我正在尝试创建一副牌,但是我在访问我使用ArrayList.get方法创建的牌组时遇到了一些问题,我不确定我是否已正确创建了牌阵列。
卡类:
package cardgame;
public class Card {
private String number;
private String suit;
public Card(String number, String suit) {
this.number = number;
this.suit = suit;
}
public void setNumber(String number) {
this.number = number;
}
public void setSuite(String suit) {
this.suit = suit;
}
public String getNumber() {
return number;
}
public String getSuit() {
return suit;
}
}
甲板课程:
package cardgame;
import java.util.*;
public class Deck {
private String[] numberList = {"Ace", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Jack", "Queen", "King"};
private String[] suitList = {"Hearts", "Diamonds", "Clubs", "Spades"};
private ArrayList cardDeck = new ArrayList();
public ArrayList Deck() {
for(int i = 0; i < 4; i++) {
for(int j = 0; j < 13; j++) {
cardDeck.add(new Card(numberList[j], suitList[i]));
}
}
return cardDeck;
}
}
主要课程:
package cardgame;
public class CardGame {
public static void main(String[] args) {
Deck newDeck = new Deck();
System.out.println(newDeck.get(1));
}
}
在主类中,我只是使用newDeck.get来检索数据,以查看是否实际填充了ArrayList。我收到以下错误指向该行:
Exception in thread "main" java.lang.RuntimeException: Uncompilable source code - Erroneous tree type: <any>
at cardgame.CardGame.main(CardGame.java:7)
Java Result: 1
答案 0 :(得分:1)
我会使Card类不可变,因为一旦分配了数字和套装,它们将永远不会改变。或者你曾经在现实生活中玩纸牌游戏吗? 你必须以这两种方式定义这两个参数。
final
关键字表示,一旦为变量指定了值,就不能再对其进行更改。您还可以保存几行代码。
以下是Card
类:
package cardgame;
public class Card {
private final String number;
private final String suit;
public Card(String number, String suit) {
this.number = number;
this.suit = suit;
}
public String getNumber() {
return number;
}
public String getSuit() {
return suit;
}
// Convenient for System.out.println()
@Override
public String toString() { return suit + " " + number; }
}
toString()是为每个Object
声明的方法,并返回该类的String表示形式。我只是重新声明了这种表示的样子。您可以在下面的主要方法中看到我为什么这样做。
现在到Deck
班级:
package cardgame;
import java.util.*;
// The Deck is a list of cards, so let it inherit from AbstractList
// Deck will be a Read-only list with all the syntactic sugar of lists then
public class Deck extends AbstractList {
private String[] numberList = {"Ace", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Jack", "Queen", "King"};
private String[] suitList = {"Hearts", "Diamonds", "Clubs", "Spades"};
private ArrayList<Card> cardDeck = new ArrayList<>();
// Constructor
public Deck() {
for(String number : numberList) {
for(String suit : suitList) {
cardDeck.add(new Card(number, suit));
}
}
}
@Override
public Card get(int i) { return cardDeck.get(i); }
@Override
public int size() { return cardDeck.size(); }
}
使用AbstractList,您可以在Deck
内的卡片列表中获得一种查看。然后,您可以像普通列表一样使用套牌(它是只读的,没有add(Card)
和set(int, Card)
操作!)。
例如,在您的主要课程中:
package cardgame;
public class CardGame {
public static void main(String[] args) {
Deck newDeck = new Deck();
// Pick the card at index 1
System.out.println(newDeck.get(1));
// Iterate over all cards in the Deck and print them.
// This is basically sugar from AbstractList.
for(Card card : newDeck) {
System.out.println(card.toString());
}
}
}
答案 1 :(得分:0)
您必须指定arraylist存储的类型。
更改此内容:private ArrayList cardDeck = new ArrayList();
:private List<Card> cardDeck = new ArrayList<>();
此外,这不是构造函数,在创建实例时不会被执行:
public ArrayList Deck()
您必须将其更改为:
public Deck()
答案 2 :(得分:0)
各种评论:
for(String number : numberList)
for(String suit : suitList) {
cardDeck.add(new Card(number, suit));