在vs调试里，一个链表节点，怎么会有不一样的值

如图，head2是一个值为2的单链表节点，head1是值为4的单个链表节点，我疑惑的是为什么cur->next和d明明是同一个变量，在调试时却是一个为2，一个没有初始化。完整代码有点长，不过不重要，是实现一个链表的归并排序。

我想要达到的结果

```c++
struct ListNode {
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
 };

 class Solution {
 public:
     ListNode* sortList(ListNode* head) {
         if (head == NULL || head->next == NULL)
             return head;
         int a = head->val;
         ListNode* fast = head;
         ListNode* slow = head;

         ListNode* brk = nullptr;
         while (fast != NULL && fast->next != NULL) {
             fast = fast->next->next;

             if (fast == NULL || fast->next == NULL)
                 brk = slow;
             slow = slow->next;
         }

         brk->next = nullptr;
         ListNode* head1 = sortList(head);
         int b = head1->val;
         ListNode* head2 = sortList(slow);
         int c = head2->val;

         ListNode dummy(0);
         ListNode* cur = &dummy;
         while (head1 != nullptr || head2 != nullptr) {
             if (head1 == nullptr || (head1 != nullptr && head2 != nullptr && head1->val >= head2->val)) {
                 cur->next = head2;
                 head2 = head2->next;
                 int e = cur->next->val;
                 cur = cur->next;
                 int d = cur->val;
             }
             else {
                 cur->next = head1;
                 head1 = head1->next;

                 cur = cur->next;
             }
         }

         return dummy.next;
     }
 };

Solution solution;
int main() {
    ListNode * four = new ListNode(3, nullptr);
    ListNode * three = new ListNode(1, four);
    ListNode  *two = new ListNode(2,three);
    ListNode* head = new ListNode(4, two);
    ListNode* temp = head;
    solution.sortList(head);
    return 0;
}

其实我觉得纠结这些东西没多大用处，但就是想搞清楚为什么