我正在使用BlueJ中的JUnit为GiftSelector类编写一个测试类。当我运行testGetCountForAllPresents()方法时，我在行上得到了一个NullPointerException:
assertEquals(true, santasSelector.getCountsForAllPresents().get(banana) == 3);
关于此NPE的奇怪之处在于，它一次运行测试时很少出现，而第二次运行测试时经常出现。有时直到我连续进行7-8次测试后，它才会出现。
我收到的错误消息是:
没有异常消息。
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，我已经添加了打印语句。有趣的是，在获得NPE的那一行之前，print语句打印出HashMap返回的getCountForAllPresents()的正确值。我注意到的唯一另一个奇怪的事情是，当我使用BlueJ的内置调试器检查
testGetCountForAllPresents()方法时，我注意到giftList3并未出现在santaMap的HashMap santasSelector中，但是print语句仍会打印正确的计数，表示它必须了解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
santaMap是HashMap，以Child对象作为键，并以GiftList对象作为值。它基本上将一个 child 映射到他们的圣诞节愿望 list 。一个santaMap只能包含一个同一个 child 的愿望 list 。我不知道为什么要获得NPE，这与我编写
getCountForAllPresents()方法的方式有关吗？如何实现测试方法/类？ 最佳答案
您的Present类不会覆盖hashCode()和equals()。这意味着banana1和banana是任何HashMap中的两个不同的键，它将使用它们作为键。
因此，让我们看看这里发生了什么。您具有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()内部，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。
您添加遇到的第一个香蕉。如果它是banana对象，则将具有以下内容:
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，获取其关联频率并对其进行递增。它还按同一对象存储。您现在拥有的是:独特的礼物
香蕉果10➞[香蕉实例]
现在的频率
[香蕉实例]➞整数(3)
请注意，
presentFrequency根本没有banana1 key 。现在，您返回该对象。当您尝试通过
banana进行检索时，它可以正常工作-断言有效。但是请记住，
santaMap本身就是HashMap。这意味着没有保证的订单。迭代器可能会给您giftList1，giftList2，giftList3，但也可能会给您giftList3，giftList1，giftList2-或任何其他顺序。那么，当它首先给您
giftList3时会发生什么呢？您最终会得到:独特的礼物
香蕉果实10➞[banana1实例]
现在的频率
[banana1实例]➞Integer(3)
为什么？因为
banana1是带有banana-fruit-10键的第一个礼物，所以从现在开始将使用它。发生这种情况时，当您尝试从返回的对象获取
banana时，该 key 在频率列表中不存在。它返回null-并且有您的NullPointerException。