULL || l == NULL)
return;
if (!(*l)->head)
return;
no_t *next = (*l)->head; //create two variables for iterating through the list
no_t *aux; //set aux to free
while (next->nxt) //the pointer for next node, in the last node, is NULL
{ //by that I believe I'm able to iterate through all nodes
aux = next;
free(aux);
next = next->nxt;
}
free(*l);
(*l) = NULL;
}
is quite a simple code, but I can't see where I'm missing here
next = next->nxt;
For the compiler it makes no difference, for sure. But for someone, even you, it is hard to read this next = next->nxt stuff. Or is it is not?
A possible alternative (using your code) and a short test program
so_list.h
#include <stdio.h>
#include <stdlib.h>
typedef int Tipo;
typedef struct st_no
{
Tipo info;
struct st_no* prev;
struct st_no* next;
} Node;
typedef struct
{
Node* head;
Node* tail;
unsigned size;
} List;
List* list_create();
List* list_destroy(List*);
int list_insert(const Tipo, List*);
In the header, only typedefs and the function prototypes.
names with only the first letter in uppercase are reserved here for defined names. An useful convention.
instead of using List** is often clearer to just return the pointer to the list. In this way it is easier for example to invalidate the pointer and to create the linked lists as in
List* my_list = list_create();
my_list = list_destroy(my_list);
and there is no need to test the two levels of indirection as you need when ** is used
main.c: a minimalist test set
#include "so-list.h"
int main(void)
{
List* my_list = list_create();
my_list = list_destroy(my_list);
my_list = list_create();
for (int i = 1; i <= 5; i += 1)
printf("insert(%d,list) returned %d\n",
i, list_insert(i,my_list)
);
my_list = list_destroy(my_list);
my_list = list_create();
for (int i = 11; i <= 15; i += 1)
printf("insert(%d,list) returned %d\n",
i, list_insert(i, my_list)
);
my_list = list_destroy(my_list);
return 0;
}
A list is created, then destroyed
using the same pointer, a list is created, values 1 to 5 are inserted ant then the list is deleted.
using the same pointer, a list is created, values 11 to 15 are inserted ant then the list is again deleted.
the output
List created!
List deleted!
List created!
insert(1,list) returned 1
insert(2,list) returned 2
insert(3,list) returned 3
insert(4,list) returned 4
insert(5,list) returned 5
1 deleted
2 deleted
3 deleted
4 deleted
5 deleted
List deleted!
List created!
insert(11,list) returned 1
insert(12,list) returned 2
insert(13,list) returned 3
insert(14,list) returned 4
insert(15,list) returned 5
11 deleted
12 deleted
13 deleted
14 deleted
15 deleted
List deleted!
code for destroy_list()
List* list_destroy(List* l)
{
if (l == NULL) return NULL;
// delete the ´size´ nodes, 1 by 1
Node* p = NULL;
for (unsigned i = 0; i < l->size; i += 1)
{
p = l->head->next; // save pointer
printf("%d deleted\n", l->head->info); // just for the demo
free(l->head); // free head
l->head = p; // advance head
}
free(l); // free list
printf("List deleted!\n\n"); // just for the demo
return NULL;
}
This function always return NULL as just a way to invalidade the pointer in the caller in the same expression as in pList = destroy_list(pList);
This is somewhat different than the code you wrote. We just delete the elements one by one as we know the list has size elements. A local pointer is used in the loop to save the address of the next element. It seems to be easier to read.
The complete code for so-list.c
#include "so-list.h"
List* list_create()
{
List* one = (List*)malloc(sizeof(List));
one->head = NULL;
one->tail = NULL;
one->size = 0;
printf("List created!\n");
return one;
}
List* list_destroy(List* l)
{
if (l == NULL) return NULL;
// delete the ´size´ nodes, 1 by 1
Node* p = NULL;
for (unsigned i = 0; i < l->size; i += 1)
{
p = l->head->next; // save pointer
printf("%d deleted\n", l->head->info);
free(l->head); // free head
l->head = p; // advance head
}
free(l); // free list
printf("List deleted!\n\n");
return NULL;
}
// just for test, insert ´info´ at the end, returns size
int list_insert(const Tipo info, List* l)
{
// insert node at the end, just for test
Node* one = (Node*)malloc(sizeof(Node));
one->info = info;
one->next = NULL;
one->prev = l->tail;
if (l->size == 0)
l->head = one; // 1st node
else
l->tail->next = one;
l->tail = one;
l->size += 1;
return l->size;
};
about your version of list_destroy()
The logic there is a bit wrong but the error is well described in another answer. I recommend not to use ** in this situations. But it can be done for sure.
