c++ - 模板化类中带有模板对象的结构

Question

我正在尝试创建一个模板化类来保存不同的对象，以学习一些我不熟悉的自写数据结构和指针。我被困在这一点上:

template <class T> class Linked_List
{

    struct Node
    {
        T data;
        Node* prev;
        Node* next;
        Node(T d, Node* p, Node* n) : data(d),
        prev(p), next(n) {}
    };

public:
    Linked_List();
    ~Linked_List();

    void push_back(T data);
    void push_front(T data);
    void insertAt(T data, int pos);

    T pop_back();
    T pop_front();

    int size();
    bool empty();

private:
    Node* tail;
    Node* head;
};

我收到此错误:'Linked_List<T>::Node::data' uses undefined class 'T' . (使用VS2013)

我做错了什么？

谢谢

---

更新:

#include "Linked_List.h"


template<class T> Linked_List<T>::Linked_List()
{
    head = 0;
    tail = 0;
}

template<class T> Linked_List<T>::~Linked_List()
{
    {
        while (head)
        {
            Node* temp(head); //pointer to head
            head = head->next; // go to next
            delete temp; //delete the pointer
        }
    }
}

template <class T>
void Linked_List<T>::push_back(T data)
{
    //create the new Node
    tail = new Node(data, tail, NULL);
    if (tail->prev) //add to tail
        tail->prev->next = tail;
    if (empty()) // if empty its also the head
        head = tail;
}

template <class T>
void Linked_List<T>::push_front(T data)
{
    head = new Node(data, 0, tail);
    if (head->next)
        head->next->prev = head;
    if (empty())
        tail = head;
}

template <class T>
bool Linked_List<T>::empty()
{
    return (!head || !tail);
}

template <class T>
T Linked_List<T>::pop_back()
{
    if (empty())
        throw("Linked_list is empty!");
    //pointer to the element we want to have
    Node* temp(tail);
    //set the tail to prev
    tail = tail->prev;
    //get the data
    T data = temp->data;

    //the prev does not have a next now
    if (tail)
        tail->next = 0;
    else
        head = 0;

    //delete the node which held our data
    delete temp;
    //return the data
    return data;
}

template <class T>
T Linked_List<T>::pop_front()
{
    if (empty())
        throw("Linked_list is empty!");
    //pointer to the element we want to have
    Node* temp(head);
    //set the tail to prev
    head = head->next;
    //get the data
    T data = temp->data;

    //the prev does not have a next now
    if (head)
        head->prev = 0;
    else
        tail = 0;

    //delete the node which held our data
    delete temp;
    //return the data
    return data;
}

新的堆栈跟踪，如果我想像这样简单地使用它的话

int main(){
    Linked_List<int> list;
    for (int i = 0; i < 20; i++)
    {
        list.push_back(i);
    }

    for (int j = 0; j < 20; j++)
    {
        cout <<  list.pop_back() << endl;
    }
}

错误:

 Error  1   error LNK2019: unresolved external symbol "public: __thiscall Linked_List<int>::Linked_List<int>(void)" (??0?$Linked_List@H@@QAE@XZ) referenced in function _main   C:\Users\...\Documents\Visual Studio 2013\Projects\Double_Linked_List\Double_Linked_List\main.obj   Double_Linked_List
    Error   2   error LNK2019: unresolved external symbol "public: __thiscall Linked_List<int>::~Linked_List<int>(void)" (??1?$Linked_List@H@@QAE@XZ) referenced in function _main  C:\Users\...\Documents\Visual Studio 2013\Projects\Double_Linked_List\Double_Linked_List\main.obj   Double_Linked_List
    Error   3   error LNK2019: unresolved external symbol "public: void __thiscall Linked_List<int>::push_back(int)" (?push_back@?$Linked_List@H@@QAEXH@Z) referenced in function _main C:\Users\...\Documents\Visual Studio 2013\Projects\Double_Linked_List\Double_Linked_List\main.obj   Double_Linked_List
    Error   4   error LNK2019: unresolved external symbol "public: int __thiscall Linked_List<int>::pop_back(void)" (?pop_back@?$Linked_List@H@@QAEHXZ) referenced in function _main    C:\Users\...\Documents\Visual Studio 2013\Projects\Double_Linked_List\Double_Linked_List\main.obj   Double_Linked_List
    Error   5   1   error LNK1120: 4 unresolved externals   C:\Users\...\Documents\Visual Studio 2013\Projects\Double_Linked_List\Debug\Double_Linked_List.exe  1   Double_Linked_List

Answer 1

链接器非常清楚:您缺少这些方法的定义。编译器不会提示，因为它看到了 Linked_List 类的声明，但在处理模板时，这还不够。

为了使模板实例化成功，所有模板定义都需要可见。由于您在 cpp 文件中实现了模板类的方法(人们通常在头文件或 impl 文件中实现)，因此您需要在实例化模板的位置包含该文件。在您的具体示例中，在 main.cpp 中:

#include <iostream>
#include "Linked_List.cpp"   // !!!


int main(){
    Linked_List<int> list;
    for (int i = 0; i < 20; i++)
    {
        list.push_back(i);
    }

    for (int j = 0; j < 20; j++)
    {
        std::cout <<  list.pop_back() << std::endl;
    }
}

---

您可以在源文件中显式地执行此操作，而不是包含植入文件并让编译器隐式实例化模板。在这种情况下，将此行添加到 Linked_List.cpp 的末尾:

template class Linked_List<int>;

但在那种情况下，您将无法使用不同的模板参数(int 除外)创建对象。

---

第三个选项是第一个选项的变体。在 header 末尾包含源文件。

标题

// header.hpp
#ifndef HEADER_HPP
#define HEADER_HPP

template< typename T >
struct A
{
    void foo();
};

#include "source.impl"

#endif

来源

// source.impl

#include "header.hpp"

template< typename T >
void A::foo()
{
}

主要

// main.cpp

#include "header.hpp"

int main()
{
  A<int> a;
  a.foo();
}