java - 使用循环的初学者二十一点游戏

Question

这个问题已经有答案了:

How do I compare strings in Java? (23 个回答)

已关闭 9 年前。

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拿到前两张牌后要求另一张牌后我的程序不断重新开始。任何帮助都会很棒。我的所有代码如下。

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);
                    }

                }

            }

        }
    }
}

Answer 1

这个表达式:

if (playAgain == "y")

永远不会为真，因为==运算符仅在两个操作数是相同对象时才为真。要比较字符串的值，请使用equals():

if (playAgain.equals("y"))

<小时/>

别难过。问题在于语言 - 使用这样的 == 运算符是一个愚蠢的选择。这里很大一部分问题都表明了这个问题的根源。