Thread: linked list (good example)

/* is the following a good example of a linked list and if not why */


#include<stdio.h>
#include<conio.h>

struct linkedlist{
int a,b,c;
struct linkedlist *next_linkedlist;
};



int main()
{
struct linkedlist list,*ptr; /* pointer to structure */
ptr=&list; /* of type linkedlist */

ptr->a=10; /* pointer points to first data member/variable a */
ptr->next_linkedlist->a=37; /* pointer now points to the next */
/* structure then it's first data member which is a */

printf("%d",ptr->a); /* output 10 */
printf("\n%d",ptr->next_linkedlist->a); /* output 37 */
getch();

return 0;
}
/* The identifier a has been used twice in the above code */
/* has it been used correctly, To create more lists would */
/* I do this in the outer most structure add more statements */
/* like so struct linklist *next1_linkedlist; */
/* struct linklist *next2_linkedlist; */
/* ... */
/* ... */
/* ... */
/* Therefor being able to use a,b,c over and over and... */
/* But if all these pointers point to abc how come there not */
/* overwriten */


thanks for all the help

Not really. It doesn't show how the list is build. And I would not call the structure linkedlist, but something like node. Since the list is build up of nodes. In the example it looks like if the pointer in the structure points to a different list, but it's just pointing to the following node in the list.

Code:

struct node
{
    int data;
    struct node *next_node;
};

A good example about linked lists should show how a list is build. So it should make clear how to create a linked list, something like this should be present.

Code:

/* The list */
struct node list;
/* A pointer required to traverse the list */
struct node *list_ptr;

/* Initialise list */
list.data = JUST_A_VALUE;
list.next_node = NULL;

/* Add a node to the list */
new_node = create_node (sample_data);
if (new_node != NULL)
    list_ptr->next_node = new_node;

This should be part of an intro to linked lists.