Question

我正在使用BlueJ中的JUnit为我的GiftSelector类编写测试类。当我运行testGetCountForAllPresents()方法时，我会在行上找到NullPointerException：

assertEquals(true, santasSelector.getCountsForAllPresents().get(banana) == 3);

关于这个NPE的奇怪之处在于，当我运行测试一次时它很少出现，但是经常在我第二次运行测试时出现。有时直到我连续7-8次测试才会出现。

我得到的错误信息是：没有异常消息。

GiftSelectortest.testGetCountForAllPresents第215行的NPE

我的测试类的代码是：

import static org.junit.Assert.*;
import org.junit.After;
import org.junit.Before;
import org.junit.Test;

/**
 * The test class GiftSelectorTest. The GiftSelector that you are 
 * testing must have testMode enabled for this class to function. 
 * This is done in the setUp() method.
 */
public class GiftSelectorTest
{
    private GiftList giftList1;
    private GiftList giftList2;
    private GiftList giftList3;
    private Child jack;
    private Child bob;
    private Child dave;
    private Child naughty1;
    private GiftSelector santasSelector;
    private Present banana1;
    private Present orange;
    private Present banana;
    private Present apple;
    private Present bike;
    private Present doll;
    private Present got;
    private Present pearlHarbour;
    private Present dog;
    private Present cat;
    private Present ball;
    private Present heineken;

    /**
     * Default constructor for test class GiftSelectorTest
     */
    public GiftSelectorTest()
    {
        //Nothing to do here...
    }

    /**
     * Sets up the test fixture.
     *
     * Called before every test case method.
     */
    @Before
    public void setUp()
    {
        santasSelector = new GiftSelector();
        santasSelector.setTestMode(true);
        jack = new Child("Jack", 20, "1 A Place", true, true, true, false);
        bob = new Child("Bob", 10, "2 A Place", true, true, true, true);
        dave = new Child("Dave", 10, "3 A Place", true, true, true, true);
        naughty1 = new Child("John", 5, "4 A Place", true, true, true, true);
        giftList1 = new GiftList(jack);
        giftList2 = new GiftList(bob);
        giftList3 = new GiftList(dave);
        banana = new Present("banana", "fruit", 10);
        orange = new Present("orange", "fruit", 10);
        banana1 = new Present("banana", "fruit", 10);
        apple = new Present("apple", "fruit", 10);
        bike = new Present("bike", "toy", 200);
        doll = new Present("doll", "toy", 40);
        got = new Present("game of thrones", "dvd", 50);
        pearlHarbour = new Present("pearl harbour", "dvd", 20);
        dog = new Present("dog", "animal", 100);
        cat = new Present("cat", "animal", 80);
        ball = new Present("ball", "toy", 5);
        heineken = new Present("heineken", "beer", 1.60);
    }

    /**
     * Tears down the test fixture.
     *
     * Called after every test case method.
     */
    @After
    public void tearDown()
    {
        //Nothing to do here...
    }


    @Test
    public void testGetCountForAllPresents()
    {
        System.out.println(santasSelector.getCountsForAllPresents());
        //Test on empty GiftSelector
        assertNull(santasSelector.getCountsForAllPresents());

        //Test on a GiftSelector with one giftlist containing one present
        giftList1.addPresent(banana);
        santasSelector.addGiftList(giftList1);
        System.out.println(santasSelector.getCountsForAllPresents());
        assertEquals(true, santasSelector.getCountsForAllPresents().get(banana) == 1);

        //Test when GiftSelector contains 2 giftlists, each containing the same present object

        giftList2.addPresent(banana);
        santasSelector.addGiftList(giftList2);
        System.out.println(santasSelector.getCountsForAllPresents());
        assertEquals(true, santasSelector.getCountsForAllPresents().get(banana) == 2);

        //Test when GiftSelector contains 3 giftlists, 2 containing the same present object and another containing an identical present but with a different present instance
        giftList3.addPresent(banana1);
        santasSelector.addGiftList(giftList3);
        System.out.println(santasSelector.getCountsForAllPresents());
        assertEquals(true, santasSelector.getCountsForAllPresents().get(banana) == 3); //This is the line I get the NPE

        //Test when GiftSelector contains 3 giftLists, the first with one with a banana, the second with a banana and apple, and the third with a banana1 and ball
        giftList2.addPresent(apple);
        giftList3.addPresent(ball);
        System.out.println(santasSelector.getCountsForAllPresents());
        assertEquals(true, santasSelector.getCountsForAllPresents().get(banana) == 3);
        assertEquals(true, santasSelector.getCountsForAllPresents().get(apple) == 1);
        assertEquals(true, santasSelector.getCountsForAllPresents().get(ball) == 1);

    }


    @Test
    public void testGetMostPopularPresent()
    {
        //Test on empty GiftSelector
        assertNull(santasSelector.getMostPopularPresent());

        //Test on a GiftSelector with one giftList and one Present
        giftList1.addPresent(heineken);
        santasSelector.addGiftList(giftList1);
        assertEquals(true, santasSelector.getMostPopularPresent().comparePresent(heineken));

        //Tset on a GiftSelector with 1 giftList and 2 presents, one more expensive than the other
        giftList1.addPresent(banana);
        assertEquals(true, santasSelector.getMostPopularPresent().comparePresent(banana));

