How exactly does a XOR Linked list work?

Question

The following link explains it.
The implementation is said to work by storing the XOR of the previous and next address(say nxp), instead of storing both(previous and

Answer 1

Let us consider the following XOR list

A->B->C->D

suppose you created nodes in this format below

Key|Link|

A|0^addr(B)| ->  B|addr(A)^addr(C)|  ->  C|addr(B)^addr(D)| -> D|addr(C)^0|

CASE #1:[Forward Traversal] Now Suppose you are in B (current_node=>B) want visit C , so you need Address of C . How you will get ?

Addressof(Next_node) = addressof(Prev_node) ^ Current_node(Link)

addr(A)^ ( addr(A)^ addr(C) )
=>(addr(A) ^ addr(A)) ^ addr(C) 
=> 0 ^ addr(C)
=>addr(C)

CASE #2: [Backward traversal] Now Suppose you are in C (current_node=> C) want visit B , so you need Address of B . How you will get ?

Addressof(Prev_node) = addressof(Next_node) ^ Current_node(Link)

addr(D) ^ ((addr(B) ^ addr(D)) 
=> (addr(D)^ addr(D)) ^ addr(B)
=> 0^addr(B)
=> addr(B)

Traversing: To traverse whole list ,You will need 3 pointers prevPtr , currPtr , nextPtr to store relative current, previous and next node's address starting with head. Then in each iteration these pointers need be move to one position ahead.

struct Node *currPtr = head;
struct Node *prevPtr = NULL;
struct Node *nextPtr;

printf ("Following are the nodes of Linked List: \n");

while (currPtr != NULL)
{
    // print current node
    printf ("%d ", currPtr->key);

    // Save the address of next node
    nextPtr = XOR (prevPtr, currPtr->link);

    //move prevPtr and currPtr one position for next iteration

    prevPtr = currPtr;
    currPtr = nextPtr;
}