Having trouble with a simple linked list in C - c

Question

I'm trying to understand linked lists. I created this code which is supposed to give me experience with linked lists. When I run this code I get 0, though the output should be 10,9,8,7,6,5,4,3,2,1,0. Could someone help me figure this out?
#include <stdio.h>
#include <stdlib.h>
typedef struct node
{
int value;
struct node *next;
}node;
node * addnode (node *ptr, int value);
void traverse(node *ptr);
int main (void)
{
int i;
node *ptr,*root;
root=(node *)malloc(sizeof(node));
ptr=root;
root->value=0;
root->next=0;
for(i=1;i>10;i++)
ptr=addnode(ptr,i);
traverse(ptr);
return(0);
}
void traverse(node *ptr)
{
printf("%d\n",ptr->value);
if(ptr->next) /*assuming that the pointer to the next node is initialized to NULL, so if its NULL then the if codeblock won't execute*/
traverse(ptr->next);
}
node * addnode (node *ptr, int value)
{
node *newnode;
newnode=(node *)malloc(sizeof(node));
newnode->value=value;
newnode->next=ptr;
return(newnode);
}

Answer 1

There are two problems with your code:
Ideally, you would do traverse(root) instead of traverse(ptr).
You are adding nodes at the beginning of your linked list, when you should add them on the end of the list.
You can do it like this (source):
void Push(struct node** headRef, int newData) {
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
new_node->data = newData;
new_node->next = *headRef;
*headRef = new_node;
}
which in your case would look like:
void addnode(struct node** headRef, int newData) {
struct node* new_node =
(struct node*) malloc(sizeof(struct node));
new_node->value = newData;
new_node->next = *headRef;
*headRef = new_node;
}
Then you add nodes like this: addnode(&root, some_value);