链表：结构中的结构？

Question

//Specifcation file for the NumberList class
#ifndef NUMBERLIST_H
#define NUMBERLIST_H

class NumberList
{
private:
  //Declare a structure for the list
  struct ListNode
  {
     double value;         //The value in this node 
     struct ListNode *next;  //To point to the next node 
  };

  ListNode *head;      //List head pointer
public:
  //Constructor 
  NumberList();
   { head = nullptr; }

  //Destructor 
  ~NumberList();

  //Linked list operations
  void appendNode(double);
  void insertNode(double);
  void deleteNode(double);
  void displayList() const;
};
#endif

谁能给我解释一下这个吗？

Answer 1

因此，我想知道在表示struct ListNode*next；的行中到底发生了什么

正如您后面提到的，它是一个指针，指向结构列表节点的下一个块。查看下图。listnode由data和next组成。next是指向下一个listnode单元格的指针。

它看起来是一个指针，那么为什么我们必须在这个指针上包含“struct”部分呢？

#include <iostream>

using namespace std;

class NumberList
{
private:
  //Declare a structure for the list
  struct ListNode
  {
     double value;         //The value in this node 
     ListNode *next;  //To point to the next node 
  };

  ListNode *head;      //List head pointer
public:
  //Constructor 
  NumberList() { head = nullptr; }
  
  NumberList(double val) {
      head = new ListNode();
      head->value = val;
  }

  //Destructor 
  ~NumberList();

  //Linked list operations
  void appendNode(double);
  void insertNode(double);
  void deleteNode(double);
  void displayList() {
      ListNode *curr = head;
      while(curr != nullptr) {
          cout << curr->value << " ";
          curr = curr->next;
      }
      cout << endl;
  }
};

int main()
{
    NumberList *nl = new NumberList(5.0);
    nl->displayList();
    return 0;
}