Question

我刚刚开始学习C ++，我需要编写一个通用的链表和迭代器。这是我写的代码（list.h），但我认为这是不正确的。它不起作用，我不确定它是通用的。

#include <iostream>        
#include <cassert>


using namespace std;        
using namespace ListExceptions;

class List;    
class Iterator;

template<class T>

class Node{    
private:    
   T data;    
   Node* previous;    
   Node* next;    
   friend class List;    
   friend class Iterator;    
public:    
   Node(T element){    
       data = element;    
       previous = NULL;    
       next = NULL;    
   }    
};    

class List{    
private:    
   Node* first;    
   Node* last;    
public:    
   List(){    
       first = NULL;    
       last = NULL;    
   }    
   void pushBack(T element);    
   void insert(Iterator iter, T element);    
   Iterator remove(Iterator i);    
   Iterator find(const Predicate& predicate);    
   void sort(const Compare& comparer);    
   int getSize() const;    
   Iterator begin();    
   Iterator end();    
};    

class Iterator{    
private:    
   Node* position;    
   Node* last;    
   friend class List;    
public:    
   Iterator();    
   void next();    
   T getElement()const;    
   bool equals(Iterator b) const;    
   bool notEquals(Iterator b) const;    
};

如果有人可以帮助我？

Answer 1

首先，List和Iterator是非模板化的类，您可能想要创建给定类型的List。您可以考虑重构代码，以便Node和Iterator都是List类型的内部类（这会使事情更简单）：

template <typename T>
class List {
public:
   typedef T value_type;

   class Iterator;
   struct Node {           // Internal to List<T>, so there will be different
                           // List<T>::Node for each instantiationg type T
                           // But you don't need the extra <T> parameter for List
                           // or Iterator
      value_type data;
      Node* next;
      Node* last;

      friend class List;       // Inside List<T>, List by itself refers to List<T>
      friend class Iterator;
   };
   //...
};

替代方案在代码中稍微复杂一些：

template <typename T> class List;
template <typename T> class Iterator;
template <typename T> class Node {
   T data;
   Node * next;
   Node * last;
   friend class List<T>;
   friend class Iterator<T>;
};
template <typename T>
class List {
   Node<T>* first;              // note <T> required
//...
};

C ++通用链接列表和迭代器

1 个答案: