Thread: help with linked lists

Hi guys, im stuck with this linked list problem, i just want to make 3 nodes all with values in them declared in main and just print out its values, but im confused on how to link them together especially with the *head and *tail.
I can implement it with out the *head and *tail pointers but with them included its more confusing!

thanks

Code:

#include <stdio.h>

typedef int data_t;
typedef struct node node_t;

struct node {
	data_t data;
	node_t *next;
};

typedef struct {
	node_t *head;
	node_t *tail;
} sequence_t;

void print_list( sequence_t * );

int main( void )
{
	sequence_t nodes;
	node_t num1, num2, num3, *start;

	num1.data = 2;
	num2.data = 10;
	num3.data = 20;

	return 0;

}


void print_list( sequence_t *seq )
{
	while ( print != NULL ) {
		printf( "head: %d\n", seq->head->data );
	}
	seq=seq->head->next;
}

How about

Code:

nodes.head = &num1;
num1.next = &num2;
num2.next = &num3;
num3.next = NULL;
nodes.tail = &num3;

....

void print_list( sequence_t *seq )
{
  node_t * ptr;

  for (ptr = seq->head; ptr != NULL; ptr = ptr->next) 
  {
    printf( "data: %d\n", ptr->data );
  }
}

it makes tail a bit redundant as is but I'm sure you can fix that if you want to...

dont i have to pass in something into the print_list function?

Code:

void insertNode( Node *n )
{
    if( n->value < head->value )
    {
        n->next = head;
        head->prev = n;
        head = n;
    }
    else
    if( n->value > tail->value )
    {
        n->prev = tail;
        tail->next = n;
        tail = n;
    }
    else
    ...loop through and insert...
}

This code assumes:
a) A double linked list. Infinately easier than single linked lists.
b) Global pointers for 'head' and 'tail'.
c) That you're sorting by some value and you want it sorted from least to greatest.

Otherwise, modify it to suit your needs. Basicly, if prepending, you make the new node point to whatever is currently the head. Then you make the head be the new node.

If appending, do the opposite.

Quzah.

thanks guys

ill have to look through the doubly linked list

cheers