Question

我编译器无法找到嵌套类的构造函数的定义。

我的嵌套类Node在中间，构造函数在最后。

错误：

错误C2244：＆＃39; CircularDoubleDirectedList :: Node :: Node＆＃39; ：无法   将函数定义与现有声明匹配请参阅   宣告＆＃39; CircularDoubleDirectedList :: Node :: Node＆＃39;

定义

＆＃39; CircularDoubleDirectedList :: Node :: Node（const T＆amp;）＆＃39;

现有声明

＆＃39; CircularDoubleDirectedList :: Node :: Node（const T＆amp;）＆＃39;

代码：

#ifndef CIRCULARDOUBLEDIRECTEDLIST_H
#define CIRCULARDOUBLEDIRECTEDLIST_H

#include "ICircularDoubleDirectedList.h"

template <typename T> class CircularDoubleDirectedList;
template <typename T> class Node;

template <typename T>
class CircularDoubleDirectedList :
    public ICircularDoubleDirectedList<T>{
public:
    //Variabels
    Node<T>* current;
    int nrOfElements;
    direction currentDirection;

    //Functions
    CircularDoubleDirectedList();
    ~CircularDoubleDirectedList();
    void addAtCurrent(const T& element) override;

private:
    template <typename T>
    class Node
    {
    public:
        T data;
        Node<T>* forward;
        Node<T>* backward;

        Node(const T& element);// The constructor
    };

};
template <typename T>
CircularDoubleDirectedList<T>::CircularDoubleDirectedList(){
    this->nrOfElements = 0;
    this->current = nullptr;
    this->currentDirection = FORWARD;
}
template <typename T>
CircularDoubleDirectedList<T>::~CircularDoubleDirectedList(){
    //TODO: Destroy all nodes
}
template <typename T>
void CircularDoubleDirectedList<T>::addAtCurrent(const T& element){
    Node<T>* newNode = new Node<T>(element);
    newNode->data = element;
    if (this->nrOfElements == 0){
        newNode->forward = newNode;
        newNode->backward = newNode;
    }
    else{
        //this->current->forward = newNode;
        //this->current->forward->backward = newNode;
    }
    //this->current = newNode;
}
template <typename T>
CircularDoubleDirectedList<T>::Node<T>::Node(const T& element){
    this->data = element;
}

#endif

Answer 1

首先，前向声明的template <typename T> class Node;与CircularDoubleDirectedList::Node不同 - 前者是全局类模板，后者是嵌套类。

其次，您不需要将CircularDoubleDirectedList::Node声明为模板（如果这样做，则必须为其使用另一个模板参数名称，而不是T）。但据我所知，对于这种情况，你应该把它设为非模板，所以：

template <typename T>
class CircularDoubleDirectedList :
    public ICircularDoubleDirectedList<T>{
private:
    class Node
    {
    public:
        T data;
        Node* forward;
        Node* backward;

        Node(const T& element);// The constructor
    };
public:
    Node* current;
    //...
};

template <typename T>
CircularDoubleDirectedList<T>::Node::Node(const T& element){
    this->data = element;
}

Answer 2

您有两个名为Node的类模板，而实际上您需要一个名为Node的非模板类。您已向前声明::Node<T>，并且您拥有嵌套的::CircularDoubleDirectedList<T>::Node<U>。

如果您真的想要这样，那么您必须在构造函数定义中添加另一个template关键字：

template <typename T>  //because CircularDoubleDirectedList is a template
template <typename U>  //because Node is a template
CircularDoubleDirectedList<T>::Node<U>::Node(const T& element) : data(element)
{}

但是，我看不出让Node成为模板的单一原因。在CircularDoubleDirectedList<T>内，您是否要使用T以外的类型的节点？如果没有，请将Node设为普通的非模板类：

template <typename T>
class CircularDoubleDirectedList :
    public ICircularDoubleDirectedList<T>{
public:
    //Variabels
    Node<T>* current;
    int nrOfElements;
    direction currentDirection;

    //Functions
    CircularDoubleDirectedList();
    ~CircularDoubleDirectedList();
    void addAtCurrent(const T& element) override;

private:
    class Node
    {
    public:
        T data;
        Node* forward;
        Node* backward;

        Node(const T& element);// The constructor
    };
};


template <typename T>
CircularDoubleDirectedList<T>::Node::Node(const T& element) : data(element)
{}

嵌套模板类构造函数

2 个答案: