C++使用模板類LinkedList需要模板參數

Question

因此，當我的鏈接列表類只使用ints而不是模板T時，它完美地工作，但是當我經歷並將所有事情都改變爲模板化時，它給我和錯誤說「使用模板類LinkedList需要模板參數「。

#ifndef LINKEDLIST_H 
#define LINKEDLIST_H 

#include <iostream> 
#include <memory> 

template<typename T> 
class LinkedList 
{ 
private: 

    struct Node 
    { 
     T data; 
     std::shared_ptr<Node> next; 

     Node(T d, std::shared_ptr<Node> n) 
     :data(d) 
     ,next(n) 
     {} 
     Node() 
     {}; 

     T getData() 
     { 
      return data; 
     } 

    }; 


    std::shared_ptr<Node> head; 
    std::shared_ptr<Node> current; 


public: 


    LinkedList() 
    :head() 
    {} 
    LinkedList(LinkedList& other) 
    :head(Clone(other.head)) 
    {} 

^This is where I think the problem is but did I just forget to replace an int somewhere? I've looked this over a lot and I haven't found anything but I could just be blind. 

    std::shared_ptr<Node> getStart() const 
    { 
     return head; 
    } 

    //Insertion 
    void InsertAt(T value, std::shared_ptr<Node> &n) 
    { 
     n = std::make_shared<Node>(value, n); 

    } 

    void Insertion(T value) 
    { 
     Insertion(value, head); 
    } 

    void Insertion(T value, std::shared_ptr<Node> &n) 
    { 
     if (!n) 
     { 
      InsertAt(value, n); 
      return; 
     } 

     if (value > n->data) 
      Insertion(value, n->next); 
     else 
      InsertAt(value, n); 

    } 

    //Deletion 
    void Remove(T value) 
    { 
     Remove(value, head); 
    } 

    void Remove(T value, std::shared_ptr<Node>& n) 
    { 
     if (!n) return; 
     if (n->data == value) 
     { 
      n = n->next; 
      Remove(value, n); 
     } 
     else 
     { 
      Remove(value, n->next); 
     } 
    } 



    void for_each(const std::shared_ptr<Node> &n) 
    { 
     if(!n) return; 

     else 
     { 
      std::cout<<n->data<<std::endl; 
      for_each(n->next); 
     } 

    } 



    std::shared_ptr<Node> Clone(std::shared_ptr<Node> n) const 
    { 
     if(!n) return nullptr; 
     return std::make_shared<Node>(n->data, Clone(n->next)); 
    } 

    LinkedList& operator = (const LinkedList &list) 
    { 
     head = list.head; 
     this->Clone(head); 
     return *this; 
    } 



}; 



#endif 

Answer 1

您是否也改變了代碼在哪裏使用您的LinkedList？

例如鏈接整數的列表：LinkedList<int> list;