试图理解指针和双指针

我正在尝试在C++中实现一个简单的链表。我可以创建节点，它们似乎可以正确链接自己。我的问题涉及 listIterate(( 函数，但我在这里附加了整个代码，以备不时之需。

#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(); 

}

现在，我将引用此方法 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;
    }

命令cout << *p;抛出错误，我相信这就是原因：此时，p 指向 head->link ，这是指向我的头节点的链接字段的另一个指针。现在，我明白如果我在程序的这一点上取消引用 p，head->link中将没有实际值，因为它指向一个变量。

对我来说，如果我取消引用 p 两次 ( **p (，它应该跟随指针两次 ( p -> head->link -> 链表中第二个节点的值 ( 10 (。但是，取消引用 p 两次会引发此错误。

LinkedListADT.cc:89：错误：与"** p"中的"运算符*"不匹配

谁能帮我理解为什么会这样？这是非法操作吗？它是否以我不熟悉的另一种方式执行？