How do I return to bottom element in a singly linked list stack?

[bmos]

How would I go about finding the first/bottom element 'pushed' into the linked list stack?

would I need to copy the stack, then pop the elements until I reach the first element?

I'm not even sure how I would go about doing this. The only thing I can think of is traversering the linked list, but how would that without traversing the entire list? Or would I copy the original list to a temp list, then pop elements off the stack until I reach the last one?

Here's what I have so far (this one attempts at traversing)

StackElement Stack::bottom(const Stack & original) const 
{
    bottomPtr = myTop, //myTop is top element on stack
    origPtr = original.myTop->next;

    if(!empty())
    {
        while(origPtr != 0)   
        {
              //THIS IS WHERE I'M STUCK, I DONT WANT TO TRAVERSE  
              //THE ENTIRE LIST
            return bottomPtr;
        }
    }
    else
    {
        cerr << "stack is empty -- returning garbage\n";
        StackElement * temp = new(StackElement);
        StackElement garbage = *temp;
        delete temp;
        return garbage;
    }
}

[onoffon]

i don't think there is another way than to traverse the entire list.as the list is dynamic, we have traverse using pointers and hence traverse each node.:)

[Ancient Dragon]

Add another pointer -- call it tail or something -- that keeps track of the last node. Then when you need the end of the list it will be easily available without taking the time to traverse the entire list.