c - reverse a linkedlist recursively -

my linkedlist reverse function gives correct result. am
confused.

linked list= 12->5->4->3  per reverse function result should 4->5->12  fortunately produces correct reversed list 3->4->5->12.  pls me understand happening      //head_ref  global variable    struct n* reverse(struct n *head){      struct n *pre,*cur;  //temp variable first , second node      if(head->next==null){         head_ref = head;   // head_ref global variable initialized head pointer         return;      }     pre = head;     cur = head->next;     reverse(cur);      cur->next = pre ;     pre->next = null;   } 

first call pre = 12 cur = 5 second call pre = 5 cur = 4 3rd call pre = 4 cur = 3

   3->next = null //base condition fullfilled    exit ( 3rd call )     reverse start     second call       pre = 5       cur = 4 

my reversed linked list should 4->5->12

but produces 3->4->5->12 (correct reversed linked list)

why gives correct result. pls explain????

let start linked list 12->5->4->3. when main (or other calling function) calls reverse function, head has data 12. following stack created after forth call reverse function.

                 |                                           |                  |___________________________________________|     gonna                  |__*head=3 *head->next=null__head_ref=3_____| => popped out                  |__*head=4 pre=4 curr=3_____________________|                     |__*head=5 pre=5 cur=4______________________|      main calls=>|__*head=12_pre=12 cur=5____________________|  

as said, clear above figure. when have head->next = null, top-most entry in stack gets popped out , 1 *head=4, pre=4 , curr=3 becomes topmost entry in stack. curr->prev evaluates 3->next=4 , 4->next->null.

                 |                                                      |                  |                                                      |     popped out=> |______________________________________________________|                  |__*head=4 pre=4 curr=3__(3->next)=4__(4->next)=null___|                  |__*head=5 pre=5 cur=4_________________________________|                  |__*head=12_pre=12 cur=5_______________________________|  

then entry *head=4 pre=4 curr=3__(3->next)=4__(4->next)=null pops out , same thing happens.

                 |                                                      |                  |                                                      |                  |                                                      |     popped out=> |______________________________________________________|                  |__*head=5 pre=5 cur=4____(4->next)=5___(5->next=null)_|                  |__*head=12_pre=12 cur=5_______________________________|  

in remaining 1 entry can see 12->next becomes null, 12 becomes tail of linked list.

                 |                                                      |                  |                                                      |                  |                                                      |                  |                                                      |     popped out=> |______________________________________________________|                  |__*head=12_pre=12 cur=5__(5->next)=12_(12->next)=null_|                     |                                                      |                  |                                                      |                  |                                                      |                  |                                                      |                  |                                                      |     popped out=> |______________________________________________________|  

stack becomes empty , final linked list has 3 head , 12 tail:3->4->5->12