c++ - C++模板类的内部类访问

Question

我有一个这样的课:

#include <iostream>

template <class T>
class LL
{
    using size_t = unsigned int;

    class Node
    {
        T m_data;
        Node* m_next;

        Node(const T& data) :m_data{ data }, m_next{ nullptr }{}

        friend std::ostream& operator<<(std::ostream& out, const Node& node)
        {
            out << node.m_data;
            return out;
        }

        friend std::ostream& operator<<(std::ostream& out, const LL& ll);

        friend class LL;
    };

    Node* m_first{ nullptr };
    size_t m_size{ 0 };

    Node* newNode(const T& data)
    {
        return new Node{ data };
    }

public:
    void push(const T& data)
    {
        Node* temp = newNode(data);
        temp->m_next = m_first;
        m_first = temp;
        ++m_size;
    }

    Node* head()
    {
        return m_first;
    }

    size_t size() const
    {
        return m_size;
    }

    ~LL()
    {
        if (m_first)
        {
            Node* trav = m_first->m_next;
            Node* foll = m_first;
            while (trav)
            {
                delete foll;
                foll = trav;
                trav = trav->m_next;
            }
            delete foll;
        }
    }

    friend std::ostream& operator<<(std::ostream& out, const LL& ll)
    {
        Node* trav = ll.m_first;
        while (trav)
        {
            out << *trav << ' ';
            trav = trav->m_next;
        }
        return out;
    }
};

我在同一文件的此类下的其他位置也有一个函数模板，该文件尝试访问Node，看起来像这样，但有两个编译器错误:

template <typename T>
int getSize(LL<T>::Node* node) //C2065: node is undeclared, C3861: node is not found
{
    if (node)
    {
        return 1 + getSize(node->m_next);
    }
    return 0;
} //does not compile

一段时间后，我再次尝试使用两个编译器:

template <typename T>
int getSize(LL<T>::Node<T>* node) //C2065 like before, C7510: use of dependent template name must be prefixed with 'template'
{
    if (node)
    {
        return 1 + getSize(node->m_next);
    }
    return 0;
} //does not compile

再过一段时间后，我尝试了以下编译良好的方法。

template <typename T>
int getSize(typename LL<T>::template Node<T>* node)
{
    if (node)
    {
        return 1 + getSize(node->m_next);
    }
    return 0;
}

现在，当我尝试从驱动程序函数调用此函数时，再次出现编译器错误:

int main()
{
    LL<int> ll;
    std::cout << getSize(ll.head()); //E0304, C2672 and C2783

    //E0304: no instance of the function template "getSize" matches the argument list
    //C2672: no matching overload function found
    //C2783: could not deduce template argument for 'T'
}

我尝试了所有可能且无法解决此问题的方法。有人可以解释一下发生了什么吗？
注意:我在这里提到的所有代码都在同一个文件中。

Answer 1

getSize(ll.head())由于non-deduced context而失败；无法自动推导模板参数T。

If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

• 1) The nested-name-specifier (everything to the left of the scope resolution operator ::) of a type that was specified using a qualified-id:

声明应为

template <typename T>
int getSize(typename LL<T>::Node* node) // using class instead of typename works for OP (MSVC)
{
    //some code
}

而且由于Node不是模板，因此不需要使用template关键字。

LIVE

有关为什么使用关键字typename和template的信息，请参见Where and why do I have to put the “template” and “typename” keywords?。