我们可以使用继承实现链表吗?

Question

可以使用继承来实现链表吗？
例如：

 class List {/* ... */}; // abstract class
 class IntList : public List {/* ... */}; // derived class

Answer 1

一种可能的解决方案是使List基类仅处理节点，即跟踪列表头、尾和添加/删除节点。List类可以有一个基本的Node类，例如 IntList专业化。

类似的东西

class List
{
public:
    virtual ~List() {}
protected:
    // Protected constructor so this class can only be inherited
    List() {}
    struct Node
    {
        Node* next;
        Node* prev;
    };
    void add_head(Node*);
    void add_tail(Node*);
    Node* pop_head();
    Node* pop_tail();
    Node* get_head();
    Node* get_tail();
private:
    Node* head;
    Node* tail;
};
class IntList : public List
{
public:
    IntList();
    ~IntList();
    void add_head(int);  // Creates an `IntNode` and calls `add_head` with that
    void add_tail(int);  // Creates an `IntNode` and calls `add_tail` with that
    int pop_head();  // Calls `pop_head` to get the node, and downcast to `IntNode`
    int pop_tail();  // Calls `pop_tail` to get the node, and downcast to `IntNode`
    int get_head();  // Calls `get_head` to get the node, and downcast to `IntNode`
    int get_tail();  // Calls `get_tail` to get the node, and downcast to `IntNode`
private:
    struct IntNode : List::Node
    {
        int value;
    };
};