Collections.shuffle(Arrays.asList(deck));
没有正确洗牌。我不知道为什么。
我正在完成一项任务，我必须编写/填写一些代码来编写简单的二十一点程序。
我定义了三个类，一个Card类，一个Hand Class（一个Cards数组和一些其他值）和一个Deck类，它也是一个Cards数组。
在Hand类中，我有一个方法可以将卡的所有值相加，每当我运行代码时，我都会得到一个java.lang.NullPointerException，并且控制台指向我的“addHand”方法：< / p>
public int addHand(){
int j=0;
for(int i=0; i<=counter-1; i++){
j += Cards[i].getValue();
}
return j;
}
我认为这是因为Cards []中的某些值为null。但“反击”不应该让任何一个加入。我将包括其他手工课程以及卡片类。
手类：
public class Hand {
private Card[] Cards;
private int counter;
//private int valueOfHand;
public Hand(){
Cards = new Card[10];
counter = 0;
}
public int numCards(){
return counter;
}
public Card getCard(int i){
return Cards[i];
}
public int getCounter() {
return counter;
}
public void setCounter(int counter) {
this.counter = counter;
}
public void addCard(Card card){
Cards[counter] = card;
setCounter(counter+1);
}
public int addHand(){
int j=0;
for(int i=0; i<=counter-1; i++){
j += Cards[i].getValue();
}
return j;
}
}
卡片类：
public class Card {
private String value;
private String suit;
public Card(String number, String shape){
value = number;
suit = shape;
}
public int getValue() {
if (value.equals("King") || value.equals("Queen") || value.equals("Jack")){
return 10;
}
else if (value.equals("Ace")){
return 11;
}
else {
return Integer.parseInt(value);
}
}
public void setValue(String value) {
this.value = value;
}
public String getSuit() {
return suit;
}
public void setSuit(String suit) {
this.suit = suit;
}
}
我不知道是否需要发布任何其他内容，如果有必要，我会很高兴。
确切的错误是：
Exception in thread "AWT-EventQueue-1" java.lang.NullPointerException
at Hand.addHand(Hand.java:39)
at BlackJackGUIPanel$CardPanel.hitMe(BlackJackGUIPanel.java:182)
at BlackJackGUIPanel$CardPanel.actionPerformed(BlackJackGUIPanel.java:161)
以下是Hitme代码：
void hitMe() {
if (!gameInProgress) {
message = "Click \"New Game\" to start a new game!";
repaint();
return;
}
player.addCard(deck.giveCard());
if(player.addHand() > 21){
message = "Game over, you busted";
gameInProgress = false;
}
message = "You clicked Hit Me!";
repaint();
和giveCard（）代码：
public Card giveCard(){
Card temp = Decks[0];
for(int i = 1; i<52; i++){
Decks[i-1] = Decks[i];
}
Decks[51] = temp;
return temp;
}
我找不到更好的方法来做这个部分。 这是我初始化我的套牌的地方：
public Deck(){
Decks = new Card[52];
//int cardNumber = 2;
for(int i = 0; i < 52; i += 4){
switch (i/4){
case 0: Decks[i] = new Card("2", "Spades");
Decks[i+1] = new Card("2", "Hearts");
Decks[i+2] = new Card("2", "Clubs");
Decks[i+3] = new Card("2", "Diamonds");
case 4: Decks[i] = new Card("3", "Spades");
Decks[i+1] = new Card("3", "Hearts");
Decks[i+2] = new Card("3", "Clubs");
Decks[i+3] = new Card("3", "Diamonds");
case 8: Decks[i] = new Card("4", "Spades");
Decks[i+1] = new Card("4", "Hearts");
Decks[i+2] = new Card("4", "Clubs");
Decks[i+3] = new Card("4", "Diamonds");
case 12: Decks[i] = new Card("5", "Spades");
Decks[i+1] = new Card("5", "Hearts");
Decks[i+2] = new Card("5", "Clubs");
Decks[i+3] = new Card("5", "Diamonds");
case 16: Decks[i] = new Card("6", "Spades");
Decks[i+1] = new Card("6", "Hearts");
Decks[i+2] = new Card("6", "Clubs");
Decks[i+3] = new Card("6", "Diamonds");
case 20: Decks[i] = new Card("7", "Spades");
Decks[i+1] = new Card("7", "Hearts");
Decks[i+2] = new Card("7", "Clubs");
Decks[i+3] = new Card("7", "Diamonds");
case 24: Decks[i] = new Card("8", "Spades");
Decks[i+1] = new Card("8", "Hearts");
Decks[i+2] = new Card("8", "Clubs");
Decks[i+3] = new Card("8", "Diamonds");
case 28: Decks[i] = new Card("9", "Spades");
Decks[i+1] = new Card("9", "Hearts");
Decks[i+2] = new Card("9", "Clubs");
Decks[i+3] = new Card("9", "Diamonds");
case 32: Decks[i] = new Card("10", "Spades");
Decks[i+1] = new Card("10", "Hearts");
Decks[i+2] = new Card("10", "Clubs");
Decks[i+3] = new Card("10", "Diamonds");
case 36: Decks[i] = new Card("Jack", "Spades");
Decks[i+1] = new Card("Jack", "Hearts");
Decks[i+2] = new Card("Jack", "Clubs");
Decks[i+3] = new Card("Jack", "Diamonds");
case 40: Decks[i] = new Card("Queen", "Spades");
Decks[i+1] = new Card("Queen", "Hearts");
Decks[i+2] = new Card("Queen", "Clubs");
Decks[i+3] = new Card("Queen", "Diamonds");
case 44: Decks[i] = new Card("King", "Spades");
Decks[i+1] = new Card("King", "Hearts");
Decks[i+2] = new Card("King", "Clubs");
Decks[i+3] = new Card("King", "Diamonds");
case 48: Decks[i] = new Card("Ace", "Spades");
Decks[i+1] = new Card("Ace", "Hearts");
Decks[i+2] = new Card("Ace", "Clubs");
Decks[i+3] = new Card("Ace", "Diamonds");
}
}
Collections.shuffle(Arrays.asList(Decks));
}
答案 0 :(得分：0)
作为调试你可以做到这一点
public int addHand()
{
int j=0;
for(int i=0; i<=counter-1; i++)
{
Card c = Cards[i];
if(c != null)
j += Cards[i].getValue();
else
System.out.println("Add hand was trying to add a null card for index = " + i);
}
return j;
}
您可能还想在addCard(Card)方法中添加支票，以便无法添加空卡
public void addCard(Card card)
{
if(card != null)
{
Cards[counter] = card;
setCounter(counter+1);
}
else
System.out.println("NO NO NO! Trying to add a null card to your hand");
}
答案 1 :(得分：0)
这取决于对addCard方法的调用。此方法是将卡设置为cards变量。假设您调用了addCard 3次，那么addHand中的循环的0,1和2迭代将是正常的，但迭代3将通过异常，因为cards[3]是null。< / p>
确保正确添加卡片。