Question

My instructor defined the function remove() as:

struct node
{
  node *next;
  int value;
}

int IntList::remove()
{
  node *victim = first;
  int result;

  if(isEmpty()) throw listIsEmpty();

  first = victim->next;
  result = victim->value;
  delete victim;
  return result;
}

Where first “points to the sequence of nodes representing this IntList.”

If victim and first both point to the same thing, and we delete victim, doesn’t this delete first as well?

Answer 1

If victim and first both point to the same thing, and we delete victim, doesn’t this delete first as well?

At the time of deletion, they do not point to the same thing.

Basically what’s happening here is that the the second node is becoming the first node, and then the old first node is being deleted.

Since first = victim->next before the delete, first == victim will never be the same unless victim->next == victim which should never happen in a linked list.

---

For a way simplified example, assume:

first is at address 0xDEAD (crazy address, but go with it)
first->next is at address 0xF00D

Then it goes down like this:

node *victim = first; //victim = 0xDEAD
first = victim->next; //first = 0xF00D, so first now = first->next
delete victim; //victim = 0xDEAD, so you deleted the original first