使用双指针链接到链表的字符串

Question

I have the following code I"m converting the string stored into the linked list. Example : ABC A->B->C->NULL

Problems : When printing the list,it is not giving the desired output.Following is the code and sample input/outputs.

Code

#include<stdio.h>
#include<stdlib.h>
typedef struct node
{
    char ch;
    struct node *next;
}node;
void create(node **head,char ch)
{
    node *new;
    new=malloc(sizeof(node));
    new->next=NULL;
    new->ch=ch;
    if(*head==NULL)
    {
        *head=new;
        printf("%c",(*head)->ch);
        return ;
    }
    while((*head)->next)
    {
        (*head)=(*head)->next;
    }
    (*head)->next=new;


}
void printList(node *head)
{
    printf("\nThe list has - ");
    while(head)
    {
        printf("%c",head->ch);
        head=head->next;
    }
    printf("\n\n");
}
int main()
{
    node *head=NULL;
    int i=0;
    char *str=NULL;
    str=malloc(sizeof(char)*15);
    printf("\nEnter the string - ");
    scanf("%s",str);

    while(str[i]!='\0')
    {
        create(&head,str[i]);
        i++;
    }
    printList(head);
    return 0;
}

Sample Input/Outputs

Input 1

Enter the string - abc 
a
The list has - bc

Input 2

Enter the string - abcde
a
The list has - de

Input 3

Enter the string - ab
a
The list has - ab

Note :

If I change my create function to this , everything just works fine!. I want to know what's the difference here? Has it something to do with the double pointer??

void create(node **head,char ch)
{
    node *new,*ptr;
    new=malloc(sizeof(node));
    new->next=NULL;
    new->ch=ch;
    ptr=*head;
    if(ptr==NULL)
    {
        ptr=new;
        return;
    }
    while(ptr->next)
    {
        ptr=ptr->next;
    }
    ptr->next=new;

}

Thanks!

Answer 1

There is a issue with your insert function in the first code snipper where you move the *head so by the time you insert the last node to the list the head is pointing to one before the last node

a->b->c->d
      |
      |

Head is at c now

So you should never move the head and just use temporary variable to get the value of head and move temp.

a->b->c->d
|     |
|     |
Head  temp

Has it something to do with the double pointer??

No it is just that in the second snippet you use ptr as temporary pointer and doesn't move head your code works as shown above.

Answer 2

Gopi has already pointed out the issue with your code. You can use that advice to insert a new node if you distinguish between the two cases of inserting the first node to an empty list (in which case you must update head ) and of appending to an existing list. (You already catch the two cases.)

But the pointer-to-pointer strategy adds one level of indirection, which you can use here to do without this distinction: head holds the pointer to the pointer to the head node. If you use head to walk through the list, head should always point to the pointer that points to the current node. When the current node is NULL , assign the new node, ie overwrite the pointer:

void create(node **head, char ch)
{
    /* create new node */
    node *nd = malloc(sizeof(*nd));
    nd->next=NULL;
    nd->ch=ch;

    /* advance to end of list */
    while (*head) {
        head = &(*head)->next;
    }

    /* assign */
    *head = nd;
}

By the way, your second function does not work just fine, because you never update the head. You'll get an empty list and memory leaks.