如何在双向链表上实现复制分配？

问：

我对如何在双链接列表上实现复制分配有点困惑。我设法让复制构造函数工作，但我不确定分配。我试图在没有复制和交换方法的情况下做到这一点。

列表.H

class List 
{
public:
    List();
    ~List();
    List(const List& c);
    List& operator= (const List& t);
private:
    List *Next;
    List *Prev;
    Node *Head;

List.cpp

List::~List()
{
    Node* move = Head;
    while (move!=NULL)
    {
        Node *temp = move->Next;
        delete move;
        move = temp;
    }
}

List::List(const List& c)
{
    name = c.name;
    Prev = c.Prev;
    Next = c.Next;
    Node* dummy, * current;
    Head= dummy = new Node();
    current = c.Head;
    while (current)
    {
        dummy->Next = new Node(*current);
        current = current->Next;
        dummy = dummy->Next;
    }

    Node* temp = Head;
    Head = Head->Next;
    delete temp;
}

List& List::operator=(const List& t)
{
    Next = t.Next;
    return *this;
}

我是否还必须遍历赋值运算符中的每个节点？

编辑 这就是我现在所拥有的。问题是当我从列表中获取数据时，它是空的。

List& List::operator=(const List& that)
{

    if (this != &that)
    {
        while (Head)
        {
            Node* temp = Head;
                Head = Head->Next;
                delete temp;
        }
        Node* dummy, * current;
        Head = dummy = new Node();
        current = that.Head;
        while (current)
        {
            dummy->Next = new Node(*current);
            current = current->Next;
            dummy = dummy->Next;
        }
    dummy->Next = nullptr;
}
return *this;
}