用模板实现双转移的问题

• Z时代
• 2024-01-10
• 分类：问答

最近我试图用模板实现续集。这是导致用模板实现双转移的问题

#pragma once 
#include <iostream> 
using namespace std; 
template<typename Type> 
struct Node { 
    Node(); 
    Node<Type>* next = NULL; 
    Node<Type>* prev = NULL; 
    Type data; 
}; 
template<typename Type> 
class deque{ 
private: 
    Node<Type>* front = NULL; 
    Node<Type>* back = NULL; 
public: 
    deque(Type key); 
    ~deque(); 
    Node<Type>* back(); 
    Node<Type>* front(); 
    int size(); 
    bool empty(); 
    void push_front(Type key); 
    void push_back(Type key); 
    void pop_front(); 
    void pop_back(); 
    void print_in_order(); 
}; 
template<typename Type> 
    deque<Type>::deque(Type key) 
{ 
    if (front != NULL || back != NULL) 
     return 
    else 
    { 
     front = new Node<Type>(); 
     front->data = key; 
     prev = front; 
    } 
} 
template<typename Type> 
    deque<Type>::~deque() 
{ 
    Node<Type>* delPtr; 
    while (front->prev != NULL) 
    { 
     delPtr = front; 
     cout << "Deleted " << front->data<<endl; 
     front = front->prev; 
     delete delPtr; 
    } 
    back = NULL; 
    delete front; 
} 
template<typename Type> 
    Node<Type>* deque<Type>::back() 
{ 
     if(back!= NULL) 
     return back; 
    else 
      return NULL 
} 
template<class Type> 
    Node<Type>* deque<Type>::front() 
{ 
    return front; 
} 
template<typename Type> 
    int deque<Type>::size() 
{ 
    int counter = 0; 
    Node<Type>* temp = front; 
    while (temp->back != NULL) 
    { 
     counter++; 
     temp = temp->back; 
    } 
    return counter; 
} 
template<typename Type> 
    bool deque<Type>::empty() 
{ 
    if (front == NULL && back == NULL) 
     return true; 
    else 
     return false; 
} 
template<typename Type> 
    void deque<Type>::push_front(Type key) 
{ 
    Node<Type>* temp = new Node<Type>(); 
    temp->data = key; 
    temp->prev = front; 
    front->next = temp; 
    front = temp; 
} 
template<typename Type> 
    void deque<Type>::push_back(Type key) 
{ 
    Node<Type>* temp = new Node<Type>(); 
    temp->data = key; 
    temp->next = back; 
    back->prev = temp; 
    back = temp; 
} 
template<typename Type> 
inline void deque<Type>::pop_front() 
{ 
    if (front != NULL) { 
     Node<Type>* delPtr = front; 
     front = front->prev; 
     front->next = NULL; 
     delete delPtr; 
    } 
    else 
     cout << "There is no front in empty deque" << endl; 
} 
template<typename Type> 
    void deque<Type>::pop_back() 
{ 
    if (back != NULL) { 
     Node<Type>* delPtr = back; 
     back = back->next; 
     back->prev = NULL; 
     delete delPtr; 
    } 
    else 
     cout << "There is no back in empty deque " << endl; 
} 
template<typename Type> 
    void deque<Type>::print_in_order() 
{ 
    if (front == NULL) 
     return; 
    Node<Type> temp = front; 
    while (temp->prev != NULL) { 
     cout << temp->data << " "; 
     temp = temp->prev; 
    } 
    cout << endl; 
} 
template<typename Type> 
inline Node<Type>::Node() 
{ 
} 

然后一些代码来检查是否正常工作

int main(){ 
    deque<int> qq(1); 
    qq.push_front(2); 
    qq.push_front(3); 
    qq.print_in_order(); 
    return 0; 
} 

此代码不希望编译。我试图寻找类似的项目或错误，并找不到有用的东西。因此，如果有人能够解释发生了什么问题，那将是非常好的。 大部分错误都是一样的。例如，不允许使用back()或front()，因为它们不是函数，也不是静态成员。或者back()和front()作为类模板的成员不能接收类型的函数。希望有人能帮忙。

回答：

Node<Type>* front = NULL; 
Node<Type>* back = NULL; 

Node<Type>* back(); 
Node<Type>* front(); 

Node<Type> temp = front; 

Node<Type>* temp = front; 

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