free linked list

[lambs4]

How do I free a linked list and would the code below be the answer to my question?

Thank you

struct node {
 int data;
 struct node* next;
};

int main()
{
 struct node* head = NULL;
 struct node* second = NULL;

 head = malloc(sizeof(struct node));
 second = malloc(sizeof(struct node));

 head->data = 1;
 head->next = second;

 second->data = 2;
 second->next = NULL;

 free(head);

 return 0;
}

[Hammer]

Remember that everything you malloc(), you must free(), so given this rule, it's clear that your program doesn't do the job.

For a linked list, you should traverse the list, free()'ing each node as you go.

[lambs4]

For a linked list, you should traverse the list, free()'ing each node as you go.

Then does this code accomplish that problem?

struct node* current = head;
while(current != NULL)
{
    free(current->next)
    current = current->next;
}

[Hammer]

struct node* current = head;
while(current != NULL)
{
    free(current->next)
    current = current->next;
}

No, this has a nasty bug. Once you free() memory, you shouldn't reference it again. So in your example, you free()'d the next node, and then pointed to it.

Try something like this:

struct Node *Current = Head;
struct Node *tmp;

while (Current)
{
	tmp = Current->Next;
	free(Current);
	Current = tmp;
}