How to display all list values without endless loop - c++

Question

I have a list and I want to display it's values.
I want to see 1 2 3 4, but I have a endless loop like 1 2 3 4 1 2 3 4 1 2..
Can't understand, why?
struct node
{
int item;
node *next;
node(int x, node *t)
{
item = x;
next = t;
}
};
int main()
{
node *firstElement = new node(1, NULL);
firstElement->next = firstElement;
node *lastElement = firstElement;
for (int i = 2; i < 5; i++)
lastElement = (lastElement->next = new node(i, firstElement));
for (node *first = lastElement; first != 0; first = first->next)
cout << first->item << " ";
delete firstElement;
return 0;
}

Answer 1

Try using this code:
struct node
{
int item;
node *next;
node(int x, node *t)
{
item = x;
next = t;
}
};
int main()
{
node *firstElement = new node(1, NULL);
node *lastElement = firstElement;
for (int i = 2; i < 5; i++)
lastElement = (lastElement->next = new node(i, nullptr));
for (node *first = firstElement; first != 0; first = first->next)
cout << first->item << " ";
return 0;
}
IdeOne live code
The problem is that you set the "next" link of your last node to this node itself, not nullptr.
Also, it's better to delete the memory allocated

Answer 2

The problem is that your data structure has an infinite loop in itself: this line
firstElement->next = firstElement;
makes firstElement point back to itself, creating a circular list. When you add more elements, your list remains circular, so exit condition first == 0 is never achieved.
If you want your list to remain linear, not circular, your insertion code should be modified as follows:
node *firstElement = new node(1, NULL);
node *lastElement = firstElement;
for (int i = 2; i < 5; i++) {
lastElement->next = new node(i, lastElement->next)
lastElement = lastElement->next;
}
The printing code should start with firstElement:
for (node *current = firstElement; current != 0; current = current->next)
cout << current->item << " ";
Finally, deleting a single firstItem is not sufficient. You need a loop to traverse the whole list. Alternatively, you could chain deletion in the destructor by calling delete next, but this is dangerous, because recursive invocation of destructors may overflow the stack.

Answer 3

You have a loop in your list, because lastElement->next always points to firstElement. This is why first will never be equal to 0.
If you really need a loop I think you should write something like this:
node* p = firstElement;
do {
cout << p->item << " ";
p = p->next;
} while (p != firstElement);

Answer 4

The problem is that you create every node with firstElement as its next.
This would make sense if you were adding nodes to the front of the list, but you're adding them at the back, so the last node will point back to the start.
Since you're adding to the back, terminate the list on every insertion instead:
lastElement->next = new node(i, nullptr))