Writing a LinkedList destructor?

Question

Is this a valid LinkedList destructor? I\'m still sort of confused by them.

I want to make sure I\'m understanding this correctly.

 LinkedList::~Link         


        

Answer 1

The condition "head;" should check that it is not null, but I'm not sure if it will work.

Yes, "head" by itself is the same as "head != null" -- but why use a meaningless typing shortcut if even you find it confusing? It's only 6 more keystrokes (and generates identical machine code), so go for the long form.

Additionally, your code is a bit more complicated than necessary because you are using a for() construct. Why not use a while()? Your code will be much cleaner.

Finally, I realize you are doing this as a learning exercise, but keep in mind that list<> is in the standard library --- Linked lists are official a "Solved Problem".

Answer 2

You can run it through a debugger or you can run it through that bit of wetware inside your skull - both will show you that it works fine. For example, let's start with the list:

head(47) -> [47]single_node -> [NULL]end-of-list.

Running that list through your statements:

• ptr = head sets ptr to 47.
• head is non-zero so enter loop.
• head = head->next sets head to NULL.
• delete ptr will delete the single_node.
• ptr = head sets ptr to NULL.
• head is now NULL (0) so exit loop.

There you go, you've deleted the only entry in the list and head is now set to NULL. That's all you need to do.

You can do a similar thing with a longer list or an empty list, you'll find it's still okay (there's no real difference between a one-element list and a fifty-element list).

As an aside, I'm not a big fan of treating pointers as booleans - I'd rather write it as something like:

for (ptr = head; head != NULL; ptr = head)

It makes the code read better in my opinion and you don't really sacrifice any performance (unless you have a brain-dead compiler). But that's a matter of taste.

Re your comment:

The thing that concerns me is reaching the very last node. The condition "head;" should check that it is not null, but I'm not sure if it will work.

It will work. A value of zero will be treated as false so you'll find that you never dereference head->next when head is NULL simply because you will have exited the loop body before that point (or not even entered the body if the list is empty).

Any other pointer value will be treated as true and you will enter or continue the loop body.

Answer 3

Why not do it much much simpler - with an elegant while-loop instead of trying to carefully analyze whether that overcompilcated for-loop is correct?

ListNode* current = head;
while( current != 0 ) {
    ListNode* next = current->next;
    delete current;
    current = next;
}
head = 0;

Answer 4

Tested OK

Destructor for class List

List::~List()
{
    Node *n = this->front, *current = NULL; //initialization part

    while(n)                               //start cleanup of nodes of the list
    {
        current = n;
        n=n->next;
        delete(current);
    }

    front = end = NULL;
}

Answer 5

Your code might be correct, you should try running it with e.g. valgrind and see what it says. However, I would write it like this:

for (ListNode *current = head, *next; current; current = next) {
    next = current->next;
    free(current);
}

Answer 6

This one is a better approach for freeing/deleting the memory using destructor of a Linked-List.

List()::~List()
            {
                for( Link* ptr= Head; Head; Head= Head->next)
                {
                    delete ptr;
                }
            }