这是什么类型的C++语法，我应该采取什么步骤来理解这一点

首先，这是我在数据结构和算法课程中的第一个项目。我在第一年用过C++，但代码呈指数级"漂亮"，从那以后我与 Swift 和 Javascript 进行了大量交互。这段代码对我来说是合法的，我很沮丧，因为我无法弄清楚如何做最简单的事情。

该项目是关于双向链表的，链表类和嵌套的 Node 类是

template <typename Type>
class Double_sentinel_list {
public:
class Double_node {
public:
//imutable reference to Type
Double_node( Type const & = Type(), Double_node * = nullptr, Double_node * = nullptr );
//Functions to return variables
Type value() const;
Double_node *previous() const;
Double_node *next() const;
//Variables
Type         node_value;
Double_node *previous_node;
Double_node *next_node;
};
Double_sentinel_list();
Double_sentinel_list( Double_sentinel_list const & );
Double_sentinel_list( Double_sentinel_list && );
~Double_sentinel_list();
// Accessors
int size() const;
bool empty() const;
Type front() const;
Type back() const;
Double_node *begin() const;
Double_node *end() const;
Double_node *rbegin() const;
Double_node *rend() const;
Double_node *find( Type const & ) const;
int count( Type const & ) const;
// Mutators
void swap( Double_sentinel_list & );
Double_sentinel_list &operator=( Double_sentinel_list );
Double_sentinel_list &operator=( Double_sentinel_list && );
void push_front( Type const & );
void push_back( Type const & );
void pop_front();
void pop_back();
int erase( Type const & );
private:
Double_node *list_head; //looks like head/tail pointers are nodes as well..
Double_node *list_tail;
int list_size;
// List any additional private member functions you author here
// Friends
template <typename T>
friend std::ostream &operator<<( std::ostream &, Double_sentinel_list<T> const & );
};

链表构造函数编写如下：

//Constructor
template <typename Type>
Double_sentinel_list<Type>::Double_sentinel_list():
// Updated the initialization list here
list_head( nullptr ),
list_tail( nullptr ),
list_size( 0 )
{
list_size = 0;
//Creating Sentinal Nodes
//Node* used so double node type points to the address that holds the new double_node()
Double_node* headSentinal = new Double_node(); //new calls constructor for double node
Double_node* tailSentinal = new Double_node();
//Interconnecting Sentinals
headSentinal->next_node = tailSentinal;
tailSentinal->previous_node = headSentinal;
//Assigning head/tail pointers to sentinal nodes
//arrow operator used to access the object that the pointer of the object is pointing to 
list_head->next_node = headSentinal;
list_tail->next_node = tailSentinal;
}

我相信我对正在发生的事情有一个大致的了解，但对于一些特定的函数调用，我不知道如何返回值。

例如，此类应该只返回列表中的下一个节点

template <typename Type>
typename Double_sentinel_list<Type>::Double_node   *Double_sentinel_list<Type>::Double_node::next() const {
// Enter your implementation here
return Double_node->next_node; ????
}

我一辈子都无法在这个函数声明中弄清楚要为返回值调用 ->next_node 的内容。如果我看得太久，整个事情看起来就像一团垃圾。见鬼，我什至如何在节点上调用此函数？