free()ing a list

consider two functions for free()ing a linked list:

Code:

---

void freelist(struct node *root)
{
  struct node *index, *save;
  if (root != NULL)
    for (index=root; index!=NULL; index=save)
    {
      save=index->next;
      free(index);
    }
}

---

second:

Code:

---

void freelist(struct node **root)
{
  struct node *save;
  if (*root != NULL)
    for ( ; *root != NULL; *root = save)
    {
      save = (*root)->next;
      free(*root);
    }
}

---

when i call the first function to free the list and then try to print it i get something like:

Code:

---

134518592
134518608
134518624
134518640
134518656
134518672
134518688
134518704
134518720

---

what's that? the memory addresses of the data in the nodes? but they were freed already.

when trying to printf the list after the call to the second freelist function, i simply get 'Empty List'.
So, why is the output different? which function should i use, or are they both ok?

>why is the output different?
Because the first function doesn't end up setting root to null, the second does. Your print function no doubt tests for a null list and prints an appropriate message if the root is null.

>which function should i use, or are they both ok?
They're both fine for releasing the memory in a list, but you should consider what happens if you try to use the root after calling freelist. A good idea would be to have a list structure rather than just a simple node as your list so that you can maintain a consistent state without having to rely on pointers to pointers, like so:

Code:

---

struct list {
  struct node *root;
  /* Other handy stuff */
};

---

Then you can define freelist like this:

Code:

---

void
freelist(
  struct list *list
  )
{
  if (list && list->root) {
    struct node *index, *save;
    for (index = list->root; !index; index = save) {
      save = index->next;
      free(index);
    }
    list->root = NULL;
    /* Reset other handy stuff */
  }
}

---

i see, it's like having a special structure to hold info about the list, not the list itself, cool.
here:

Code:

---

  if (!list && !list->root) {

---

you meant

Code:

---

  if (list && list->root) {

---

right?
thanks Prelude :) .

>you meant
Yes, I did. Thank you. That's the result of writing the function my way originally and then hastily changing it to mirror how you wrote it. ;)