Java-扑克牌游戏输出问题

我一直在想为什么我的班级没有以正确的格式打印。当四只手中的每一只手都应该由不同的卡组成时，它只是为每只手打印相同的卡对象。

一个可运行的示例会很好，但从快速阅读中，我认为您的

if

语句实际上应该是

if then else

，例如

//If there is a suit with 3 cards or more, return 0 points.
if ( c >=3 || d >= 3 || s >=3 || h >=3) {
    retVal = 0;
}  
//If there is a suit with 2 cards, return 1 points.
else if ( c == 2 || d == 2 || s == 2 || h == 2) {
    retVal = 1;
} 
//If there is a suit with 1 card, return 2 points.
else if ( c == 1 || d == 1 || s == 1 || h == 1) {
    retVal = 2;
}     
//If there is a suit with 0 cards, return 3 points.
else {
    retVal = 3;
}        

请注意结尾的

else

——因为您已经涵盖了大于0的所有内容，所以不需要在结尾测试0

编辑

你的代码可以被清理很多。代码越干净，就越容易阅读。一些建议：

评论

您有很多注释，但是您正在使用这些注释来补偿错误的变量名！比如说,

//Variable to hold number of cards that contain the club suit.
int c = 0;

public int countDistributionPoints() {
    //Variable to hold the number of distribution points.
    int retVal = 0;

    // a whole bunch of stuff that doesn't use retVal

    //If there is a suit with 3 cards or more, return 0 points.
    if ( c >=3 || d >= 3 || s >=3 || h >=3) {
        retVal = 0;
    }  
    //etc

    return retVal;
}

这很好，但是当您稍后在代码中使用

c

时呢？这不是不言自明的。首先有意义地命名变量更有用

int clubs = 0;

枚举

不要对套装使用字符，而是使用枚举。它更好地体现了含义，并减少了打字错误破坏代码的机会，这在测试字符串/字符时可能会发生

public enum Suit
{
    CLUBS,
    DIAMONDS,
    HEARTS,
    SPADES
}

对于每个循环

除非您需要索引变量，否则for each循环更易于读取

 for (int i = 0; i < theHand.length; i++) {
     ...
 }

缩小范围

在需要时声明变量，而不是在此之前-这减少了变量的范围，从而减少了错误使用它们的机会。比如说,

//Variable to hold number of cards that contain the club suit.
int c = 0;

public int countDistributionPoints() {
    //Variable to hold the number of distribution points.
    int retVal = 0;

    // a whole bunch of stuff that doesn't use retVal

    //If there is a suit with 3 cards or more, return 0 points.
    if ( c >=3 || d >= 3 || s >=3 || h >=3) {
        retVal = 0;
    }  
    //etc

    return retVal;
}

您可以通过将retVal移动到使用它的位置来缩小它的范围

public int countDistributionPoints() {
    // a whole bunch of stuff that doesn't use retVal

    //Variable to hold the number of distribution points.
    int retVal = 0;

    //If there is a suit with 3 cards or more, return 0 points.
    if ( c >=3 || d >= 3 || s >=3 || h >=3) {
        retVal = 0;
    }  
    //etc

    return retVal;
}

把它放在一起

重新编写您的

Hand

类，以及一些

switch

语句和重命名的变量为我们提供了这一点

public class Hand {
    private final Card[] cards;

    public Hand(Card[] cards) {
        this.cards = cards;
    }

    /**
     * Looks through each Card in the hand array and
     * adds its points (if it has any) to a sum.
     * @return the sum of the hand
    */
    public int countHighCardPoints() {
        int points = 0;
        for (Card card : cards) {
            points += card.getPoints();
        }
        return points;
    }

    /**
     * Count the number of Cards in each suit:
     * a suit with 3 cards or more counts for zero points
     * a suit with 2 cards counts one point (this is called a doubleton)
     * a suit with 1 card counts 2 points (this is called a singleton)
     * a suit with 0 cards counts 3 points (this is called a void)
     * @return the sum of the points
    */
    public int countDistributionPoints() {
        int clubs = 0;
        int diamonds = 0;
        int spades = 0;
        int hearts = 0;

        for (Card card : cards) {
            switch (card.getSuit()) {
                case CLUBS:
                    clubs++;
                    break;
                case DIAMONDS:
                    diamonds++;
                    break;
                case SPADES:
                    spades++;
                    break;
                case HEARTS:
                    hearts++;
                    break;
            }
        }

        final int result;
        if (clubs >= 3 || diamonds >= 3 || spades >= 3 || hearts >= 3) {
            result = 0;
        }
        else if (clubs == 2 || diamonds == 2 || spades == 2 || hearts == 2) {
            result = 1;
        }
        else if (clubs == 1 || diamonds == 1 || spades == 1 || hearts == 1) {
            result = 2;
        }
        else {
            result = 3;
        }

        return result;
    }

    public String toString() {
        String club = "";
        String diamond = "";
        String heart = "";
        String spade = "";

        for (Card card : cards) {
            switch (card.getSuit()) {
            case CLUBS:
                club.append(card.toString().replace(",", " "));
                break;
            case DIAMONDS:
                diamond.append(card.toString().replace(",", " "));
                break;
            case HEARTS:
                heart.append(card.toString().replace(",", " "));
                break;
            case SPADES:
                spade.append(card.toString().replace(",", " "));
                break;
        }
    }

    //Concatenates all of the string values of the clubs, diamonds, hearts, and spades into one string.
    return club + "\n" + diamond + "\n" + heart + "\n" + spade;
}

---

}您需要再次声明对象

classname objectname=新的classname（）

---

在制作双手的东西里面

在过去几天里，我看到了这个问题的几个变体。我想桥牌和其他纸牌游戏都有家庭作业。你希望有人读这个问题吗？！你好我不太明白你的意思。