understanding pointers

[cppnoob25]

So I've been playing around with linked list adding functions here and there just to test my conceptual understanding of them.

This is just a snippet of code from the list class I've made.

void List::push_back(int d)
{
    Node * n = new Node;
    n->next = 0;
    n->data = d;

    if(head != 0)
    {
        current = head;
        while(current->next != 0)
        {
            current = current->next;
        }
        current->next = n;
    }
    else{
        head = n;
    }

}

In this code, how exactly does this work.

n->next = 0;

to my understanding this is the same as this:

(*n).next = 0;

In my node class next is declared as follows:

Node * next;

I'm having a hard time wrapping my head around how this is working. What I believe is that since next is a pointer of type node, that next stores the address 0 inside next which is inside node n, right?

[Disch]

A pointer holds an address. Hopefully, what is at that address is an object of the desired type... so if you do things properly, a Node* pointer will always point to a Node object.

The notable exception is if the pointer contains a "null" address, or an address of 0. In which case it is assumed to point to "nothing". For a linked list, this is useful because you can check to see if a node is null to know whether or not there are any nodes after it in the list.

With that in mind:

void List::push_back(int d)
{
    Node * n = new Node;    // This creates a new unnamed node.  Let's call it "foo".
                            //   Our pointer 'n' now points to the 'foo' node.
    n->next = 0;            // Set foo's 'next' member to point to nothing, indicating that 'foo'
                            //   is to be the last node in our list.
    n->data = d;            // Set foo's data member to hold the appropriate data.

    // Now that we have a node prepped, we need to actually place it at the end of our existing list
    if(head != 0)       // if there are ANY nodes in the list
    {
        // Then find the end of the list.  This loop walks through the list until it finds
        //   the last node.  We know we found the last node once we find the null 'next' pointer.
        current = head;
        while(current->next != 0)
        {
            current = current->next;
        }
        
        // at this point, 'current' now points to the last node in the list.  We no longer
        //   was that to be the last node in the list... we want 'foo' to be the last node.
        //   So change the last node's 'next' pointer to point to foo:
        current->next = n;
        
        // to recap, where we once had this in our list:
        //      first -> second -> third -> 0
        //
        // 'third' would have been the last node in the list.  And above... 'current' would be pointing to third.
        //    But now that we've set third.next to point to foo, we now have:
        //      first -> second -> third -> foo -> 0
        //
        // Making 'foo' the new last node.
    }
    else    // else, if head is null (indicating there are no nodes in the list)
    {
        // simply make 'foo' the head node.
        head = n;
    }

}

In this code, how exactly does this work.

n->next = 0;

to my understanding this is the same as this:

(*n).next = 0;

You are correct. When you use the arrow operator (->) you are accessing whatever the pointer points to (in this case, 'foo') rather than accessing the pointer itself.

[cppnoob25]

Thanks so much for that thorough explanation. Best one I've ever had anyone give me. I understand everything now.