初學者使用循環的二十一點遊戲

我正在嘗試爲學校創建一個二十一點程序，我不明白爲什麼我的程序在重新獲得我的前兩張卡之後再次請求另一張卡後重新開始。任何幫助都會很棒。我的代碼一切都在下面。初學者使用循環的二十一點遊戲

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); 
        } 

       } 

      } 

     } 
    } 
}

來源

2014-01-28 user3242607

閱讀[如何比較Java中的字符串？]（http://stackoverflow.com/questions/513832/how-do-i-compare-strings-in-java） – PakkuDon

使用'String''s equals '比較字符串值的方法，而不是'=='運算符。 – rgettman

如果'playAgain'是單個字符，則使用'char'。 –

這個表達式：

if (playAgain == "y")

將永遠是真實的，因爲如果兩個操作數是相同對象==運營商是唯一的真實。比較字符串的值，使用equals()：

if (playAgain.equals("y"))

不要難過。問題出在語言上 - 使用這樣的==運算符是一個愚蠢的選擇。這個問題的根源在於很大一部分問題。

來源

2014-01-28 00:50:53 Bohemian

初學者使用循環的二十一點遊戲

回答

相關問題