Implementing pop function with doubly linked list - c++

Question

void Stack::pop()
{
Node* temp;
if(top == NULL)
return;
temp = top;
top = top->prev;
top = top->next;
delete temp;
}
So I am having trouble implementing the pop function in a doubly linked list. Whenever I build my compiler crashes. The code above is what I'm struggling with. I am really unsure on how to actually unlink my top node from my list. If you guys can explain the pop function in a doubly linked list that would be great.
void Stack::ToString()
{
//needs to be implemented
Node* cursor = top; //create a temporary pointer for top
while(cursor->next != NULL) //When debugging this is where it crashes.
cursor = cursor->next;
while(cursor != NULL)
{
cout << cursor->content << endl;
cursor = cursor->prev;
}
}

Answer 1

The logic in the function is flawed.
Let' say you have two node in the linked list:
top -> +--+
| |
+--+
(next) ^ | (prev)
| v
n2 -> +--+
| |
+--+
Then, top->prev = n2 and n2->next = top.
When you execute:
top = top->prev; // Now top is n2
top = top->next; // Now top is what it used to be start with.
You did nothing in the list to pop top from the top of the list.
The fix to this problem depends on whether the list is circular or not.
If the list is not circular, you can do:
Node* prev = top->prev;
Node* next = top->next;
if ( prev != NULL )
{
prev->next = next;
}
if ( next != NULL )
{
next->prev = prev;
}
top->next = NULL;
top->prev = NULL;
If the list is circular, the logic needs to be a little bit more elaborate.

Answer 2

Imagine the list like this pic:
And the code is clear to understand.
struct Node {
int data;
struct Node* next;
struct Node* prev;
};
void Pop(struct Node *&head) {
if (head != NULL) {
// if list hase just one item, set head to null
if (head->next == NULL) {
head = head->next;
}
else {
// copy head to temp node
Node *tmp = new Node;
tmp = head;
// traverse list to raech end of it
while (tmp->next->next != NULL)
tmp = tmp->next;
// deallocate last node
delete tmp->next;
tmp->next = NULL;
}
}
}
For printing the list:
void Print(struct Node *&head) {
if (head == NULL) {
std::cout << "list is empty!\n";
}
else {
while (head != NULL) {
std::cout << head->data << " ";
head = head->next;
}
}
}