当用作返回值时，临时堆栈分配的指针会发生什么？

Question

I'm doing some basic leetcode questions. Here I'm trying to swap pairs of a singly-linked list using recursion. The code below passes the tests but some point eludes me. new_head is a pointer created on the stack. I understand it means that once the function returns it is cleaned up and can potentially point to garbage. Is it correct to assume that here it works "by accident" and is not correct way to do it or is my understanding wrong?

/**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     struct ListNode *next;
     * };
     */
    
        struct ListNode* swapPairs(struct ListNode* head){
            
        if (head == NULL || head->next == NULL) {
            return head;
        }    
            
            struct ListNode* new_head;
        
            new_head = head->next;       
            head->next = swapPairs(head->next->next);
            new_head->next = head;
          
            return new_head;
        
                        
        }

Another question related to the code above:

if I change the order of assignments I get a stack overflow but I can't wrap my head around the reason why

        new_head = head->next;
        new_head->next = head;       
        head->next = swapPairs(head->next->next);

Nothing that is touched in this line new_head->next = head; has an effect on what happens inside the recursion no (well it must have but I missing it)?

Answer 1

First Question

Return return new_head; does not return the object new_head to the caller. It returns the current value of new_head to the caller. That is fine.

Second Question

With:

new_head = head->next;       
head->next = swapPairs(head->next->next);
new_head->next = head;

at the time swapPairs is called, the value passed to it, head->next->next , is the address of some node beyond head->next in the list.

With:

new_head = head->next;
new_head->next = head;       
head->next = swapPairs(head->next->next);

at the time swapPairs is called, the value passed to it, head->next->next , is head , because new_head->next = head; just set head->next->next to head .