我正在尝试用 C++ 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,但我已在此处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;
}
命令cout << *p;
抛出错误,我相信这就是原因:此时,p 指向 head->link
,这是另一个指向我的头节点的链接字段的指针。现在,我知道如果我在程序中的此时取消引用 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: error: no match for ‘operator*’ in ‘** p’
谁能帮我理解为什么会这样?这是非法操作吗?是否以我不熟悉的其他方式执行?
最佳答案
cout << *p
尝试打印 node
目的。由于没有为节点对象定义打印操作(即没有 operator<<
用于输出流),因此尝试打印失败。您可能正在寻找的是:
cout << p->data;
对于你的第二点,陈述可以这样分解:
**p == *(*p)
所以第一个星取消引用 p
, 返回 node
.第二颗星尝试取消引用该操作的结果,但由于节点是 struct
而不是指针,编译器会提示。
希望这对您有所帮助。
关于c++ - 试图理解指针和双指针,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11531371/