C: How to free nodes in the linked list?

Question

An iterative function to free your list:

void freeList(struct node* head)
{
   struct node* tmp;

   while (head != NULL)
    {
       tmp = head;
       head = head->next;
       free(tmp);
    }

}

What the function is doing is the follow:

1. check if head is NULL, if yes the list is empty and we just return

2. Save the head in a tmp variable, and make head point to the next node on your list (this is done in head = head->next

3. Now we can safely free(tmp) variable, and head just points to the rest of the list, go back to step 1

Simply by iterating over the list:

struct node *n = head;
while(n){
   struct node *n1 = n;
   n = n->next;
   free(n1);
}

One function can do the job,

void free_list(node *pHead)
{
    node *pNode = pHead, *pNext;

    while (NULL != pNode)
    {
        pNext = pNode->next;
        free(pNode);
        pNode = pNext;
    }

}

struct node{
    int position;
    char name[30];
    struct node * next;
};

void free_list(node * list){
    node* next_node;

    printf("\n\n Freeing List: \n");
    while(list != NULL)
    {
        next_node = list->next;
        printf("clear mem for: %s",list->name);
        free(list);
        list = next_node;
        printf("->");
    }
}

You could always do it recursively like so:

void freeList(struct node* currentNode)
{
    if(currentNode->next) freeList(currentNode->next);
    free(currentNode);
}