C - linked list reverse recursively: why use double pointer?

Question

I understand the logic when the struct Node *head is a global variable.

However, when I do the reverse using struct Node *head as a local variable in main() , I have to use double-pointer and pointer of Head , and I don't understand where exactly I have to place them.

void Reverse(struct Node *Head, struct Node **headpointer) {
    struct Node *first;
    // when the list is empty
    if (Head == NULL)
        return;

    first = Head;
    // when there is one node left
    if (first->next == NULL) {
        *headpointer = first;
        return;
    }

    Reverse(first->next, headpointer);

    first->next->next = first;
    first->next = NULL;
}

I am unclear why I have to use...

first = Head;
*headpointer = first;

why can't I just use

Head = first? 

Also, if first = Head line in front of the recursive function call Reverse(first->next, headpointer) , don't the first->next value also equal to Head which points to the first node?

Is there any good logical diagram/pic/explanation/examples that can explain this difference?

Answer 1

I am unclear why I have to use... first = Head

Actually, the first variable isn't needed in this code at all. It is just a copy of Head , so to simplify you can just replace the variable with Head . Everything will work the same without it:

if (Head == NULL)
  return;

if (Head->next == NULL) {
  *headpointer = Head;
  return;
}

Reverse(Head->next, headpointer);

Head->next->next = Head;
Head->next = NULL;

I am unclear why I have to use... *headpointer = Head

The Head variable is the pointer to the head of the list you want to reverse, and the function stores the head of the newly reversed list into *headpointer . It does this because Head is just a copy of the pointer that was passed in to the Reverse() function, so updating its value won't change the original pointer; that's why a separate double pointer variable is used.

why can't I just use Head = first ?

Head is a copy of the pointer that was passed to the function. Updating the Head pointer will not update the original list pointer that you passed in. Also, as I said before, Head is the same as first , so an assignment like that does nothing.

Also, if first = Head line in front of the recursive function call Reverse(first->next, headpointer) , don't the first->next value also equal to Head which points to the first node?

first->next (or Head->next ) is just the next node in the list. The purpose of the function call is to reverse the rest of the list first, and then place Head (or first ) at the end of the list (which is what the last two lines do).

Answer 2

Let's consider the function signature was like this:

void Reverse(struct Node* Head, struct Node* headpointer) {

And you call it like this

Reverse(myList);

myList is just an address to a Node, for example 0x1234 . So it's equivalent to do:

Reverse(0x1234);

The address is copied to a new variable headpointer . When we modify headpointer we are modifying a local variable, not the myList we passed.

It's like if we did this:

struct Node* myList = 0x1234;
struct Node* headpointer = myList;
headpointer = 0xABCD;
// at this point myList is still 0x1234

So after the function returns, myList is still equal to 0x1234 . That's not what we want because is should now point to the last node.

So how do we allow, the function to modify myList ? We have to tell the function "hey, here's the address where you have to write to".

In C, in order to take the address of something we use the '&' operator:

Reverse(&myList);

And we must change the signature of the function accordingly:

void Reverse(struct Node* Head, struct Node** headpointer) {

Now headpointer is an address to an address to a Node . Or, as we say in C, a pointer to a pointer to a Node .

Here's a final example that can help understand.

struct Node* myList = 0x1234;
struct Node** headpointer = &myList; // now headpointer points to myList
*headpointer = 0xABCD;
// at this point myList is still 0xABCD
// we haven't changed headpointer, but the thing headpointer is pointing to!

Answer 3

It is necessary to be able to return the updated head node, but this could be done with a return value in lieu of using a double pointer.

Also, I believe that the function needs to be passed a pointer to the previous node so it can set that as the new next pointer.

So, I think your function would need three arguments (eg) cur , prev , headpointer .

If we return the head pointer, we can reduce this to two .

---

Here's a version that does that.

I've tried to annotate it as much as possible. I've added debug printf and some cross checking [which I used to debug it myself].

If you need to have a double pointer, this example could be adapted without too much trouble.

