struct clist {
int pos;
char* color;
struct clist *next;
};
typedef struct clist sl_clist;
sl_clist *head = NULL;
sl_clist* link[5];
So i am trying to create multiple single linked circular lists and put them into a stack, every list will have the same types of data. In this case i am using the stack as an array. But i just cannot figure out how to create multiple linked lists from single type. I am a student so i am not very experienced on C. Thaks for the help in advance.
void create_list (int N, sl_clist* head){
int i;
sl_clist *new;
sl_clist *old;
if(N == 0){
head = NULL;
}
srand(time(0));
for(i = 0; i < N; i++){
new = (sl_clist*)malloc(sizeof(sl_clist));
if(i == 0){
head = new;
new -> color = color[(rand()%10)];
new -> pos = pos[i];
new -> next = head;
}
else{
new -> color = color[(rand()%10)];
new -> pos = pos[i];
old -> next = new;
new -> next = head;
}
old = new;
}
}
I have also tried creating multiple "head" variables but for some reason when i use them in this function(just imagine there are arrays for color and pos) they always return NULL.
Do not use global variables. Remove them.
sl_clist *head = NULL;
sl_clist* link[5];
Using local varaibles will "force" you to use a modular design that supports multiple lists.
Variables are passed by value:
head = new;
modifies a copy sl_clist* head of the original pointer passed as the argument. The original pointer is unaffected.
There are multiple ways you can solve that problem. You can return the new value:
sl_clist *create_list (int N, sl_clist* head){
...
return new; // or old value
}
int main() {
sl_clist *head = NULL;
head = create_list(5, head);
}
You can take the pointer by reference:
int create_list (int N, sl_clist **head){
...
*head = new; // set new value
(*head)->something = something; // be aware of operator precedence
}
int main() {
sl_clist *head = NULL;
create_list(5, &head); // head is getting modified
}
But I recommend doing a separate type for the head. That way the function is clear - it takes the head, specifically, not any list element. Be verbose:
struct sl_head {
struct clist *head;
};
int create_list(int N, struct sl_head *head) {
// ^^^^^^^^^^^^ - verbose, this is the head, not some element, less mistakces
head->head = new; // a bit more to type
head->head->something = something;
}
int main() {
struct sl_head head = {0}; // no longer a pointer
create_list(5, &head); // head is getting modified
}
Move srand(time(0)); to main(). It's not a function that you call when creating a list.
I'm currently doing an assignment for uni and I need to find the sum of a graph.
To do this I believe I need a linked list that I can use to remember which nodes have been visited. I have the linkedlist working correctly but I can't get a contains function to work. This is the code I have:
struct listnode
{
struct N *val;
struct listnode *next;
};
int contains(struct listnode *head,struct N* value)
{
struct listnode *current = head;
while (current)
{
if ((current -> val) == value)
{
return 1;
}
current = current -> next;
}
return 0;
}
note: N is a node of the graph.
Can anyone see any problems with what I'm doing?
EDIT: contains function should return 1 when N *value is in the list, 0 otherwise
EDIT2:
I have a push function:
void push(struct listnode *head,struct N *value)
{
if (head)
{
struct listnode *current = head;
while (current->next)
{
current = current -> next;
}
current->next = malloc(sizeof(struct listnode*));
current->next->val = value;
current->next->next = NULL;
}
else
{
head = malloc(sizeof(struct listnode*));
if (head)
{
head -> val = value;
head -> next = NULL;
}
else
{
printf("error");
exit(0);
}
}
}
and I want the following line to return 1:
contains(push(visited,p),p);
where p is a pointer to a struct N and visited is my global linked list
EDIT3:
this is my final sum function that I believe should work, but doesnt because of contains.
long sum(struct N *p)
{
if (p)
{
if (contains(visited,p) == 0) //if p hasnt been visited
{
push(visited,p); //make it visited
return (p -> data) + sum(p -> x) + sum(p -> y) + sum(p -> z);
}
else
{
return 0;
}
}
else
{
return 0;
}
}
Your contains function appears to be fine. The issue is that you are always passing a NULL list to it, which is caused by a faulty push function. You need a return in push, or to pass in a pointer with one more level of indirection, so you can assign to head outside of push. One more possible improvement is to notice that no matter what you pass in, the malloc and initialization of a new node is actually the same.
