question on linked lists(stack with linked lists)

Hi Guys.I have some questions for this program.
This program uses linked lists to construct a stack (lifo)
I dont understand the bold part of the code:
In general i dont understand why
Pop() and Push() use a double pointer .... :confused:

Code:

---

* lifo-list.c */
#include <stdio.h>
#include <stdlib.h>

struct node {
        int data;
        struct node *next;
};

---

Code:

---

void Push(struct node **headRef, int newData)
{
        struct node *newNode = (struct node*)malloc(sizeof(struct node));
  newNode->data = newData;
        newNode->next = (*headRef);
       
  (*headRef) = newNode;
        return;
}

---

Code:

---

int Pop(struct node **headRef)
{
        struct node *head;
        int result;
       
  head = *headRef;
        if(head != NULL) {
                result = head->data;
                *headRef = head->next;
                free(head);
                return(result);
        } else {
                printf("Pop() - Underflow !!\n");
                return(0);
        }
}

---

Code:

---

int Length(struct node *head)
{
        int count = 0;
        struct node* current = head;
       
  while(current != NULL) {
                count++;
                current = current->next;
        }
        return(count);
}

---

Code:

---

int main()
{
        struct node *stack = NULL;
        int x, i, n;
        printf("\n Enter sequence of integers (EOF to stop): ");
        while(scanf("%d", &x) == 1) {
                printf("  %d values\n",&stack);
                Push(&stack, x);
        }

        n = Length(stack);
        printf("\t Stack contains %d values\n", n);
        printf("\n Stack values (in reverse order) :\n");
        for(i = 0; i < n; i++) {
                printf("\t value #%d: [%d]\n", i, Pop(&stack));
        }
printf("\n Calling Pop() with empty stack: value=%d\n", Pop(&stack));
        return(0);
}

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They're being passed a pointer to the list itself, so they can update the head of the list. Remember, any time you want to change something in a function, you need a pointer to it. Thus, if you want to change the pointer that stored your list, you need a pointer to that list so it can be changed inside the function call itself. (You can of course do it this way...)

Code:

---

Node *popanddestroy( Node *list )
{
    Node *n;
    n = list->next; /* point to the top->next node so we can keep track of it */
    free( list ); /* free the top node */

    return n; /* return the new "top" */
}

...

Node *list = constructabiglistfunction( );

...

list = popanddestroy( list ); /* get rid of the top one, assign the list the next one */

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This would inside the function, strip the top node of the list. It then returns the "next" one, which is assigned via a return value to the "list" pointer.

Or, by using a pointer to a pointer, you do it in one shot:

Code:

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Node *list = makeabiglist( );

pop( &list ); /* strips the top one off, does something to it, updates the list itself */

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Quzah.