Thread: Question on sample code

Originally Posted by john.c

Why do you think that's what would happen?
Where's your test code?

I added a few lines to the main to declare a new list2 and pushed 10,20 and 30 to it. I expected list to be still printing out 1, 2 and 3. But I am seeing the output for traverse as 1,2,3,10,20,30.

below is the code added in main()

Code:

//Driver program
int main()
{
    LinkedList<int> list;
    LinkedList<int> list2;

    // Add few items to the end of LinkedList
    list.push_back(1);
    list.push_back(2);
    list.push_back(3);

    list2.push_back(10);
    list2.push_back(20);
    list2.push_back(30);

    cout << "Traversing LinkedList through method" << endl;
    list.Traverse();
    cout << endl;

    cout << "Traversing LinkedList through Iterator" << endl;
    for (LinkedList<int>::Iterator iterator = list.begin();
        iterator != list.end(); iterator++)
    {
        cout << *iterator << " ";
    }

    cout << endl;

    return 0;
}

Originally Posted by john.c

You're right. It's because they made m_spRoot static for some reason.
Just remove the word static from in front of the declaration of m_spRoot,
and remove the bit of code after the class that initializes m_spRoot to nullptr.
Just initialize it to nullptr in the ctor (which is already being done anyway).
There should probably be a destructor.

Thanks for the response. Follow up question:

How does this work?

Code:

// Root of LinkedList wrapped in Iterator type
    Iterator begin()
    {
        return Iterator(m_spRoot);
    }

the above call looks like a call to the parameterized constructor which is defined as below

Code:

Iterator(const Node* pNode) noexcept :

        m_pCurrentNode (pNode) { }

Which means all it does is set m_pCurrentNode to a node*. But the Iterator begin() function expects a return value of type Iterator. We are just setting a private variable value to something and returning void. How is this working? Why isn't the compiler complaining when we are not returning an Iterator?