Finally, the main issue, that is really the most likely to cause a segfault is the fact that you are allocating enough space for a pointer to a node, not for the node itself.
Here is an example:
#ifdef BY_INDIRECTION
#define RET_TYPE void
#define IN_TYPE struct listnode **
#else
#define RET_TYPE struct listnode *
#define IN_TYPE struct listnode *
#endif
RET_TYPE push(IN_TYPE head, struct N *value)
{
struct listnode *current, **next;
if(head)
{
for(current = head; current->next; current = current->next) ;
next = &(current->next);
}
else
{
#ifdef BY_INDIRECTION
next = head;
#else
next = &head;
#endif
}
*next = malloc(sizeof(struct listnode));
if(!*next) {
printf("error");
exit(0);
}
(*next)->val = value;
(*next)->next = NULL;
#ifndef BY_INDIRECTION
return head
#endif
}
I have included both suggestions here. If you want to read the one where we use indirection (pass in a listnode ** and have void return), choose the path where BY_INDIRECTION is defined. If you want to have head returned (and pass in just a regular listnode *) read the path where BY_INDIRECTION is not defined.
The latter approach has a return value, so it can be used to write a shortened form like if(contains(push(head, value), value)) { ... }. The former approach does not, so you would have to do
push(&head, value);
if(contains(head, value)) { ... }
I would recommend using the indirect approach regardless because there are very few instances that you would want to check for containment after putting in a value.
This comparison:
if ((current -> val) == value)
it's comparing pointers. If you call your contains() function this way...
...
struct N val_to_find;
...
result = contains (list, &val_to_find);
You will never find the value, even if the contents of val_to_find are the same as the contents of any struct whose pointer is stored in the list.
If your intention for contains() is to find nodes that have the same data, and not just the same pointers, I'd suggest you something like this:
if (struct_n_comparing_function (current -> val, value) == EQUAL) ...
Where struct_n_comparing_function should have the following prototype:
int struct_n_comparing_function (struct N *a, struct N *b);
which compares the contents of the two structs pointed by a and b and return EQUAL if all the fields of the struct pointed by a have the same value as the fields of struct pointed by b.
I am trying to create a function splitlist(), which will split a singly linked list into two sublists – one for the front half, and one for the back half. I have come up with a code below which will work for the first time that I call the function, but when I call the function repeatedly, the program crashes. Any advice on how I can change my code to prevent such an error? The function splitlist() is void as it prints two lists which contains frontList and backList.
typedef struct _listnode {
int item;
struct _listnode *next;
} ListNode;
typedef struct _linkedlist {
int size;
ListNode *head;
} LinkedList;
void splitlist(LinkedList* list1, LinkedList * firsthalf, LinkedList *secondhalf)
{
ListNode *cur = list1->head;
ListNode *front = firsthalf->head;
ListNode *back = secondhalf->head;
int totalnodes = list1->size;
int i;
if (totalnodes % 2 != 0) //if odd number of elements, add 1 to make it easier for traversal of list
{
totalnodes = totalnodes + 1;
}
int halfnodes = totalnodes / 2;
{
for (i = 0; i < halfnodes; i++)
{
if (firsthalf->head == NULL) //initialise the head
{
firsthalf->head = malloc(sizeof(ListNode)); //create first node
front = firsthalf->head;
}
else
{
front->next = malloc(sizeof(ListNode));
front = front->next;
}
front->item = cur->item; // insert value from list1 into firsthalf
cur = cur->next; //point to next node in list1
}
front->next = NULL; //last node
for (i = halfnodes; i < totalnodes; i++)
{
if (secondhalf->head == NULL)
{
secondhalf->head = malloc(sizeof(ListNode));
back = secondhalf->head;
}
else
{
back->next = malloc(sizeof(ListNode));
back = back->next;
}
back->item = cur->item;
cur = cur->next;
}
back->next = NULL;
}
}
There are many things wrong with this code. First of all malloc return values are not checked, malloc can fail. And i strongly suspect that because of malloc fail your programm stops. You repeatedly allocate the memory inside the function, but do you free it when you do not need it anymore? Why do yo use malloc at all?
