为什么我会在程序运行时偶尔出现NPE？

Question

我正在GiftSelectorBlueJ中使用JUnit 为我的类编写测试类.当我运行该testGetCountForAllPresents()方法时,我得到一个NullPointerException在线:

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.它有时直到我连续7-8次运行测试才出现.

我得到的错误消息是:没有异常消息.

NPE在GiftSelectortest.testGetCountForAllPresents的第215行

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));
    }
}

我只包括引用该getCountsForAllPresents()方法的测试方法.你会注意到我在每次调用assertEquals()包含该getCountForAllPresents()方法的方法之前都添加了print语句.有趣的是,在我获得NPE的行之前,print语句打印出HashMap返回的正确值getCountForAllPresents().

我注意到的另一个奇怪的事情是,当我testGetCountForAllPresents()使用BlueJ的内置调试器时,我注意到giftList3它没有出现在santaMap HashMapin中santasSelector,但print语句仍然打印正确的计数,这意味着它必须知道关于giftList3.

代码getCountForAllPresents()是:

/**
 * 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;
    }
}

我应该解释santaMap是HashMap,用一个Child对象作为重点和GiftList对象的值.它基本上将孩子映射到他们的圣诞愿望清单.A santaMap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,是否与我编写getCountForAllPresents()方法有关？我是如何实现测试方法/类的？

Answer 1

您的Present课程不会覆盖hashCode()和equals()。这意味着banana1和banana是任何HashMap将使用它们作为密钥的两个不同的密钥。

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那么让我们看看这里会发生什么。你有banana和banana1对象 - 第一个是两个，第二个是一个。

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里面getCountsForAllPresents()有两个哈希图。第一个是对象的比较字符串，第二个是对象本身。

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你添加你遇到的第一个香蕉。如果它是banana对象，你会得到这样的东西：

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uniquePresents \nbanana-fruit-10 \xe2\x9e\x9e [香蕉实例]\n\n PresentFrequency \n[香蕉实例] \xe2\x9e\x9e Integer(1)\n

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你继续迭代。你遇到下一个banana物体。这是同一个对象。你会得到：

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uniquePresents \nbanana-fruit-10 \xe2\x9e\x9e [香蕉实例]\n\n PresentFrequency \n[香蕉实例] \xe2\x9e\x9e Integer(2)\n

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现在你开始接触banana1对象了。这是一个不同的对象，但它具有相同的比较字符串！会发生什么？

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这个条件为真：uniquePresents.containsKey(present.toComparisonString())。这意味着它进入了 的真实部分if。

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

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这意味着它将获取当前指向的对象banana-fruit-10（即banana对象，而不是banana1对象），获取其关联的频率，并递增它。它也按同一对象存储。你现在拥有的是：

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uniquePresents \nbanana-fruit-10 \xe2\x9e\x9e [香蕉实例]\n\n PresentFrequency \n[香蕉实例] \xe2\x9e\x9e Integer(3)\n

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

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当您尝试通过 检索时banana，它工作正常 - 断言有效。

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但请记住，santaMap它本身就是一个HashMap. 这意味着没有保证订单。迭代器可能会给出giftList1, giftList2, giftList3，但也可能给出giftList3, giftList1, giftList2- 或任何其他顺序。

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giftList3那么当它首​​先给你时会发生什么呢？你最终会得到：

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uniquePresents \nbanana-fruit-10 \xe2\x9e\x9e [香蕉 1实例]\n\n PresentFrequency \n[香蕉 1实例] \xe2\x9e\x9e 整数(3)\n

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为什么？因为banana1这是第一份带有钥匙的礼物banana-fruit-10，从现在起它就将用它。

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发生这种情况时，当您尝试从返回的对象中获取时banana，频率列表中不存在该键。它返回null- 这就是你的NullPointerException.

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