I found this code online for reversing a linked list using only 2 pointers using xor operation :
void reverse(struct Node** head_ref)
{
struct Node* prev = NULL;
struct Node* current = *head_ref;
// at last prev points to new head
while (current != NULL) {
current = (struct Node*)((ut)prev ^ (ut)current ^ (ut)(current->next) ^ (ut)(current->next = prev) ^ (ut)(prev = current));
}
*head_ref = prev;
}
Can you please explain how this code works ?
Have you read this: Iteratively Reverse a linked list using only 2 pointers?
while (current != NULL) {
// This expression evaluates from left to right
// current->next = prev, changes the link fron
// next to prev node
// prev = current, moves prev to current node for
// next reversal of node
// This example of list will clear it more 1->2->3->4
// initially prev = 1, current = 2
// Final expression will be current = 1^2^3^2^1,
// as we know that bitwise XOR of two same
// numbers will always be 0 i.e; 1^1 = 2^2 = 0
// After the evaluation of expression current = 3 that
// means it has been moved by one node from its
// previous position
current = (struct Node*)((ut)prev ^ (ut)current ^ (ut)(current->next) ^ (ut)(current->next = prev) ^ (ut)(prev = current));
}
(ut)(current->next = prev) ^ (ut)(prev = current), is storing addresses (uintptr_t is an integer type capable of holding a pointer) in a temporary using xor operations while assigning the previous node (stored in the previous iteration). Is obfuscated code and add nothing to the conventional way of doing it: geeksforgeeks.org/reverse-a-linked-list/?ref=rpNULL (0) with the address of the first node: 0 ^ 1 (xoring any value with 0 is a NOOP), then 1 ^ 2, then 2 ^ 3 and so on ...^ is swapping addresses not values
prev/prev = current, andcurrent->next/current->next = prev. geeksforgeeks is not a reliable source of knowledge.