在我的 while 循环中使用此开关时遇到问题

Question

所以我试图复制我以前玩过的这款游戏。在这个游戏中，您会看到一个数字，并且有一个隐藏的数字。您需要猜测这个隐藏的数字是更小、更大还是与显示的数字相同。我在让输入正常工作时遇到问题。我似乎无法让 switch 语句正常工作。我的扫描仪也有问题。虽然它在 while 循环的外部它可以工作，但在它内部却没有。

import java.util.Scanner;
import java.util.Random;
public class Jamey {

    /**
     * @param args
     */
    public static void main(String[] args) {

            //This will give us the first random shown number
        Random yourRandom = new Random();
        int y = 1+yourRandom.nextInt(10);

        //Here is the introduction text
        System.out.println("Welcome to the guessing game!");
        System.out.println("The objective of this game is simple.");
        System.out.println("You will be shown one of two numbers which range between one and ten.");
        System.out.println("You have to gues if the number shown is larger, smaller, or equal to the hidden number.");
        System.out.println("If you believe the number you see is larger enter 1.");
        System.out.println("If you believe the number you see is smaller enter the 3.");
        System.out.println("If you believe the number you see is the same enter the 2.");
        System.out.println("Good luck, your number is "+y+".");

        boolean isDone = false;
        while(isDone=false){            

            //This allows the user to guess
            Scanner guess = new Scanner(System.in);
            int g = guess.nextInt();

            //This will help us to keep score later.
            int score = 0;

            //This will give us the new random number
            Random newRandom = new Random();
            int n = 1+newRandom.nextInt(10);

            //This will give us the new hidden number
            Random hiddenRandom = new Random();
            int r = 1+hiddenRandom.nextInt(10);

            //This is to allow multiple different inputs
            switch(score){
            case 1 :
                score +=1;
                if(y>r){
                    System.out.println("Correct");
                    System.out.println("Your new number is "+n+".");
                }
                else{
                    score +=1;
                    System.out.println("Inccorect, your overall score was "+score+".");
                    isDone = true;
                }
                break;

            case 2 :
                score +=1;
                if(y==r){
                    System.out.println("Correct");
                    System.out.println("Your new number is "+n+".");
                }
                else{
                    System.out.println("Inccorect, your overall score was "+score+".");
                    isDone = true;
                }
                break;

            case 3 :
                score +=1;
                if(y<r){
                    System.out.println("Correct");
                    System.out.println("Your new number is "+n+".");
                }
                else{
                    System.out.println("Inccorect, your overall score was "+score+".");
                    isDone = true;
                }
                break;
            default:
                System.out.println("Invalid input.");
                isDone = true;
                break;

            }       
        }
    }
}

Answer 1

您的 while 语句正在使用赋值

while(isDone=false) // Incorrect
while (isDone == false) OR while(!isDone) // Better

注意== 而不是=。您没有进行比较，只是将 isDone 设置为 false 每次迭代。

您的 switch 语句失败，因为您正在检查 score 变量而不是猜测变量。

switch(score) // Incorrect
switch(g) // Better

此外，当您只需要创建一个实例时，您正在创建大量 Random 对象，然后相应地命名每个变量。例如

Random rand = new Random();
int yourRandom = 1 + rand.nextInt(10);
int newRandom = 1 + rand.nextInt(10);
int hiddenRandom = 1 + rand.nextInt(10);

Answer 2

while(isDone=false){ 应该 while(isDone==false){ 甚至更好 while(!isDone){