Thread: linked list

hi .i am wirtting a linked list and i tried testing my add method.I observed that my values are not beeing written in the right place.for e.g my test value 5 should be in node 1 node 1 and my test value 6 should be in node 2

Code:

/*  * File:   main.c
 * Author: onwugbej
 *
 * Created on November 23, 2011, 10:24 PM
 */


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


/*
 * 
 */
struct node{
    int key;
    struct node *next;
    
};


struct listset{
    int length;  
    struct node *head;
};
struct listset * litset_new(){
    struct listset *p = malloc(sizeof(*p));
    struct node *dummy;  // creating a dummy for my node to point to
   
    p->head=malloc(sizeof(*dummy)); //creating space for my node head in the list
    dummy=malloc(sizeof(*dummy));
   
    p->head->next=dummy;
    p->head->key=0;
    dummy->next=dummy;  //pointing to itself
    
    
    return p;
    
}
void listset_add(struct listset * t, int item){
    //struct listset *a=t;
    struct node *p= malloc(sizeof(*p));
   
    p->key=item;
    p->next = t->head;
    t->head->next=p;    //let my nlist set head point to my my new node
    //p->key=item;  //settinh my item
}


/* Given a linke list head pointer 
 *  this function to counts the number of nodes in a list
 */
int Length(struct node *head) {


struct node* current =head; 
int count = 0;
while (current != NULL) 
{ count++;
current = current->next;
}
return count;
}


int main(int argc, char** argv) {
    struct listset  *s;
     s=litset_new();
     listset_add(s,5);
     listset_add(s,6);
     printf("%d\n",s->head->next->key);
      printf("%d\n",s->head->next->next->key);


    return (EXIT_SUCCESS);
}

The problem is you "listset_add" function. In fact, what you do is attaching the last node to the first (which is ok if you want to do circular linked list), and then attaching the first one to the last, which means that your linked list will only be able to contains 2 elements, since you're always overwriting the "->next" of the head. So actually when you're printing "s->head->next->next->key" you are printing the first element. What you have to do in your "listset_add" function is :
->iteratate to the last element (for a circular linked list you iterate until the "->next" is equal to the head).
->link the last element to the new node.
->link the new element to the head (for circular list again).

Also, why are you using this "dummy" node in your initialization? Do you really want to insert this useless node each time you create a list set? You could set the head to NULL and check in you're "list_add" function if the head is NULL, and create it if yes.

Originally Posted by Tibo-88

The problem is you "listset_add" function. In fact, what you do is attaching the last node to the first (which is ok if you want to do circular linked list), and then attaching the first one to the last, which means that your linked list will only be able to contains 2 elements, since you're always overwriting the "->next" of the head. So actually when you're printing "s->head->next->next->key" you are printing the first element. What you have to do in your "listset_add" function is :
->iteratate to the last element (for a circular linked list you iterate until the "->next" is equal to the head).
->link the last element to the new node.
->link the new element to the head (for circular list again).

Also, why are you using this "dummy" node in your initialization? Do you really want to insert this useless node each time you create a list set? You could set the head to NULL and check in you're "list_add" function if the head is NULL, and create it if yes.

hi i have changed my initiliasation of my listset struct to

Code:

struct listset * litset_new(){    struct listset *p = malloc(sizeof(*p)); 
    p->head=malloc(sizeof(p->head)); //creating space for my node head in the list
    p->head->next=NULL;
    p->head->key=0;
    
    
    return p;
     
}