Question

这个程序假设是一个游戏，我向用户询问2个数字的低位和高位，它可以是任何顺序。生成器将在这两个数字之间创建一个随机数，并将继续向用户询问一个数字，直到他们得到答案。我唯一的问题是提示用户是否想要玩游戏：这是我现在的代码。

import java.util.*;

public class tester
{
    public static void main(String [] args)
    {
        Scanner S = new Scanner(System.in);                                                                                                                             //Introduction Statement to game
        System.out.println("This is a guessing game.  I will ask you for two numbers.");
        System.out.println("I will choose a random number between your two numbers for you " + "\n" + "to try and guess.  As you guess, I will give you hints.");

        System.out.print("Choose two numbers to bound your range: ");                                                                                                   //Scanner to store user's input for the range
        int u1 = S.nextInt();
        int u2 = S.nextInt();
        int guess;
        int count = 0; // = 0 new
        String playAgain = "Y";

        int high;
        int low;
        if(u1 < u2)
        {
            high = u2;
            low = u1;
        }
        else
        {
            high = u1;
            low = u2;
        }
        Random gen = new Random();
        int rand = gen.nextInt(high-low+1) + low;
        System.out.println("Now guess a number between " + low + " and " + high + ": ");
        guess = S.nextInt();

        while (playAgain.equalsIgnoreCase("Y"))
        {
                System.out.println("Now guess a number between " + low + " and " + high + ": ");
                guess = S.nextInt();
                while(guess != rand)
                {
                count++;
                if (guess > high || guess < low)
                {
                    System.out.println("Out of range. Please follow the directions dumb-ass.");
                }
                else if(guess > rand)
                {
                    high = guess;
                    System.out.println("Too high!");
                }
                else
                {
                    low = guess;
                    System.out.println("Too low!");
                }
                System.out.print("Now guess a number between " + low + " and " + high + ": ");
                guess = S.nextInt();
                if(guess == rand)
                {
                    System.out.println("you got it!.");
                }
                    else if(count == 10)
                {
                    System.out.println("You lost! So Sorry.");
                    //break;
                }
            System.out.println("Would you like to play again? Enter Y to play or any other key to quit: ");
            S.nextLine();
            playAgain = S.nextLine();
                }
        }
}
}

Answer 1

使用2个嵌套循环而不是1个。

String playAgain = "Y";
while(playAgain.equalsIgnoreCase("Y"))
{
    Scanner S = new Scanner(System.in);                                                                                                                                  
    //Introduction Statement to game
    System.out.println("This is a guessing game.  I will ask you for two numbers.");
    System.out.println("I will choose a random number between your two numbers for you " + "\n" + "to try and guess.  As you guess, I will give you hints.");

    System.out.print("Choose two numbers to bound your range: ");                                                                                                   //Scanner to store user's input for the range
    int u1 = S.nextInt();
    int u2 = S.nextInt();
    int guess;
    int count = 0; // = 0 new

    int high;
    int low;

    int gameHasEnded = 0;
    while(!gameHasEnded)
    {
        // do game stuff here; after he won or lost, set gameHasEnded = 1
    }
    System.out.println("Would you like to play again? Enter Y to play or any other key to quit: ");
    S.nextLine();
    playAgain = S.nextLine();

通过这种方式，您可以区分这两个事件（1）游戏已经结束（无论出于何种原因）和（2）再次播放它。

Answer 2

你只需要在函数中放置代码。

执行新函数后，如果用户输入等于Y，则再次执行

否则你只需返回^^

无法弄清楚如何进入playAgain循环

2 个答案: