Implement a doubly linked list in C - c

Question

I want to insert a node at the beginning of linked list, whenever insertAtBeginning method is called. My code builds well, but i don't get the desired output.
I get the following output:
0------>NULL
The desired output is:
9------>8------>7------>6------>5------>4------>3------>2------>1------>0------>NULL
Following is my code:
#include<stdio.h>
#include<stdlib.h>
struct dll{
int data;
struct dll* previous;
struct dll* next;
};
struct dll* insertAtBeginning(int a, struct dll* head){
if(head == NULL){
head->data = a;
head->previous = NULL;
head->next = NULL;
return head;
}
else{
struct dll *first;
first = (struct dll*) malloc( sizeof(struct dll));
first->data = a;
first->next = head;
head->previous = first;
first->previous = NULL;
head = first;
free(first);
return head;
}
}
void display_from_first(struct dll* head){
struct dll *temp;
temp = head;
printf("\nThe linked list contains: ");
while(temp != NULL) {
printf("%d------>",temp->data);
temp = temp->next;
}
printf("NULL\n");
free(temp);
}
int main(){
int i = 0;
struct dll *head1, *tail1;
head1 = (struct dll*) malloc( sizeof(struct dll));
head1->next = NULL;
head1->previous = NULL;
for(i=0; i<10; i++){
insertAtBeginning(i, head1);
}
display_from_first(head1);
return 0;
}

Answer 1

The code for a doubly linked list is much cleaner if you start with an empty list of two nodes as shown below.
That way you don't have to deal with special cases like if(head==NULL). There's always a node before and after the node that is being inserted (or deleted), so you just hook things up and you're done.
#include <stdio.h>
#include <stdlib.h>
typedef struct s_node Node;
struct s_node
{
Node *prev;
Node *next;
int data;
};
Node *insertAtBeginning( Node *head, int value )
{
// allocate memory for the new node
Node *node = malloc( sizeof(Node) );
if ( node == NULL )
return NULL;
// insert the node at the beginning of the list
Node *temp = head->next;
head->next = node;
temp->prev = node;
// fill in the fields of the node
node->prev = head;
node->next = temp;
node->data = value;
return node;
}
void showList( Node *head )
{
Node *node;
printf( "The list contains: " );
for ( node = head->next; node->next != NULL; node = node->next )
printf( "%d--->", node->data );
printf( "NULL\n" );
}
int main( void )
{
// create an empty list with two nodes
Node head = { NULL , NULL, 0 };
Node tail = { &head, NULL, 0 };
head.next = &tail;
// insert more nodes
for ( int i = 0; i < 10; i++ )
insertAtBeginning( &head, i );
// display the list
showList( &head );
}

Answer 2

There are mainly two problems here :
free(first) : This is not required as you wish to save the memory you just allocated, not delete it.
Your insertAtBeginning() function returns a pointer to head, so in main(), where you are calling this function change it to head1=insertAtBeginning(i, head1); This way your head is also saved.
Here's the code with the two edits :
http://ideone.com/nXwc8z

Answer 3

You have several mistakes here.
1) Your function insertAtBeginning returns pointer to changed list, but you do not update pointer to head of list in the main function.
2) You are freeing just allocated pointer to new node in the insertion function. You think that you are freeing pointer, but actually you say that this place in memory is not needed more and so your node can't be there.

Answer 4

You can't free(first); in insertAtBeginning().
code here.
And btw when you have empty list your display_from_first() prints The linked list contains: 0------>NULL because of
head1 = (struct dll*) malloc( sizeof(struct dll));
head1->next = NULL;
head1->previous = NULL;
in main(). Remove it from main to have correct output