How does `operator delete` deallocate the memory block for node of LinkedList

Question

struct node
{
    node(int _value, node* _next) : value(_value), next(_next) {}

    int   value;
    node* next;
};

node *p = new node(10, NULL);
delete p;

Based on my understanding, operator delete will first call the destructor of node, then deallocate the raw memory block originally allocated for p.

Since struct node doesn't provide a customized destructor, the compiler will provide a default one.

Question1>what does the default destructor look like? For example,

node::~node()
{
  next = NULL;
}

Question2>should we define a destructor for struct node or not? I assume that we don't have to explicitly provide such a destructor. The reason is that the member variable next doesn't own the pointed resource and it behaves as a weak_ptr. Is that correct?

score 3 · Accepted Answer · edited May 23 '17 at 10:08

Sounds correct to me.

Because struct node is a "Plain Old Datatype" the compiler won't actually create any destructor code at all, there's no need to specially destruct ints or pointers.

Even for non-POD types, the compiler default destructor will just take the form:

~node() { }

Individual members will still have their destructors called, but if you didn't allocate any non-automatic resources, you generally don't need to provide a destructor. There are situations where you might, but they're somewhat special purpose.

You may want to refer to this question on the rule of three.

How does `operator delete` deallocate the memory block for node of LinkedList

1 Answers1