Thread: Memory adress as output

Also this might be a pointer arthmetic question, but how would I access the next pointer of the node before the one im inserting. I assume it I would do it by using the tail pointer before I dereference it. But how would I do that? can I do something like this? link->tail->next as I think you mentioned.

Okay Im getting it to work!! This is how the code looks like now:

Code:

#include "stdio.h"
#include "stdlib.h"


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


typedef struct list
{
	node_t *head;
	int size;
	node_t *tail;
}list_t;


int IsEmpty(list_t *list)
{
	if (list->size == 0)
	{
		printf("0");
		return 0;
	}
	else
	{
		printf("1");
		return 1;
	}
}


list_t *createList(void)
{
    list_t *list = malloc(sizeof(*list));
    if (list)
    {
        list->head = NULL;
        list->size = 0;
		list->tail = NULL;
    }
    return list;
}
node_t * createNode(int n)
{
	node_t *newnode = malloc(sizeof(node_t));
	if (!newnode)
    {
        return NULL;
    }
	newnode->value = n;
	newnode->next = NULL;
	newnode->prev = NULL;
	return newnode;	
}
void printlist(list_t* list)
{
	node_t *current = list->head;
	while (current != NULL)
	{
		printf("%d ", current->value);
		current = current->next;
	}
	printf("\n");
}


int insertTail(list_t *list, node_t *node)
{	
	if (list->head == NULL)
	{
		list->head = node;
		node->next = NULL;
		node->prev = NULL;
		list->tail = node;
		list->size += 1;
		return 1;
	}
	else
	{
		node->prev = list->tail;
		list->tail->next = node;
		list->tail = node;
		node->next = NULL;
		list->size += 1;
		return 1;
	}
}




int main()
{
	list_t *list2;
	list2 = createList();
	node_t *node1;
	node_t *node2;
	node_t *node3;
	node_t *node4;
	node_t *node5;
	node1 = createNode(6);
	node2 = createNode(7);
	node3 = createNode(8);
	node4 = createNode(4);
	node5 = createNode(9);
	insertTail(list2, node1);
	insertTail(list2, node2);
	insertTail(list2, node3);
	insertTail(list2, node4);
	insertTail(list2, node5);
	printlist(list2);
	free(node5);
	free(node4);
    free(node3);
	free(node2);
    free(node1);
    free(list2);
}

I can't thank you enough! Your help has been invaluble for me to actually understand this whole ordeal. I really appreciate that you made me conceptualize the problem. I really feel like drawing out the problem really made the whole concept click. Thanks!!