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 ....

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);
}

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 */

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:

Node *list = makeabiglist( );

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

Quzah.