具有相同字段的自定義對象集不計算爲「相等」

Question

我正在寫一個紙牌遊戲並擁有一個Card類和一個Playable類。可玩牌代表玩家可能選擇爲給定遊戲狀態玩的牌的集合。我正在測試一個獲得合法Playables集合的函數。我的方法是手動構造正確的答案，然後聲明它等於我的函數返回的集合。

這是我的設置功能。 cards是List<Card>類型的私人成員。

@Before 
    public void setup() { 
     cards = new ArrayList<Card>(); 
     cards.add(new Card(Suit.HEARTS, CardValue.ACE)); 
     cards.add(new Card(Suit.DIAMONDS, CardValue.ACE)); 
     cards.add(new Card(Suit.CLUBS, CardValue.ACE)); 
     cards.add(new Card(Suit.SPADES, CardValue.ACE)); 
     cards.add(new Card(Suit.CLUBS, CardValue.EIGHT)); 
     cards.add(new Card(Suit.HEARTS, CardValue.EIGHT)); 
     cards.add(new Card(Suit.SPADES, CardValue.EIGHT)); 
     cards.add(new Card(Suit.DIAMONDS, CardValue.SIX)); 
     cards.add(new Card(Suit.SPADES, CardValue.SIX)); 
     hand = new Hand(cards); 
    } 

這裏是測試用例。

@Test 
public void testGetLegalDoubles() throws Exception { 
    Card toBeat = new Card(Suit.CLUBS, CardValue.NINE); 
    Set<Playable> correct = new HashSet<Playable>(); 
    Collection<Card> doubles1 = Arrays.asList(new Card[] {cards.get(0), cards.get(1)}); 
    Collection<Card> doubles2 = Arrays.asList(new Card[] {cards.get(0), cards.get(2)}); 
    Collection<Card> doubles3 = Arrays.asList(new Card[] {cards.get(0), cards.get(3)}); 
    Collection<Card> doubles4 = Arrays.asList(new Card[] {cards.get(1), cards.get(2)}); 
    Collection<Card> doubles5 = Arrays.asList(new Card[] {cards.get(1), cards.get(3)}); 
    Collection<Card> doubles6 = Arrays.asList(new Card[] {cards.get(2), cards.get(3)}); 
    Playable move1 = new Playable(doubles1, PlayMode.DOUBLES, cards.get(0)); 
    Playable move2 = new Playable(doubles2, PlayMode.DOUBLES, cards.get(0)); 
    Playable move3 = new Playable(doubles3, PlayMode.DOUBLES, cards.get(0)); 
    Playable move4 = new Playable(doubles4, PlayMode.DOUBLES, cards.get(1)); 
    Playable move5 = new Playable(doubles5, PlayMode.DOUBLES, cards.get(1)); 
    Playable move6 = new Playable(doubles6, PlayMode.DOUBLES, cards.get(2)); 
    correct.add(move1); 
    correct.add(move2); 
    correct.add(move3); 
    correct.add(move4); 
    correct.add(move5); 
    correct.add(move6); 
    Set<Playable> doubles = hand.getLegalNOfAKind(2, toBeat); 
    assertEquals(doubles, correct); 
} 

這個測試用例失敗，但我已經試過打印出的doubles和correct內容和它們匹配 - 我toString()爲Playable實施顯示了這些類的所有領域。我也覆蓋equals()和hashCode()爲Card和Playable如下所示（由Eclipse自動生成）。

Card：

@Override 
public int hashCode() { 
    final int prime = 31; 
    int result = 1; 
    result = prime * result + ((suit == null) ? 0 : suit.hashCode()); 
    result = prime * result + ((value == null) ? 0 : value.hashCode()); 
    return result; 
} 

@Override 
public boolean equals(Object obj) { 
    if (this == obj) 
     return true; 
    if (obj == null) 
     return false; 
    if (getClass() != obj.getClass()) 
     return false; 
    Card other = (Card) obj; 
    if (suit != other.suit) 
     return false; 
    if (value != other.value) 
     return false; 
    return true; 
} 

Playable：

@Override 
public int hashCode() { 
    final int prime = 31; 
    int result = 1; 
    result = prime * result + ((cards == null) ? 0 : cards.hashCode()); 
    result = prime * result + ((highCard == null) ? 0 : highCard.hashCode()); 
    result = prime * result + ((playmode == null) ? 0 : playmode.hashCode()); 
    return result; 
} 

@Override 
public boolean equals(Object obj) { 
    if (this == obj) 
     return true; 
    if (obj == null) 
     return false; 
    if (getClass() != obj.getClass()) 
     return false; 
    Playable other = (Playable) obj; 
    if (cards == null) { 
     if (other.cards != null) 
      return false; 
    } else if (!cards.equals(other.cards)) 
     return false; 
    if (highCard == null) { 
     if (other.highCard != null) 
      return false; 
    } else if (!highCard.equals(other.highCard)) 
     return false; 
    if (playmode != other.playmode) 
     return false; 
    return true; 
} 

我懷疑是東西了與我是如何構建correct - 也許怎麼我引用cards？

Answer 1

根據其他意見，在這裏得到完整的圖片是棘手的。由於看起來你正在向可播放構造函數傳遞一系列卡片，你是否將它們作爲數組存儲在Playable中？如果是這樣，這可能是在Playable.equals您的問題

} else if (!cards.equals(other.cards)) 
    return false; 

將永遠是假的（例子）;

final Integer[] is1 = new Integer[]{1}; 
final Integer[] is2 = new Integer[]{1}; 
Assert.assertFalse(is1.equals(is2)); 

集合將爲您節省一些工作，因此您可以將卡作爲ArrayList存儲在Playable中;

public class Playable { 
    List<Card> cards; 

    public Playable (Card[] cards, other params....) { 
     .... 
     this.cards = new ArrayList(); 
     for (Card card : cards) { this.cards.add(card); } 
    } 

，並在平等有一些像（你需要首先進行排序，如果你想在名單上使用.equals）：

public boolean equals() { 
     ..... 
     ..... 
     if (cards.containsAll(other.cards) && other.cards.containsAll(cards) { 
      return true; 

我會推薦一個簡單的測試用例你的card.equals（）是雙重保證。 作爲一個觀察，它在這裏並不重要，您使用的枚舉，但它總是用.equals而不是==

​​

會更好寫成

if (!suit.equals(other.suit)) 

希望這有助於一個好主意！