How does head connected to tail when deleting the list - c

Question

I have a function which creates a list based on a given array,
this is the function:
typedef struct Item
{
int num;
struct Item* next;
}*PItem;
int main()
{
int Arr[N] = { 3, 4, 1, 0, 8 }, i;
PItem list = NULL, tail = NULL;
CreateListFromArray(&list, &tail, Arr);
}
void CreateListFromArray(PItem* head, PItem* tail, int *Arr)
{
int i;
PItem temp;
for (i = 0; i<N; i++)
{
temp = (PItem)malloc(sizeof(struct Item));
if (temp == NULL)
{
DeleteList(head);
Error_Msg("Memmory!");
}
temp->num = Arr[i];
temp->next = NULL;
if (*head == NULL)
*head = temp;
else
(*tail)->next = temp;
*tail = temp;
}
}
I understand that if List is empty, then head's null is initialized to the first allocated temp (arr[0]). But after that, for these arrays arr[1],..,arr[N], I update only the tail, meaning that all the tails from arr[1] to arr[N] are connected. but how does the head (arr[0]) POINTS/connected to arr[1]?
I ask this because, when I try to print the list, I use temp = head, and advance head until temp is null, but when I advance head, how does it know that it has to advance to arr[1]?
Here's the full code: http://pastebin.com/VPCfMU4X

Answer 1

After the first iteration of the loop, head and tail point to the same element which contains arr[0]. After the second iteration, (*tail)->next (which is the same as (*head)->next) point to the new element that contains arr[1], and tail is moved up to this value. Subsequent iterations keep appending to the end of the list.
So after one iteration, you have this:
head tail
| |
v v
---------------
| 3 | NULL |
---------------
After the second iteration, you have this:
head tail
| |
v v
--------------- ---------------
| 3 | .-----|--->| 4 | NULL |
--------------- ---------------
And the third:
head tail
| |
v v
--------------- --------------- ---------------
| 3 | .-----|--->| 4 | .-----|--->| 1 | NULL |
--------------- --------------- ---------------