so-list.c
This is just a minimum to have a running test
#include "so-list.h"
List* list_create()
{
List* one = (List*)malloc(sizeof(List));
one->head = NULL;
one->tail = NULL;
one->size = 0;
printf("List created!\n");
return one;
}
List* list_destroy(List* l)
{
if (l == NULL) return NULL;
// delete the ´size´ nodes, 1 by 1
Node* p = NULL;
for (unsigned i = 0; i < l->size; i += 1)
{
p = l->head->next; // save pointer
printf("%d deleted\n", l->head->info);
free(l->head); // free head
l->head = p; // advance head
}
free(l); // free list
printf("List deleted!\n\n");
return NULL;
}
// just for test, insert ´info´ at the end, returns size
int list_insert(const Tipo info, List* l)
{
// insert node at the end, just for test
Node* one = (Node*)malloc(sizeof(Node));
one->info = info;
one->next = NULL;
one->prev = l->tail;
if (l->size == 0)
l->head = one; // 1st node
else
l->tail->next = one;
l->tail = one;
l->size += 1;
return l->size;
};
This has an issue
no_t *next = (*l)->head;
no_t *aux;
while (next->nxt)
{
aux = next; // aux point to the same object as next
free(aux); // free aux, which is the same as next
next = next->nxt; // deference next, which just got free'd. OOPS!
}
You invoke free on aux, which is also aliasing next. Then you try to deference next->nxt. Well, next just got released in the previous statement. Also, as I called out in the comment, you are leaking the last element in the list.
Fixed:
no_t* aux = (*l)->head;
while (aux)
{
no_t* next = aux->nxt;
free(aux);
aux = next;
}
You should look to your "free" and your "next->nxt" statements. May it can help you solve it.
I am attempting to alphabetize two separate lists in my add function: one that sorts the nodes by first names, and another that sorts by last names. I also have some logic that checks if a name is already in the list and if it is an error is printed and the list is returned unchanged. Like the title says, I am getting a segmentation fault here and am not sure why. It may be a pretty basic problem but I am new to C and especially new to linked lists.
Here is how the nodes are defined:
typedef struct node {
char *first;
char *last;
long number;
struct node *nextFirst;
struct node *nextLast;
} Node;
typedef struct mlist {
Node *headFirstName;
Node *headLastName;
} MultiLinkedList;
And here is my add function:
MultiLinkedList *add(MultiLinkedList *list, char *first, char *last, long num) {
// allocate a new node
Node *newNode = malloc ( sizeof(Node) );
newNode->first = malloc ( strlen(first) + 1 );
strcpy(newNode->first, first);
newNode->last = malloc ( strlen(last) + 1 );
strcpy(newNode->last, last);
newNode->number = num;
//make placeholder nodes
Node *a = list->headFirstName;
Node *b = list->headLastName;
// add this new node at the head of the "byFirst" list
if (strcmp(newNode->first, a->first) < 0) {
newNode->nextFirst = list->headFirstName;
list->headFirstName = newNode;
}
for (Node *i = list->headFirstName; i; i = i->nextFirst) {
// add after less alphabetical nodes
if (strcmp(newNode->first, i->first) > 0) {
newNode->nextFirst = i->nextFirst;
i->nextFirst = newNode;
}
// return error for duplicate name
if (strcmp(newNode->first, i->first) == 0 && strcmp(newNode->last, i->last) == 0) {
printf("That person is already in the list! Please try with a different name.\n");
}
}
// add this new node at the head of the "byLast" list
if (strcmp(newNode->last, b->last) < 0) {
newNode->nextLast = list->headLastName;
list->headLastName = newNode;
}
for (Node *j = list->headLastName; j; j = j->nextLast) {
// add after less alphabetical nodes
if (strcmp(newNode->last, j->last) > 0) {
newNode->nextLast = j->nextLast;
j->nextLast = newNode;
}
}
// return the multi-list object with updated or original head pointers
return list;
}
I figured out what my problem was. I had to add return list; to the end of each if statement otherwise the function attempts to perform every true statement; which causes the seg fault. In hindsight I'm surprised I didn't figure this out sooner.
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;
}
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
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.
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.
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.
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.
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.
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.
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;
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.
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/
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.
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.
#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.
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