// revlist -- reverse singly linked list recursively

#include <stdio.h>
#include <stdlib.h>

typedef struct node {
    struct node *next;
    int data;
#ifdef CHECK
    int visited;                    // 1=node visited (check for looped list)
#endif
} Node;

#ifdef DEBUG
int dbglvl;                         // nesting level -- debug only
#endif

int
prt(Node *node)
{
    int data;

    if (node != NULL)
        data = node->data;
    else
        data = -1;

    return data;
}

#ifdef DEBUG
#define dbgprt(_fmt ...) \
    do { \
        printf("DBG/%d: ",dbglvl); \
        printf(_fmt); \
    } while (0)
#else
#define dbgprt(_fmt ...) \
    do { } while (0)
#endif

// reverse -- recursively reverse list
// RETURNS: pointer to head of reversed list
Node *
reverse(Node *cur,Node *prev)
// cur -- pointer to current node
// prev -- pointer to previous node
{
    Node *next;
    Node *head = NULL;

    do {
        // empty list
        if (cur == NULL)
            break;

        next = cur->next;

        dbgprt("BEG cur=%d prev=%d next=%d\n",
            prt(cur),prt(prev),prt(next));

        // at end of list -- set new head from tail node
        if (next == NULL) {
            head = cur;
            head->next = prev;
            dbgprt("SETHEAD head=%d head->next=%d\n",
                prt(head),prt(head->next));
            break;
        }

#ifdef DEBUG
        ++dbglvl;
#endif
        // process the next node and give the current node as the previous one
        head = reverse(next,cur);
#ifdef DEBUG
        --dbglvl;
#endif

        // set the link pointer to the previous node
        cur->next = prev;
        dbgprt("POST cur->next=%d\n",prt(cur->next));
    } while (0);

    dbgprt("EXIT head=%d\n",prt(head));

    return head;
}

// addnode -- add node to tail of list
Node *
addnode(Node *head,int data)
{
    Node *newnode = malloc(sizeof(*newnode));
    Node *cur;
    Node *prev;

    newnode->next = NULL;
    newnode->data = data;
#ifdef CHECK
    newnode->visited = 0;
#endif

    // find the tail of the list
    prev = NULL;
    for (cur = head;  cur != NULL;  cur = cur->next)
        prev = cur;

    // add new node to list
    if (prev != NULL)
        prev->next = newnode;
    else
        head = newnode;

    return head;
}

// list_print -- print list
void
list_print(Node *head,const char *reason)
{
    Node *cur;

    printf("%s:",reason);

    for (cur = head;  cur != NULL;  cur = cur->next) {
        printf(" %d",cur->data);
#ifdef CHECK
        if (cur->visited) {
            printf(" DUP\n");
            exit(1);
        }
        cur->visited = 1;
#endif
    }

    printf("\n");

#ifdef CHECK
    for (cur = head;  cur != NULL;  cur = cur->next)
        cur->visited = 0;
#endif
}

// list_destroy -- destroy the list
void
list_destroy(Node *cur)
{
    Node *next;

    for (;  cur != NULL;  cur = cur->next) {
        next = cur->next;
        free(cur);
    }
}

// dotest -- test reverse function
void
dotest(int max)
{
    Node *head = NULL;

#ifdef DEBUG
    dbglvl = 0;
#endif

    // create a list
    for (int idx = 1;  idx <= max;  ++idx)
        head = addnode(head,idx);

    // show the original list
    list_print(head,"Forward");

    // reverse the list
    head = reverse(head,NULL);

    // show the reversed list
    list_print(head,"Reverse");

    // destroy the test list
    list_destroy(head);
}

int
main(void)
{

    dotest(8);

    return 0;
}

---

Here's the program output:

Forward: 1 2 3 4 5 6 7 8
Reverse: 8 7 6 5 4 3 2 1

---

Here's the program output with the debug printf enabled:

Forward: 1 2 3 4 5 6 7 8
DBG/0: BEG cur=1 prev=-1 next=2
DBG/1: BEG cur=2 prev=1 next=3
DBG/2: BEG cur=3 prev=2 next=4
DBG/3: BEG cur=4 prev=3 next=5
DBG/4: BEG cur=5 prev=4 next=6
DBG/5: BEG cur=6 prev=5 next=7
DBG/6: BEG cur=7 prev=6 next=8
DBG/7: BEG cur=8 prev=7 next=-1
DBG/7: SETHEAD head=8 head->next=7
DBG/7: EXIT head=8
DBG/6: POST cur->next=6
DBG/6: EXIT head=8
DBG/5: POST cur->next=5
DBG/5: EXIT head=8
DBG/4: POST cur->next=4
DBG/4: EXIT head=8
DBG/3: POST cur->next=3
DBG/3: EXIT head=8
DBG/2: POST cur->next=2
DBG/2: EXIT head=8
DBG/1: POST cur->next=1
DBG/1: EXIT head=8
DBG/0: POST cur->next=-1
DBG/0: EXIT head=8
Reverse: 8 7 6 5 4 3 2 1