Question

我刚创建了LinkedList的实现（仅用于自我教育目的。）。我让它运行，但输出结果有点奇怪...... 这是代码：

#include "stdafx.h"
#include <iostream>
#include <stdio.h>

using namespace std;

template <class T>
class Node{
T datum;
Node<T> *_next;
public:
 Node(T datum)
{
    this->datum = datum;
    _next = NULL;
}
 void setNext(Node* next)
 {
     _next = next;
 }
 Node* getNext()
 {
     return _next;
 }
 T getDatum()
 {
     return datum;
 }          
};

template <class T>

class LinkedList{
Node<T> *node;
Node<T> *currPtr;
Node<T> *next_pointer;
int size;
public:
LinkedList(T datum)
  {
      node = new Node<T>(datum);
      currPtr = node;  //assignment between two pointers.
      next_pointer = node;
      size = 1;
  }
LinkedList* add(T datum)  // return pointer type.
{
   Node<T> *temp = new Node<T>(datum);
   currPtr->setNext(temp);
   currPtr = temp;
   size++;
   cout<<datum<<" is added.";
   return this; //pointer type specification
}
T next()
{
   T data = (*next_pointer).getDatum();
   cout<<data<<" is visited.";
   next_pointer = next_pointer->getNext();
   return data;
}
int getSize()
{
   return size;
}   
};

现在我尝试使用LinkedList：

int main()
{
LinkedList<int> *list = new LinkedList<int>(1);
list->add(2)->add(3)->add(4);
cout<<endl;

printf("%d %d %d %d",list->next(),list->next(),list->next(),list->next());  \\One

cout<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<"\n"<<list->next()<<endl; \\Two

cout<<list->next()<<endl;\\Three
cout<<list->next()<<endl;
cout<<list->next()<<endl;
cout<<list->next()<<endl;
}

输出One将显示数据：4 3 2 1.两个将显示4 3 2 1.三个将显示1 2 3 4。我不知道运行时发生了什么。所有这些都应该以1 2 3 4顺序输出数据...我很感激你的帮助！谢谢！

Answer 1

未指定参数的评估顺序，因此：

printf("%d %d %d %d",list->next(),list->next(),list->next(),list->next());

可以评估最后一个list->next()，或者中间一个......

编辑：只是解决我认为的问题，因为我怀疑这是真正的代码：http://ideone.com/avEv7

Linked List C ++实现

1 个答案: