无法为多个玩家创造手牌

Question

我已将程序设置为要求用户输入卡的数量以及玩家的数量。我只想打印出每个玩家卡片。恩。 （玩家1：黑桃王牌，两个红心球员。玩家2：两个俱乐部等等。）我已将它打印出来，但它似乎只打印出一组卡片，尽管有2个输入到该号码球员这是我的代码。

手类：

public class Hand 
{
	private int handSize;			//how many cards in the hand
	private int cardsInHand;		//counter
	private Card [] hand;
	
	
	public Hand ()
	{
		hand = new Card[52];
		handSize = 5;
		cardsInHand = 0;
	}
	
	public Hand (int handSize)
	{
		hand = new Card [handSize];
		this.handSize = handSize;
		
	}

	public void addCard (Card card)
	{
		if (cardsInHand >= handSize) 
		{
			Card[] temp = new Card[hand.length*2];
			for (int i=0; i < cardsInHand; i++)
			{	
				temp[i] = hand[i];
				hand = temp;
			}
		}
		
	}

甲板类：

public class Deck 
{
	
	private Card [] deck;
	private int nextCard;
	
	
	public Deck()
	{
		deck = new Card[52];
        int iCardCount=0; 	// Holds how many cards have been created.
        for ( int suit = 0; suit <= 3; suit++ ) 
        {
            for ( int face = 1; face <= 13; face++ ) 
            {
                deck[iCardCount] = new Card(iCardCount);
                iCardCount++;
            }
        }
        nextCard = 0;
	}
 
    public Card dealACard ()
	{
		if (nextCard < 52)
		{
			System.out.println( deck[nextCard++]);
		}
		else
		{
			System.out.print("\nError, out of cards." );	
		}
		return (null);
	}
   
	public Hand dealAHand (int n)
	{	
		Hand hand = new Hand(n);
		Card deal = new Card(n);
		
	
		for (int i = 0; i < n; i++) 
		{ 
			hand = dealACard();
		}
		return hand;
	}

我的司机：

System.out.println ("How many cards in the hand? ");
int iHand = kb.nextInt();
Hand newHand = new Hand(iHand);
System.out.println("How many players are playing? ");
int iPlayers = kb.nextInt();
Deck secondDeck = new Deck();
secondDeck.shuffle();
secondDeck.dealAHand(iHand);

我得到的是输出：

How many cards in the hand? 
5
How many players are playing? 
2
the TWO of SPADES
the ACE of HEARTS
the TEN of DIAMONDS
the QUEEN of CLUBS
the NINE of HEARTS

我想要打印出两套卡片而不是一张卡片，但我在逻辑上无法弄清楚如何做到这一点。任何帮助都会很棒，谢谢。

Answer 1

你只有一个手的实例，就像D.Wallace说的那样，你只召唤过一次。在开始编码之前，您应该研究解决方案的更多结构。