        //Test on a GiftSelector with 1 giftList and 3 presents. Banana and Apple are equal in price, and are both in the top3, 
        //therefore it should return the present closest to the start of the list
        giftList1.addPresent(apple);
        assertEquals(true, santasSelector.getMostPopularPresent().comparePresent(banana) || santasSelector.getMostPopularPresent().comparePresent(apple));

        //Test on a GiftSelector with 2 giftLists, the second list containing banana1, an indentical present to banana
        giftList2.addPresent(banana1);
        santasSelector.addGiftList(giftList2);
        assertEquals(true, santasSelector.getMostPopularPresent().comparePresent(banana));

        //Test on a GiftSelector with 2 giftLists, the first containing four presents and the second containing 2 presents.
        //This tests to see if top3 is working.
        giftList1.addPresent(bike);
        giftList2.addPresent(bike);
        assertEquals(true, santasSelector.getMostPopularPresent().comparePresent(bike));
    }
}

我只包含了引用getCountsForAllPresents()方法的测试方法。您会注意到，在每次调用包含assertEquals()方法的getCountForAllPresents()方法之前，我都添加了打印语句。有趣的是，在我获得NPE的行之前，print语句打印出HashMap返回的getCountForAllPresents()的正确值。

我注意到的另一个奇怪的事情是，当我使用BlueJ的内置调试器进行testGetCountForAllPresents()方法时，我注意到giftList3没有出现在{{1}中} santaMap在HashMap中，但print语句仍会打印正确的计数，这意味着它必须知道santasSelector。

giftList3的代码是：

getCountForAllPresents()

我应该解释/** * For each present, calculate the total number of children who have asked for that present. * * @return - a Map where Present objects are the keys and Integers (number of children requesting * a particular present) are the values. Returns null if santaMap is empty. */ public HashMap<Present, Integer> getCountsForAllPresents() { if(!santaMap.isEmpty()) { //This HashMap contains a mapping from each unique real world present, represented by it's toComparisonString(), to a Present object representing it HashMap<String, Present> uniquePresents = new HashMap<String, Present>(); //This HashMap contains a mapping from each Present object in uniquePresents to the number of times it's toComparisonString() is equal to another in santaMap HashMap<Present, Integer> presentFrequency = new HashMap<Present, Integer>(); for(GiftList wishlist: santaMap.values()) { for(Present present: wishlist.getAllPresents()) { //Have we already seen this present? if(uniquePresents.containsKey(present.toComparisonString())) { //If so, update the count in presentFrequency Integer tmp = presentFrequency.get(uniquePresents.get(present.toComparisonString())); tmp++; presentFrequency.put(uniquePresents.get(present.toComparisonString()), tmp); } else { //If not, add it to the maps uniquePresents and presentFrequency (with a frequency of 1) uniquePresents.put(present.toComparisonString(), present); presentFrequency.put(present, 1); } } } //Return a map with unique presents as keys and their frequencies as values return presentFrequency; } else { //If there are no mappings in Santa's map, return null return null; } }是santaMap，其中HashMap个对象为关键字，Child个对象为值。它基本上将孩子映射到他们的圣诞愿望清单。 GiftList只能包含同一个孩子的一个心愿单。

我不知道为什么我要获得NPE，这与我编写santaMap方法的方式有关吗？我是如何实现测试方法/类的？

Answer 1

您的Present课程未覆盖hashCode()和equals()。这意味着banana1和banana是HashMap中将用作关键字的两个不同键。

让我们看看这里发生了什么。你有banana和banana1个对象 - 第一个中的两个，第二个中的一个。

在getCountsForAllPresents()内，您有两个哈希映射。第一个是对象的比较字符串，第二个是对象本身。

添加您遇到的第一个香蕉。如果它是banana对象，您将拥有以下内容：

uniquePresents
banana-fruit-10 ➞ [banana instance]

presentFrequency
[banana instance] ➞ Integer(1)

你继续迭代。您遇到下一个banana对象。这是同一个对象。你会得到：

uniquePresents
banana-fruit-10 ➞ [banana instance]

presentFrequency
[banana instance] ➞ Integer(2)

现在进入banana1对象。它是一个不同的对象，但它具有相同的比较字符串！会发生什么？

这种情况属实：uniquePresents.containsKey(present.toComparisonString())。这意味着它进入了if的真实部分。

Integer tmp = presentFrequency.get(uniquePresents.get(present.toComparisonString()));

这意味着它将采用banana-fruit-10当前指向的对象，即 banana 对象 - 而不是banana1对象，获取其关联频率，并增加它。它也存储在同一个对象中。你现在拥有的是：

uniquePresents
banana-fruit-10 ➞ [banana instance]

presentFrequency
[banana instance] ➞ Integer(3)

请注意，presentFrequency根本没有banana1密钥。现在你返回这个对象。

当您尝试按banana检索时，它可以正常工作 - 断言有效。

但请记住，santaMap本身就是HashMap。这意味着没有保证订单。迭代器可能会为您提供giftList1，giftList2，giftList3，但它也可能会为您提供giftList3，giftList1，giftList2 - 或任何其他顺序。

那么当它首先给你giftList3时会发生什么？你最终会得到：

uniquePresents
banana-fruit-10 ➞ [banana1 instance]

presentFrequency
[banana1 instance] ➞ Integer(3)

为什么呢？因为banana1是第一个带有密钥banana-fruit-10的礼物，因此它将从现在开始使用。

发生这种情况时，当您尝试从返回的对象中获取banana时，该密钥在频率列表中不存在。它返回null - 并且有NullPointerException。

为什么我会在程序运行时偶尔出现NPE？

1 个答案: