Thread: Linked lists

Hello,
I created a structure with two nodes in it, one that points to the next node and the other points to previous node:

Code:

#include<stdio.h>

typedef struct node
{
        int         value;
        struct node *next;
        struct node *prev;
}node;

int main(void)
{
    node *head = NULL;
    int  i;
    
    head = malloc(sizeof(node));
    head->value = 1;
    head->prev = NULL;
    
    for (i = 2; i < 11; i++)
    {
        head->next = malloc(sizeof(node));
        head->next->prev = malloc(sizeof(node));
        
        head->next->prev = head;
        head = head->next;
        
        head->value = i;
    }
    head->next = NULL;
    
    while(head != NULL)
    {
          printf("%d\n",head->value);
          head = head->prev;
    }
    
    /*Not working*/
    while(head)
    {
          printf("%d\n",head->value);
          head = head->next;
    }
    
    system("pause");
    return 0;
}

Why is the second printing not working???

Originally Posted by laserlight

Notice that you changed what head points to.

Ya I changed so that head goes back where it started from; printing out numbers in reverse order, and then print the numbers again from start to end.

Originally Posted by oogabooga

There are other anomalies in your linked list code.
Look at these two lines. Can you see a problem?

Code:

        head->next->prev = malloc(sizeof(node));
        head->next->prev = head;

And you shouldn't be mallocing two nodes each iteration of the loop.
Just malloc one new node and attach it to the list by adjusting pointers.

Okay, I removed

Code:

head->next->prev = malloc(sizeof(node));

, but still its not working.

The program is advancing head through the list as it creates the list. The program should use a separate pointer to node to build the list with, so that head isn't modified. The other option is to follow the list backwards using head until head->prev == NULL to set head back to the beginning of the list.

Originally Posted by rcgldr

The other option is to follow the list backwards using head until head->prev == NULL to set head back to the beginning of the list.

That's what I have done, haven't I?

Originally Posted by laserlight

What is your current code and how does it not work?

Code:

#include<stdio.h>

typedef struct node
{
        int         value;
        struct node *next;
        struct node *prev;
}node;

int main(void)
{
    node *head = NULL;
    int  i;
    
    head = malloc(sizeof(node));
    head->value = 1;
    head->prev = NULL;
    
    for (i = 2; i < 11; i++)
    {
        head->next = malloc(sizeof(node));
        
        head->next->prev = head;
        head = head->next;
        
        head->value = i;
    }
    head->next = NULL;
    
    /*this one works well*/
    while(head != NULL)
    {
          printf("%d\n",head->value);
          head = head->prev;
    }
    
    /*This is not working: its not printing anything*/
    while(head != NULL)
    {
          printf("%d\n",head->value);
          head = head->next;
    }
    
    system("pause");
    return 0;
}

One thing to do is print the adresses of where you are in the list. This is as simple as just printing the pointer.

Originally Posted by laserlight

Well, it looks like you still have the same problem: notice that head is changed, hence in the second loop to print, head is already NULL, so nothing is printed.

A solution is to avoid changing head when printing by introducing another variable that starts life as a copy of head. One way to do this is to implement a function to print your linked list.

Okay I did as you said:

Code:

#include<stdio.h>

typedef struct node
{
        int         value;
        struct node *next;
        struct node *prev;
}node;

int main(void)
{
    node *head = NULL;
    node *cpy;
    node *cpy2;
    int  i;
    
    head = malloc(sizeof(node));
    head->value = 1;
    head->prev = NULL;
    
    for (i = 2; i < 11; i++)
    {
        head->next = malloc(sizeof(node));
        //head->next->prev = malloc(sizeof(node));
        
        head->next->prev = head;
        head = head->next;
        
        head->value = i;
    }
    head->next = NULL;
    
    cpy = head;
    while(cpy != NULL)
    {
          printf("%d\n",cpy->value);
          cpy = cpy->prev;
    }
   
    while(cpy)
    {
          printf("%d\n",cpy->value);
          cpy = cpy->next;
    }
    
    system("pause");
    return 0;
}

Now how can I bring the head to the starting, and print by going next???

Just before that for(i = 2; ...), insert cpy = head; into the code, then use cpy in the for loop instead of head, so the line after the for loop would be cpy->next = NULL; .

Originally Posted by rcgldr

Just before that for(i = 2; ...), insert cpy = head; into the code, then use cpy in the for loop instead of head, so the line after the for loop would be cpy->next = NULL; .

Ya, I know this method, just wanted to try the next and previous method.

Originally Posted by rcgldr

Just before that for(i = 2; ...), insert cpy = head; into the code, then use cpy in the for loop instead of head, so the line after the for loop would be cpy->next = NULL; .

Originally Posted by Harith

Ya, I know this method, just wanted to try the next and previous method.

I meant this change:

Code:

/* ... */
    cpy = head;
    for (i = 2; i < 11; i++)
    {
        cpy->next = malloc(sizeof(node));
        cpy->next->prev = cpy;
        cpy = cpy->next;
        cpy->value = i;
    }
    cpy->next = NULL;