Java遊戲。需要停止代碼而不使用break語句或system.exit

以下代碼，我需要通過鍵入單詞「quit」來停止代碼，但不使用「break」或「system.exit」語句。任何人都可以幫助我？我認爲布爾可以解決這個問題，但我不知道如何使用它。我把quit語句放在循環的前兩行，，但我不知道它是否屬於那裏。即時消息在我的學習階段，所以不要太嚴格:)）Java遊戲。需要停止代碼而不使用break語句或system.exit

import java.util.*; 

public class Game { 


public static void main (String[]args){ 

    Game guessing = new Game(); 
    guessing.start(); 

} 
public void start() { 

    System.out.println("Welcome to guessing game!"); 
    System.out.println("Please enter the number between 1 and 1000"); 

    Scanner input = new Scanner(System.in); 
    String playerName; 
    String currentGuess; 
    String quit = "quit"; 
boolean playGame = true; 
int tries = 0; //number of times player guessed 

int guess = 0; //number that player inputs 

long startTime = System.currentTimeMillis(); //start timer after first guess 

int randomNumber = (int) (Math.random() * 999 + 1); // generating random number 
System.out.println(randomNumber);  // to be deleted after game is finished 

currentGuess = input.nextLine(); 


do{ 

if (currentGuess.equalsIgnoreCase(quit)) { 

     System.out.println("Thanx for playing");} 


    if (currentGuess.matches("[0-9]+")) { 
     int PlayerGuessInt = Integer.parseInt(currentGuess); 
    } 
    else { 
    System.out.println("You have netered non-numeric value,please try again"); 
    currentGuess = input.nextLine(); 
    continue; 
    } 

    guess = Integer.parseInt(currentGuess); 


    if(guess<1 || guess>1000){ 

     System.out.println("The number is out of range! Please try again"); 
     currentGuess = input.nextLine(); 
     continue; 
    } 

    if (guess>randomNumber){ 
     System.out.println("Oops,too high!"); 
     currentGuess = input.nextLine(); 
     tries++; 
    } 
    else if (guess<randomNumber){ 
     System.out.println("Sorry, to low!"); 
     currentGuess = input.nextLine(); 
     tries++; 
    } 



} 

while (guess!=randomNumber); 


    if (guess==randomNumber){ 
     tries++; 
     long endTime = System.currentTimeMillis(); 
     long gameTime = endTime - startTime; 
     System.out.println("Well done! You won the game in " + tries + " guesses " + "and " + gameTime/1000 + " seconds!"); 
     System.out.println("Please enter your name: "); 
     playerName = input.nextLine(); 

     } 




} 
}

來源

2012-10-03 G.M

你的意思是'return;'？ –

因爲在這段時間內沒有'break'語句，所以'while'後面的'if'不需要'if'。它永遠是'真實的'。 – techfoobar

設置循環條件的布爾部分，並在循環停止時將其設置爲false。 –

更改while循環，這樣它會檢查當前的猜測是「跳槽」的價值。這樣，當給出quit命令時，它將停止循環。

while(guess!=randomNumber && !currentGuess.equalsIgnoreCase(quit))

來源

2012-10-03 10:13:53

把，而內部的一個（或全部若）

if (currentGuess.equals("quit") { 
    playGame = false 
}

和改變，同時爲：

while (guess!=randomNumber && playGame)

來源

2012-10-03 10:15:22

這裏的答案：全球布爾變量：

bool userWantsToQuit = false;

所以，如果有人類型int退出你改變底部（shouln't它可以「跳槽」，因爲它是字符串？）

if (currentGuess.equalsIgnoreCase(quit)) 
{ 
    userWantsToQuit = true; // we mark that someone wants to quit 
    System.out.println("Thanx for playing"); // we output message 
    continue; // skip execution of the rest (or you could use else if) 
}

while語句停止比賽，如果userWantToQuit是真實的：

(guess!=randomNumber && !userWantsToQuit)

從使用java開始已經很長時間了，我沒有測試它，但它絕對是一個很好的方向。

來源

2012-10-03 10:29:43

Java遊戲。需要停止代碼而不使用break語句或system.exit

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