Question

我是C ++模板的新手。编译时出错：

main.cpp(17) : error C2923: 'Iterator' :
    'myWords' is not a valid template type argument for parameter 'T'

main.cpp(17) : error C2133: 'iterator' : unknown size error C2512:
    'Iterator' : no appropriate default constructor available

代码：

#include "MyList.h"
#include "MyList.cpp"
#include <string>
using namespace std;

int main()
{
    LinkedList<string> myWords;
    myWords.addAtBeginning("Data");

    Iterator<myWords> iterator;

    //iterator.advance();
    return 0;
}

标头文件

      #ifndef MYLIST_H
#define MYLIST_H

template <class T>
class Iterator;

template <class T>
struct link
{
   link * previous;
   link * next;
   T s;
};

template <class T>
class LinkedList 
{
   link<T> * head;   // No longer a global variable 
 public:
   void addAtBeginning(T word);
   void printAll();
   void printString(link<T> *p);  // prints the string in the link pointed to by p.
   void deleteFromBeginning(); // deletes the first link in the list.    
   LinkedList();  // default constructor
   friend class Iterator<T>;
};

template <class T>
class Iterator
{
private:
   link<T> * p;

public:
    // construct the iterator by having it refer to a linked list
    Iterator(LinkedList<T> whatIamIterating);   

    // returns the string in the current link
    T current();                        

    // move the Iterator to the next link
    void advance();                          

    // returns true when we are finished with list
    int atEnd(); 
};

#endif

CPP文件

      #include "MyList.h"

// Default constructor of linkedlist
template <class T>
LinkedList<T>::LinkedList()  
{
   head=0;
}

// Function to add at the beginning
template <class T>
void LinkedList<T>::addAtBeginning(T word)
{
   link<T> * tmp;
   tmp = new link<T>;
   tmp->s = word;
   tmp->next=head;
   tmp->previous=0;
   head=tmp;
}

// Function to print the all strings
template <class T>
void LinkedList<T>::printAll()
{
   link<T> * tmp;
   tmp = head;
   while (tmp != 0)
   {
      cout << tmp->s;
      tmp = tmp->next;
   }
   cout << endl;
}

// Prints the string in the link pointed to by p.
template <class T>
void LinkedList<T>::printString(link<T> *p)
{                          
    cout << p->s << endl;
}

// Deletes the first link in the list.
template <class T>
void LinkedList<T>::deleteFromBeginning()
{
    link<T> *oldhead = head;                    
    head = head->next;

    delete oldhead;
}

// Function definitions for Iterator class
// Constructor for iterator class
template <class T>
Iterator <T>::Iterator(LinkedList<T> whatIamIterating)
{
   p = whatIamIterating.head;
}

// Function to get the current string of the node
template <class T>
T Iterator<T>::current()
{
    return p->s;
}

template <class T>
void Iterator<T>::advance()
{
    p = p->next;
}

template <class T>
int Iterator<T>::atEnd()
{
    if (p == 0)
        return 1;
    else
        return 0;
}

Answer 1

需要Iterator< LinkedList<string> >。 类型名称是模板参数，而不是对象。

Answer 2

在

Iterator<myWords> iterator;

您使用myWords作为类型，但它是一个变量。你想要

Iterator<string> iterator(myWords);

由于string是您用于LinkedList的类型（解决第一个编译器错误），Iterator cosntrcutor会使用LinkedList<T>个对象（在您的情况下T为string）（解决第二个编译器错误）

链表迭代器模板中的编译错误

2 个答案: