模板类定义中的模板方法与声明不匹配 [英] Template Method within Template class definition does not match declaration

问题描述

我已经创建了模板类T的模板类LinkedList，并且在该类中，我想实现一个函数入队，该入队采用通用类型D的数据，并以数据D作为参数调用T构造函数.

I have made a template class LinkedList of template class T and within this class, I want to implement a function enqueue that takes data of a general type D and calls the T constructor on with the data D as the parameter.

这是我班级的定义:

template<class T>

struct Node {
    struct Node *_nextNode;
    struct Node *_prevNode;
    T *_value;
    int _location;
};

template<class T>
class LinkedList    {

private:
    Node<T> *_firstNode;
    Node<T> *_lastNode;
    int _size;

public:
    LinkedList<T>();
    LinkedList<T>(const int size);
    ~LinkedList<T>();

    template<typename D>
    bool enqueue(D &newData);
    bool dequeue();
    T* find(const int location);

};

这是我声明函数入队的地方:

template<class T, typename D>
bool LinkedList<T>::enqueue(D &newData) {
    Node<T> *newNode = new Node<T>;
    newNode->_value = new T(newData);
    newNode->_location = _lastNode->_location + 1;
    _lastNode->_nextNode = newNode;
    newNode->_prevNode = _lastNode;
    _lastNode = newNode;
    _lastNode->_nextNode = NULL;
    _size++;

    return true;
}

尝试编译时，我得到:

LinkedList.cpp:76:6: error: prototype for ‘bool LinkedList<T>::enqueue(D&)’ does not match any in class ‘LinkedList<T>’
 bool LinkedList<T>::enqueue(D &newData) {
      ^~~~~~~~~~~~~
LinkedList.cpp:29:7: error: candidate is: template<class T> template<class D> bool LinkedList<T>::enqueue(D&)
  bool enqueue(D &newData);

忽略入队函数的实际内容，与以前的实现相比，我还没有更改.任何帮助，将不胜感激.谢谢.

Ignore the actual content of the enqueue function, I have not changed that yet from my previous implementation. Any help would be appreciated. Thanks.

推荐答案

您需要将函数体定义为:

You need to define your function body as:

template <typename T>
template <typename D>
bool LinkedList<T>::enqueue (D& newData) {
  // ...
}

此外， const D& 可能更干净.最好使用完美的转发，以允许传递任何类型的引用类型:

Also, const D& is probably cleaner. It would be even better to use perfect forwarding, to allow passing any kind of reference type:

template <typename T>
template <typename D>
bool LinkedList<T>::enqueue (D&& newData) {
  // ...
  newNode->_value = new T (std::forward<D>(newData));
}

这也可以与构造函数的任意数量的参数一起使用:

This can also be made to work with an arbitrary number of parameters to the constructor:

template <typename T>
template <typename... D>
bool LinkedList<T>::enqueue (D&&... newData) {
  // ...
  newNode->_value = new T (std::forward<D>(newData)...);
}

此外，您的代码也不是异常安全的.如果 T 构造函数引发异常，则将永远不会释放 newNode 实例，从而导致内存泄漏.

Additionally, your code isn't exception-safe. If the T constructor throws an exception, the newNode instance is never freed, causing a memory leak.

此外，请勿使用 NULL ，但请尽可能使用 nullptr (即，如果您可以使用C ++ 11).

Also, don't use NULL, but use nullptr if possible (i.e. if you can use C++11).

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