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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，但我在此处附上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，我将参考这个方法，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;

    }

该命令cout << *p;引发错误，我相信这就是原因：此时， p 指向head->link，这是另一个指向我的头节点的链接字段的指针。head->link现在，我明白如果我在程序中取消引用 p ，因为它指向一个变量 ，所以不会有实际值。

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: '** p' 中的'operator*' 不匹配</pre>


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？
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3 回答 3

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cout << *p试图打印一个node对象。由于没有为节点对象定义打印操作（即没有operator<<为输出流定义），因此尝试打印它会失败。您可能正在寻找的是：

cout << p->data;

对于您的第二点，该声明可以分解为：

**p == *(*p)

所以第一个星号取消引用p，返回一个node. 第二颗星试图取消引用该操作的结果，但由于节点是 astruct而不是指针，编译器会抱怨。

希望这可以帮助。

于 2012-07-17T21:54:40.467 回答
1

您的节点类缺少一个operator *，因此类型的构造**pwhen pis 在node *语义上是错误的。要查看重载operator *示例，请查看实现智能指针的示例。

于 2012-07-17T21:54:55.050 回答
1

**p不会“跟随指针两次”。该操作只是尝试取消引用p两次。

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)

于 2012-07-17T22:02:21.330 回答