How to print a linked list located inside another linked list struct?

[pheF2bru]

Hi, I've been staring at the screen for 30 hours it seems and haven't gotten a single "A-ha!" light bulb, so I'm reaching out to the pro's. I have a couple of linked lists inside a node struct and can't figure out how to print this list.

struct Node
{
   int data;
   Node * listA;
   Node * listB;
   Node * next;
   Node (int);
};

How do I print out the contents inside *listA and *listB? I can print out the contents of my main list this way:

Node * firstlist;

..... // some code filling up *firstlist here.

const Node * temp (firstlist);

while (temp != NULL)
{
   cout << temp -> data << endl;
   temp = temp -> next;
}

I tried doing this to print the contents of listA:

cout << temp -> listA -> data << endl;

But apparently the compiler doesn't understand this. I've also searched all over the web and documentation for this is almost non-existent (or difficult to find). Any thoughts much appreciated!

[kbw]

Node * firstlist;

..... // some code filling up *firstlist here.

for (const Node* temp(firstlist); temp != NULL; temp = temp->next)
{
   cout << temp->data << endl;

   for (const Node* p = temp->listA; p!= NULL; p = p->next)
      cout << p->data << endl;

   for (const Node* p = temp->listB; p!= NULL; p = p->next)
      cout << p->data << endl;
}

[pheF2bru]

Thanks kbw, I'll test this out.

Is there a way to assign a Node * temp to listA? After filling a Node * temp linked list can I just do temp = listA and print out contents of listA?

[kbw]

I thought you had a linked list (connected by next), and each node was the head of two distinct linked lists (listA and listB).

I think you're asking how to print a node. Well, you'd make a generic print function:

void print_node(std::ostream &os, const Node* node)
{
    for (const Node* temp = node; temp != NULL; temp = temp->next)
    {
        os << temp->data << endl;

        for (const Node* p = temp->listA; p!= NULL; p = p->next)
            os << p->data << endl;

        for (const Node* p = temp->listB; p!= NULL; p = p->next)
            os << p->data << endl;
    }
}

Then you could call it like:

print_node(std::cout, firstlist);  // print entire list
if (firstlist)
    print_node(std::cout, firstlist->listA);  // print listA of first node
if (firstlist)
    print_node(std::cout, firstlist->listB);  // print listB of first node 

[pheF2bru]

This looks right but my brain is no longer working, haven't slept all night. Will come back again tonight and test this out! Thanks!

[pheF2bru]

Actually just realized--I need to set listA equal to another linked list. The other linked list was created elsewhere in the program, for example Node * somestuff. How can I set listA equal to *somestuff? Is it possible to simply do listA = somestuff? Basically, I'd like to copy the contents of Node * somestuff into listA.

[kbw]

Is it possible to simply do listA = somestuff?

Yes.