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，我在该行收到了一个 NullPointerException:

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

这个NPE奇怪的是，我跑一次测试的时候很少出现，第二次跑的时候经常出现。有时直到我连续运行测试 7-8 次后才会出现。

我得到的错误信息是: 没有异常消息。

NPE at line 215 in GiftSelectortest.testGetCountForAllPresents

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，我在每次调用包含 getCountForAllPresents() 方法的 assertEquals() 方法之前添加了打印语句。有趣的是，在我获得 NPE 的那一行之前，print 语句打印出 getCountForAllPresents() 返回的 HashMap 的正确值。

我注意到的另一件奇怪的事情是，当我使用 BlueJ 的内置调试器执行 testGetCountForAllPresents() 方法时，我注意到 giftList3 没有没有出现在 santasSelector 的 santaMap HashMap 中，但是打印语句仍然打印正确的计数，这意味着它必须知道 giftList3 。

getCountForAllPresents() 的代码是:

/**
 * 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;
    }
}

我应该解释一下 santaMap 是一个 HashMap，以 Child 对象作为键和一个 GiftList 对象作为值(value)。它基本上将 child 映射到他们的圣诞愿望 list 。 santaMap 只能包含同一个 child 的一个心愿单。

我不知道为什么我会收到 NPE，这与我编写 getCountForAllPresents() 方法的方式有关吗？我是如何实现测试方法/类的？

最佳答案

您的 Present 类不会覆盖 hashCode() 和 equals()。这意味着 banana1 和 banana 在任何将使用它们作为键的 HashMap 中都是两个不同的键。

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

在 getCountsForAllPresents() 中，您有两个 HashMap 。第一个通过对象的比较字符串，第二个通过对象本身。

您添加遇到的第一根香蕉。如果它是 banana 对象，您将得到如下内容:

uniquePresents
banana-fruit-10 ➞ [banana instance]

presentFrequency
[banana instance] ➞ Integer(1)

You continue to iterate. You encounter the next banana object. It's the same object. You'll get:

uniquePresents
banana-fruit-10 ➞ [banana instance]

presentFrequency
[banana instance] ➞ Integer(2)

Now you get to the banana1 object. It's a different object, but it has the same comparison string! What happens?

This condition is true: uniquePresents.containsKey(present.toComparisonString()). This means it goes into the true part of the if.

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

这意味着它将获取当前由 banana-fruit-10 指向的对象，即 banana 对象 - 而不是 banana1 对象，获取其关联的频率，并递增它。它也由同一个对象存储。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 键。现在你返回这个对象。

当您尝试通过 banana 检索时，它工作正常 - 断言工作。

但请记住，santaMap 本身就是一个HashMap。这意味着没有保证订单。迭代器可能会给你giftList1,giftList2,giftList3，但它也可能给你giftList3, giftList1、giftList2 - 或任何其他订单。

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？因为 banana1 是第一个带有 key banana-fruit-10 的礼物，从现在开始它将使用它。

发生这种情况时，当您尝试从返回的对象中获取 banana 时，该键不存在于频率列表中。它返回 null - 并且存在您的 NullPointerException。

关于java - 为什么我的 NPE 只是在程序运行时偶尔出现？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27804652/