Question regarding multiple singly linked lists - c

Question

I'm trying to create multiple singly linked lists by using a list struct and a node struct. What I am thinking is each new list is told apart by a different head. But what makes me confused is how do I initialize a new head for each list and then add on nodes?
When I worked with just one list I was able to just allocate memory for the one head pointer and set it to NULL and then just add on new nodes.
My structs looks like this for reference:
typedef struct node
{
int value;
struct node *next;
}node_t;
typedef struct list
{
struct list *head;
int size;
}list_t;
If you feel like I've been unclear on something just ask and I will clarify!
Thanks

Answer 1

In your list structure head should actually be a node_t - first node in the list.
And then you will chain the nodes by assigning another node_t to next
typedef struct node
{
int value;
struct node *next;
}node_t;
typedef struct list
{
node_t *head;
int size;
}list_t;
Simple example for reference:
node_t node1;
node_t node2;
node1.value = 1;
node1.next = &node2;
node2.value = 1;
node2.next = NULL;
list_t list;
list.head = &node1;
list.size = 2;

Answer 2

After re-reading your question multiple times, I probably finally understand what you are trying to achieve. You want list_t to contain multiple "lists" of type node_t.Then you need to make head an array of node_t pointers like this:
typedef struct node
{
int value;
struct node *next;
}node_t;
typedef struct list
{
node_t *head[2];
int size;
}list_t;
(i've used array of static size of 2 for simplicity. If you want to make it dynamic, you will need to take care of memory allocations on your own)
And example code:
// list1
node_t node11;
node_t node12;
node11.value = 1;
node11.next = &node12;
node12.value = 1;
node12.next = NULL;
//list2
node_t node21;
node_t node22;
node11.value = 2;
node11.next = &node22;
node12.value = 2;
node12.next = NULL;
list_t list;
list.head[0] = &node11;
list.head[1] = &node21;
list.size = 2;