As posted earlier you do not need to.
Please post how the function is called, because it is really unclear how LinkedList* list1, LinkedList * firsthalf, LinkedList *secondhalf are used. Also it is unclear what is the structure of LinkedList is.
why use malloc?It will create a new list.But we want to split the list.
I guess firsthalf and second half are NULL
void splitlist(LinkedList* list1, LinkedList * firsthalf, LinkedList *secondhalf)
{
ListNode *cur = list1->head;
ListNode *front;
int totalnodes = list1->size;
int i;
if (totalnodes % 2 != 0) //if odd number of elements, add 1 to make it easier for traversal of list
{
totalnodes = totalnodes + 1;
}
int halfnodes = totalnodes / 2;
firsthalf->head=list1->head;
front=firsthalf->head;
for(i=0;i<halfnode;i++)
front=front->next;
secondhalf->head=front->next;
front->next=NULL;
}
At first glance I can't see much wrong with your code (assuming the assignment is to create copies of the list nodes in the new half lists), so the error could be in how you call the function, as an exmple, that could be:
LinkedList mainlist= {0};
LinkedList firsthalf= {0}, secondhalf= {0};
//mainlist got filled somehow; we now want to split
firsthalf->List= malloc(sizeof(ListNode));
secondthalf->List= malloc(sizeof(ListNode));
memset(firsthalf->List, 0, sizeof(ListNode));
memset(secondhalf->List, 0, sizeof(ListNode));
splitlist(&mainlist, &firsthalf, &secondhalf);
#include<stdio.h>
struct llist {
int data;
struct llist *next;
};
typedef struct llist list;
struct graph {
int V;
list **adj;
};
typedef struct graph graph;
graph* create_graph(int V) {
list **adj = (list**)malloc(sizeof(list*)*V);
graph *a = (graph*)malloc(sizeof(graph));
a->V = V;
a->adj = adj;
return a;
}
void insert_list(list **head, int v) {
list *new_node, *temp_node, *last_node;
new_node = (list*)malloc(sizeof(list));
new_node->data = v;
new_node->next = NULL;
printf("\n hi %d", head);
/*
* head is empty, point the head to temp and return.
*/
if (*head == NULL) {
*head = new_node;
return;
}
temp_node = *head;
do{
last_node = temp_node;
temp_node = temp_node->next;
}while(temp_node);
last_node->next = new_node;
}
void addedge(graph *g, int u, int v) {
insert_list(&(g->adj[u]), v);
}
int
main() {
int V = 10;
graph *a = create_graph(V);
printf("\n graph created");
addedge(a, 1,2);
addedge(a,1,5);
addedge(a,2,1);
addedge(a,2,5);
addedge(a,2,3);
addedge(a,2,4);
addedge(a,3,2);
addedge(a,3,4);
addedge(a,4,2);
addedge(a,4,5);
addedge(a,4,3);
addedge(a,5,4);
addedge(a,5,1);
addedge(a,5,2);
return 0;
}
In this code, by printing messages, I have found is create_graph function executes properly and returns a graph. Then addedge is being called, In it, if(*head==NULL) part always returns false (Don't know why, first time it should return true). Then it goes ahead in do while loop and that keep executing till infinity and code terminates.
What I am trying to do is I have a structure called graph, with a integer variable V and array of linked list variable adj (represents adjacency list). And then create graph will initialise a graph variable and return it.
Addedge(V,u,v) will add edge v to u (means v is adjacent to u). so adj[0....V-1] is a array of linked list. so if edge 1 is adjacent to 2 and 3, then list will look like 1->2->3.
Comment If more info needed. I don't know what why *head is not null the first time and why the while loop never terminates.
Thanks a lot in advance.
By trail and error, I found the mistake.
In function create graph, after initialising the list, each individual list also must be initialised to NUll, otherwise it will have junk stuff.
Below is correct code :
graph* create_graph(int V) {
list **adj = (list**)malloc(sizeof(list*)*V);
int i;
for (i = 0; i < V; ++i) {
adj[i] = NULL;
}
graph *a = (graph*)malloc(sizeof(graph));
a->V = V;
a->adj = adj;
return a;
}
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