使用 XOR 操作仅使用 2 个指针反转链表

Question

我在网上找到了这段代码，用于仅使用 xor 操作使用 2 个指针来反转链表：

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; 
} 

你能解释一下这段代码是如何工作的吗？

Answer 1

你读过这个吗：仅使用 2 个指针迭代反转链表？

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)); 
}