Question

我正忙于学习Java，我的任务是制作一个安全破解游戏。我需要用类和方法来做这个游戏。但是我来到了一点，我无法走得更远。下面我分享我的代码和我的问题。如果你能看一眼，我会非常感激。

import java.util.Random; import java.util.Scanner;

public class Main {

public static void main(String[] args) {
    entrance();
    playGame();
    quitGame();
}

 private static void entrance() {
    System.out.println("Welcome to the SafeCracker!\nI need your help to open the safe box." +
            "\nThe code is with 3 digits and we need to find it out as quick as possible.\nLet's write your guess!");
}

 private static int playGame() {
    int[] safeCode = {takeRandomSafeCode(), takeRandomSafeCode(), takeRandomSafeCode()};
    int guess = takeGuess();

    //Below I need to use a for each loop but I don't get the logic of it. I stuck here. I need to check every numbers one by one but how?  

    for (int safeDigit : safeCode) {
        if (safeDigit == guess) {
            System.out.println("Your number is correct");

        }
    }
    return playGame(); // with this return type I also have a problem.

如果我返回此方法，它会一次又一次地继续播放。但我也不知道我需要给哪种返回类型。 }

private static int takeGuess() {
    Scanner keyboard = new Scanner(System.in);
    int userGuess = keyboard.nextInt();
    return userGuess;
}

private static int takeRandomSafeCode() {
    Random random = new Random();
    int result = random.nextInt(10);
    return result;
}

private static int quitGame() {
    System.out.println("Do you want to play again?\nPress 1 for play again\nPress 2 for quit the game!");
    Scanner key = new Scanner(System.in);
    int userWannaPlay = key.nextInt();

    if(userWannaPlay == 1) {
        System.out.println(playGame());
    } else if (userWannaPlay == 2) {
        System.out.println(quitGame());
    } else {
        System.out.println("You entered an invalid number. If you want to play again or quit, you need to click 1 or 2!");
    }
return userWannaPlay; //And also quitGame method. I want to ask the users that if they want to play or not and according to answer I would like to run "playGame" method again or quit game.
}

Answer 1

尝试为游戏使用循环。

您可以从playGame方法设置quitGame变量，也可以为用户决策创建新方法。

public static void main(String [] args){

   entrance();
   do{
      playGame();

   }while(!quitGame)

}

public void playGame(){
   //Your code is here
}

Answer 2

如果我返回此方法，它会一次又一次地继续播放。但是我我也不知道我需要给哪种返回类型。

您的playGame*(方法在其最后一行return playGame()中递归调用自身。我想你这样做可以归还任何东西。如果您考虑到您的问题，您可能会得出结论，您根本不想归还任何东西（因为您不知道如何处理它）。在这种情况下，您可能不会像在void方法中那样返回任何名为main的内容。

还有quitGame方法。我想问用户是否需要是否打球，根据回答，我想跑＃34; playGame＆＃34; 方法再次或退出游戏

你必须考虑你想要的东西。您希望根据条件一次又一次地调用方法。为此，您可以使用循环或递归。例如，您可以稍微更改main方法并添加一个do-while-loop。

public static void main(String[] args) {
    entrance();
    int condition;
    do {
        playGame();
        condition = quitGame();
    } while (condition == 1);

不要忘记改变你的quitGame方法，因为你试图递归地解决你的问题（删除if子句）。如果您想以递归方式执行此操作，但忽略上述内容并查看此代码段：

private static int quitGame() {
    System.out.println("Do you want to play again?\nPress 1 for play again\nPress 2 for quit the game!");
    Scanner key = new Scanner(System.in);
    int userWannaPlay = key.nextInt();

    if(userWannaPlay == 1) {
        playGame(); // you dont need a println here
    } else if (userWannaPlay == 2) {
       // you dont need to anything here
       System.out.println("Quitting...");
    } else {
        System.out.println("You entered an invalid number. If you want to play again or quit, you need to click 1 or 2!");
       // call quitGame again to ask for the choice again
       quitGame();
    }
return userWannaPlay; // if you do it like this, this return is also unnecessary and you could use a void method without returning anything
}

使用安全破解游戏的方法

2 个答案: