Question

  1 // This program helps staff manage customers'
  2 // orders and decide who should be given a ready dish.
  3 
  4 import java.util.*;
  5 
  6 // This class represents all orders of customers
  7 class ListOrder {
  8 
  9     // Data member
 10     private int numDishes;
 11     // All dishes which the restaurant offers
 12     private String[] dishes;
 13     // Each dish has a queue of customers who ordered this dish
 14     // All such queues are put inside an ArrayList called dishQueues
 15     private ArrayList<Queue<Integer>> dishQueues;
 16 
 17     // Constructor
 18     public ListOrder(int numDishes, Scanner sc) {
 19         this.numDishes = numDishes;
 20         dishes = new String[this.numDishes];
 21         ArrayList<Queue<Integer>> dishQueues = new ArrayList<Queue<Integer>>();
 22         for(int i=0;i<this.numDishes;i++)   {
 23             dishes[i] = new String(sc.nextLine());
 24             dishQueues.add(new LinkedList<Integer>());
 25         }
 26     }
 27 
 28     // Add new order to the list
 29     public void addNewOrder(int dishID, int tag) {
 30         dishQueues.get(dishID-1).offer(new Integer(tag));
 31     }   
 32     
 33     // Process food when it is ready
 34     // Return the customer who currently ordered for the dish
 35     // if there is no customer order for this dish return -1
 36     public int processReadyFood(int dishID) {
 37         int ready=0;
 38         if (dishQueues.get(dishID-1).peek() == null)
 39             return -1;
 40         else    
 41             ready = dishQueues.get(dishID-1).poll();
 42             
 43         return ready;
 44     }   
 45     
 46     // Get dish's name
 47     public String getDishName(int dishID) {
 48         return dishes[dishID-1];
 49     }   
 50     
 51 }
 52 
 53 public class QuickEat {
 54 
 55     public static void main(String [] args) {
 56     
 57         Scanner sc = new Scanner(System.in);
 58         int numDishes = sc.nextInt();
 59         String command; 
 60         sc.nextLine();
 61 
 62         // Create the list order of food
 63         ListOrder listOrder = new ListOrder(numDishes, sc);
 64 
 65         int noOfCommands = sc.nextInt();
 66         sc.nextLine();
 67 
 68         // Process commands
 69         for(int i=0; i<noOfCommands;i++)    {
 70             command = sc.next();
 71             if(command.equals("Order")) {
 72                 int tag = sc.nextInt();
 73                 int numOfDishes = sc.nextInt();
 74                 for(i=0;i<numOfDishes;i++)
 75                     listOrder.addNewOrder(sc.nextInt(),tag);
 76             }
 77             else if(command.equals("Ready"))    {
 78                 int dishID = sc.nextInt();
 79                 int customer = listOrder.processReadyFood(dishID);
 80                 if(customer == -1)
 81                     System.out.println("Throw away " + listOrder.getDishName(dishID));
 82                 else
 83                     System.out.println(listOrder.getDishName(dishID) + " ready to be served to Tag " + customer);
 84             }
 85         }
 86     }
 87 }

我对代码的其他部分没有太大问题，但如果有，请向我强调！我得到的主要问题是当我进入

时

3
Fish n Chips
Chicken Chop
Grilled Salmon
9
Order 1 2 1 3

我收到此错误。

Exception in thread "main" java.lang.NullPointerException
        at ListOrder.addNewOrder(QuickEat.java:30)
        at QuickEat.main(QuickEat.java:75)

我已经检查过了dishID-1但是提供方法没什么问题，但我还是不知道问题是什么！：（（（（HELP！:(

Answer 1

您已创建private ArrayList<Queue<Integer>> dishQueues;的引用，但尚未创建对象

这样做private ArrayList<Queue<Integer>> dishQueues=new ArrayList<Queue<Integer>> ;

Answer 2

dishQueues是null。像这样初始化列表

private ArrayList<Queue<Integer>> dishQueues = new ArrayList<Queue<Integer>>();

问题更新后，现在问题应该是

dishQueues.get(dishID-1)为null

更改addNewOrder方法代码以检查null并初始化新的队列实例（如果不存在）

     public void addNewOrder(int dishID, int tag) {
         if(dishQueues.get(dishID-1) == null)
           dishQueues.get(dishID-1) = new LinkedList<Integer>();
         dishQueues.get(dishID-1).offer(new Integer(tag));
     }

您刚刚创建了一个包含队列对象的ArrayList实例。我没有看到任何Queue个实例的创建或对Queue个实例的任何引用都要添加到列表中。因此，为列表中的每个索引创建实现Queue的任何类的新/引用实例。否则，null会导致 NPE

更新2

今天早上再次看到你的问题时，我看到构造函数中有成员变量阴影。

ArrayList<Queue<Integer>> dishQueues = new ArrayList<Queue<Integer>>();。您正在创建一个局部变量，该变量仅在构造函数中具有作用域，而该构造函数不是成员变量。使用this始终表示实例成员。这将有助于维护代码并提高可读性。您需要像这样更改构造函数

// Constructor
 18     public ListOrder(int numDishes, Scanner sc) {
 19         this.numDishes = numDishes;
 20         dishes = new String[this.numDishes];
            //this will initialize the member variable `dishQueues` in ListOrder
 21         this.dishQueues = new ArrayList<Queue<Integer>>();
 22         for(int i=0;i<this.numDishes;i++)   {
 23             dishes[i] = new String(sc.nextLine());
 24             dishQueues.add(new LinkedList<Integer>());
 25         }
 26     }

希望你现在明白了！

链表中的NullPointerException

2 个答案: