Trying to understand pointers and double pointers - c++

Question

I'm trying to implement a simple linked list in C++. I can create nodes and they seem to link themselves correctly. My question involves the listIterate() function, but I've attached the entire code here in case it is needed.
#include <iostream>
using namespace std;
//Wrapper class which allows for easy list management
class LinkedList {
//Basic node struct
struct node {
int data;
node *link;
};
node *head; //Pointer to the head (also referred to as root) node, or the first node created.
node *current; //Pointer to the /latest/ node, or the node currently being operated on.
node *tail; //Pointer to the tail node, or the last node in the list.
public:
//Default constructor. Creates an empty list.
LinkedList() {
head = NULL;
current = NULL;
tail = NULL;
cout << "*** Linked list created. Head is NULL. ***\n";
}
//Default destructor. Use to remove the entire list from memory.
~LinkedList() {
while(head != NULL) {
node *n = head->link;
delete head;
head = n;
}
}
/*
appendNode()
Appends a new node to the end of the linked list. Set the end flag to true to set the last node to null, ending the list.
*/
void appendNode(int i) {
//If there are no nodes in the list, create a new node and point head to this new node.
if (head == NULL) {
node *n = new node;
n->data = i;
n->link = NULL;
head = n;
//head node initialized, and since it is ALSO the current and tail node (at this point), we must update our pointers
current = n;
tail = n;
cout << "New node with data (" << i << ") created. \n---\n";
} else {
//If there are nodes in the list, create a new node with inputted value.
node *n = new node;
n->data = i;
cout << "New node with data (" << i << ") created. \n";
//Now, link the previous node to this node.
current->link = n;
cout << "Node with value (" << current->data << ") linked to this node with value (" << i << "). \n---\n";
//Finally, set our "current" pointer to this newly created node.
current = n;
}
}
/*
listIterate()
Iterates through the entire list and prints every element.
*/
void listIterate() {
//cursor
node *p;
//Start by printing the head of the list.
cout << "Head - Value: (" << head->data << ") | Linked to: (" << head->link << ") \n";
p = head->link;
cout << *p;
}
};
int main() {
LinkedList List;
List.appendNode(0);
List.appendNode(10);
List.appendNode(20);
List.appendNode(30);
List.listIterate();
}
Now, I'll refer to this method, listIterate().
void listIterate() {
//cursor
node *p;
//Start by printing the head of the list.
cout << "Head - Value: (" << head->data << ") | Linked to: (" << head->link << ") \n";
p = head->link;
cout << *p;
}
The command cout << *p; throws an error and I believe this is why: At this point, p is pointing to head->link, which is another pointer which points to the link field of my head node. Now, I understand if I dereference p at this point in the program, there would be no actual value in head->link as it points to a variable.
To me, if I dereference p twice (**p), it should follow the pointer twice (p -> head->link -> The value of the second node in the linked list (10). However, dereferencing p twice throws this error.
LinkedListADT.cc:89: error: no match for ‘operator*’ in ‘** p’
Can anyone help me understand why this is the case? Is this an illegal operation? Is it performed in another way that I'm not familiar with?

Answer 1

cout << *p tries to print a node object. Since no print operation is defined for node objects (ie. no operator<< for output streams), the attempt to print it fails. What you're probably looking for is:
cout << p->data;
For your second point, the statement can be broken down thus:
**p == *(*p)
So the first star dereferences p, returning a node. The second star attempts to dereference the result of that operation, but since a node is a struct and not a pointer, the compiler complains.
Hope this helps.

Answer 2

Your node class is missing an operator *, so the construct **p when p is of type node * is semantically ill-formed. To look at an example of overloading operator *, look at examples of implementing smart pointers.

Answer 3

**p Does not "follow the pointer twice". The operation simply tries to dereference p twice.
p is a pointer to a node. The first dereference (*p) will evaluate to the node that was pointed to by p. The second dereference (**p) will cause an error, because a node is a struct and not a pointer and has no overloaded operator* defined.
If you wish to dereference the pointer to the next node:
*(p->link)