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我正在尝试为学校创建一个二十一点程序，但我不明白为什么在我获得前两张卡后我要求另一张卡后我的程序继续重新开始。任何帮助都会很棒。我的所有代码都在下面。

import java.util.Scanner;
import java.util.*;

public class BlackjackGame {

    public static void main(String[] args) {

        String anotherCard, playAgain = "y", ctn;
        int nextCard, card1, card2, dCard1, dCard2;
        int cardTotal = 0, dTotal = 0;

        Scanner keyboard = new Scanner(System.in);

        Random random = new Random();

        // Begin dealing the players first two cards

        while (playAgain == "y")
        {
            //dealers first two random cards
            dCard1 = random.nextInt(10) + 1;
            dCard2 = random.nextInt(10) +1;

            //players first two random cards and card total
            card1 = random.nextInt(10) + 1;
            card2 = random.nextInt(10) + 1;
            cardTotal = card1 + card2;

            //Dealers two card total and display only one dealer card
            dTotal = dCard1 + dCard2;
            System.out.println("Dealer shows: " + dCard1);

            //Display players first two cards & card total
            System.out.println("First Cards: " + card1 + ", " +card2);
            System.out.println("Total: "+ cardTotal);

            System.out.print("Another Card (y/n)?: ");
            anotherCard = keyboard.nextLine();

            while (anotherCard == "y")
            {
                nextCard = random.nextInt(10) + 1;
                cardTotal += nextCard;
                System.out.println("Card: " + nextCard);
                System.out.println("Total: " + cardTotal);

                if (cardTotal > 21)
                {
                System.out.println("You busted, Dealer Wins");
                System.out.println("Do you want to play again? (y/n): ");
                playAgain = keyboard.nextLine();
                }   
                if (cardTotal < 21)

                System.out.print("Another Card (y/n)?: ");
                anotherCard = keyboard.nextLine();
                if (anotherCard == "n")
                System.out.print("Press c to continue dealers cards");
                ctn = keyboard.nextLine();


                while (ctn == "c" && dTotal < 17)
                {
                    nextCard = random.nextInt(10) + 1;
                    dTotal += nextCard;

                    if (dTotal > 21)
                    {
                    System.out.println("Dealer Busts, You Win!");
                    System.out.println("Play Again? (y/n): ");
                    playAgain = keyboard.nextLine();
                    if (playAgain.equalsIgnoreCase("y"))
                            playAgain = "y";
                        else
                            System.exit(0);
                    }

                }

            }

        }
    }
}
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1 回答 1

3

这个表达式：

if (playAgain == "y")

永远不会为真，因为==运算符只有在两个操作数都是同一个对象时才为真。要比较字符串的值，请使用equals()：

if (playAgain.equals("y"))

不要心情不好。问题出在语言上——像这样使用运算符是一个愚蠢的选择==。很大比例的问题在这里说明了这个问题的根源。

于 2014-01-28T00:50:53.997 回答