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java - 使用安全破解游戏的方法

问题描述

我只是忙于学习 Java，我的任务是制作一个破解游戏。我需要用类和方法来做这个游戏。但我到了一个点，我不能再进一步了。下面我分享我的代码和我的问题。如果你能看看，我将不胜感激。

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. 
If I return this method, it keeps going to play again and again.
But I don't know also which return type I need to give.
    }

    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.
    }
}

标签： javamethods

解决方案

如果我返回这个方法，它会一次又一次地继续播放。但我也不知道我需要给出哪种返回类型。

您的playGame*(方法在其最后一行以递归方式调用自身return playGame()。我猜你这样做是为了返回任何东西。如果您考虑您的问题，您可能会得出一个结论，即您根本不想返回任何东西（因为您不知道如何处理它）。void在这种情况下，您可能不会像在您的main方法中那样返回任何内容。

还有 quitGame 方法。我想问用户他们是否想玩，根据回答我想再次运行“playGame”方法或退出游戏

你必须考虑你想要什么。您想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main稍微更改您的方法并添加一个 do-while-loop。

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

不要忘记更改您quitGame的方法，因为您正在尝试递归地解决您的问题（删除 if 子句）。如果您想递归地执行此操作，请忽略上述内容并查看以下代码段：

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
}

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