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Kye Singh
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Asked: 2022-05-18T16:41:19+00:00 In:

How do i write destructor for linked list c++?

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How to write destructor for linked list c++? Below is my code

LinkedList::~LinkedList()
  {
  ListNode *ptr;
 
  for (ptr = head; head; ptr = head)
  {
  head = head->next
  delete ptr;
  }
 }

Please give me a few practical suggestions.

linked list c++
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2 Answers

  1. Best Answer
    Expert

    Instead of attempting to carefully analyze whether that overcompilated for-loop is right, why not make it much simpler with an elegant while-loop?

    ListNode* current = head;
while( current != 0 ) {
    ListNode* next = current->next;
    delete current;
    current = next;
}
head = 0;
    • 14

  2. It can be run through a debugger, or through the wetware in your skull. Both will confirm that it works. Let’s take, for example, the following list. 

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

    This list can be used to generate statements.

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

    You’ve now deleted the last entry from the list. head is now set as NULL. This is all there is to it.

    It’s possible to do the same thing with a longer or empty list. You’ll still be able to use it. There’s no difference between a one-element and fifty-element listing.

    A side note: I don’t like pointers being treated as booleans. I would rather have it written as follows: 

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

    This makes code more readable and doesn’t sacrifice performance (unless you are a brain-dead programmer). It’s all a matter taste.

    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. It will work. A zero value will be considered false. You’ll see that head->next is never used when head is NULL. This is because you have already exited the loop body (or not entered the body, if it is empty).

    Any other pointer value you provide will be considered true, and you will enter the loop body or continue it.

    • 16

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