我在c++中创建了一个链表。首先我创建了一个类Node，然后创建了另一个名为List的类。我的程序运行得很好。问题是，我不明白为什么Node和List之间没有继承，这样List类就可以访问Node的公共方法
我做了一些更改，并从Node及其工作中派生出List。那么，它怎么能在有继承和没有继承的情况下运行呢？
#include <iostream>
#include <stdlib.h>
using namespace std;
/* The Node class */
class Node
{
private:
int object;
Node * nextNode;
public:
//mutator and Accessor for Node values
void set(int object)
{
this->object = object;
}
int get()
{
return object;
}
//mutator and Accessor for Node Address
void setNext(Node * nextNode)
{
this->nextNode = nextNode;
}
Node * getNext()
{
return nextNode;
}
};
/* The List class */
class List : public Node
{
private:
int size; // List Size ( number of nodes )
Node * headNode; // address of starting node of list
Node * currentNode; // address of Current node of list
Node * lastCurrentNode; // address of previous node of list
public:
/* Constructor */
List()
{
headNode = new Node(); // creating new node and store its address in headNode as its start of list
headNode->setNext(NULL); // the headNode is not connecting to any other node
currentNode = NULL; // as theres only head node so currentNode is empty
lastCurrentNode = NULL; // Previous Node is also empty because there's only headNode
size = 0; // Lisrs Size = 0 because theres no Value/data/object inside the list
}
/* add() class method */
add (int addObject)
{
Node * newNode = new Node(); // creating/adding new node and store its address in newNode Pointer
newNode->set(addObject); // Add Value/data/object in the node just created
if( currentNode != NULL ) // at first time when Current Node pointer is not pointing to any node in the list
{
newNode->setNext(currentNode->getNext()); // get adddress of node where current node will go and store that in the nextNode Pointer which is now called by our new node pointer so the addres that currentNode had now os taken and given to the new node
currentNode->setNext( newNode ); // address of new node we just created is now stored in current node
lastCurrentNode = currentNode; // move Lastcurrent node to the current node position
currentNode = newNode; // move currentNode pointer to the newNode we created;
}
// if current node is not pointing to any node (first time)
else
{
newNode->setNext(NULL); // new node we created will not point to any other next node because there's no one
headNode->setNext(newNode); // head node now is pointing to the new node we created
lastCurrentNode = headNode; // lastCurrent node is now position to headNode so we can go back
currentNode = newNode; // current node is now position to the new node we created
}
size ++; // as there's new new in the list increase its size to +1
}
/* get() class method */
get()
{
if (currentNode != NULL)
return currentNode->get(); // if current node is not null give the value where the current node is
}
/* next() class method */
next()
{
if (currentNode == NULL)
{
return false;
}
lastCurrentNode = currentNode; // move lastCurent node tot he position of current node
currentNode = currentNode->getNext(); // move current node to the next node
if (currentNode == NULL || size == 0)
{
return false;
}
else
{
return true;
}
}
friend void traverse(List list);
friend List addNodes();
};
/* Friend function to traverse linked list */
void traverse(List list) // friend function will get the object of List class
{
Node* savedCurrentNode = list.currentNode; // create new node pointer and assign it the address of current node
list.currentNode = list.headNode; // move current node to the headNode ( starting of the list)
for(int i = 1; list.next(); i++) // while we dnt reached to the end of list or not get false from next function
{
cout << "n Element " << i << " = " << list.get(); // traverse every node and display its value
}
list.currentNode = savedCurrentNode; // after traversing the whole nodes in the list move the current node to the position where it was befor e
}
/* Friend function to add Nodes into the list */
List addNodes()
{
List list;
list.add(2);
list.add(6);
list.add(8);
list.add(7);
list.add(1);
cout << "n List size = " << list.size <<'n';
return list;
}
int main()
{
List list = addNodes();
traverse(list);
}
改编自Asker的评论：
有人能解释一下"在没有继承的情况下，第二类如何能够使用第一类的
public方法？第二类是否不必从第一类派生，从而能够指向基类(1st(的方法？
在这种情况下，继承是错误的。正如评论中所提到的，继承为"is-a"关系建模：
法拉利是汽车
梅赛德斯就是。。。两者都有四个轮子和n门。因此
一辆普通汽车有四个车轮和n车门，这是法拉利和梅赛德斯继承的。
但是列表(容器(肯定不是节点。根据您的代码，列表将包含：
值CCD_ 11。。。列表的值是多少？毫无意义
一个指针nextNode。。。列表的"下一个节点"是什么？另一份名单？毫无意义
函数set(int object)和int get()。。。设置/获取什么？列表的值？毫无意义
函数setNext(Node * nextNode)和Node* getNext()。。。如上所述：列表的"下一个节点"是什么？毫无意义。
你想要应用的模式是合成，它为"has-a"关系建模：
法拉利有方向盘
梅赛德斯的方向盘一般汽车都有
您的
列表有一个第一个节点。
o o(你的问题让我想起了一位德国作家，他坚持认为WurstBrot应该继承Wurst和Brot，后者反过来继承Supermarket[原文如此](
btw：你在问题中显示的代码不应该编译。几个函数缺少返回类型。
你应该遵守利斯科夫替代原则，该原则指出，
在计算机程序中，如果
S是[=继承自]T的子类型，则类型T的对象可以用类型S的对象替换(即类型T的对象可以被子类型S的任何对象替换(，而不改变程序的任何所需属性(正确性、执行的任务等(
至于你的代码，List的c-tor已经让我栽跟头了：
List()
{
headNode = new Node(); // why does an empty list have a Node thats its head?
headNode->setNext(NULL); // that is something the c-tor of Node should do
currentNode = NULL; // not sure what currentNode is supposed to be
lastCurrentNode = NULL; // contradicts my assumption from above that lastCurrentNode
// was a pointer to the last node of the List;
// you created a first Node so the last node should be
// pointing to headNode
size = 0; // hm. Now you say your list is empty but you created a Node?
}