Problem in linked list while using struct variable - c

Question

Code for linked list using memory allocation:
#include<stdio.h>
struct node
{
int val;
struct node *point;
};
int main()
{
struct node *head;
head=(struct node* ) malloc(sizeof(struct node ));
void create(struct node* start);
void print(struct node* start);
create(head);
print(head);
return 0;
}
void create(struct node* start)
{
int test,a=1;
struct node* cur;
cur=start;
while(scanf("%d",&test)!=EOF)
{ if(a!=1)
{
cur=cur->point;
}
cur->point=(struct node* ) malloc(sizeof(struct node ));
cur->val=test;
a++;
printf("---------------");
printf("%d\n",cur);
printf("%d\n",cur->val);
printf("%d\n",cur->point);
printf("%d\n",a);
printf("---------------");
}
cur->point=NULL;
return;
}
void print(struct node* start)
{
struct node* cur;
cur=start;
while(cur!=NULL)
{
printf("%d ",cur->val);
cur=cur->point;
}
return;
}
Code for linked list using struct variable:
#include<stdio.h>
struct node
{
int val;
struct node *point;
};
int main()
{
struct node *head,link;
head=&link;
void create(struct node* start);
void print(struct node* start);
create(head);
print(head);
return 0;
}
void create(struct node* start)
{
int test,a=1;
struct node* cur;
cur=start;
while(scanf("%d",&test)!=EOF)
{ if(a!=1)
{
cur=cur->point;
}
struct node link;
cur->point=&link;
cur->val=test;
a++;
printf("---------------");
printf("%d\n",cur);
printf("%d\n",cur->val);
printf("%d\n",cur->point);
printf("%d\n",&link);
printf("%d\n",a);
printf("---------------");
}
cur->point=NULL;
return;
}
void print(struct node* start)
{
struct node* cur;
cur=start;
while(cur!=NULL)
{
printf("%d ",cur->val);
cur=cur->point;
}
return;
}
When I use 2nd code it doesn't work correctly. It prints just two values.
In while loop of create() I use struct node type variable declaration.
That's why it should it would declare a new struct node type link variable and a new memory (struct node link) address should be seen in every execution.
So what's the problem?
I am a newbie learning data structure using C. Please explain in simple language.