Question

我是这里的新手，并且通常会进行编码。我要完成的工作是创建一个简单的猜谜游戏，提示用户输入数字，并对照计算机生成的1到100之间的数字检查该数字。我试图做到这一点，以便玩家可以继续猜测直到他们会得到正确的答案，并显示一个计数器，让玩家知道他们进行了多少次猜测。问题是，给出正确答案后程序不会终止，而且我无法弄清楚自己在做什么错。我将整个代码粘贴在底部以供参考，但是我觉得问题出在“ determineAnswer”方法的以下语句中：

} else if (userAnswer == computerNumber) {
        message = "Correct"
                + "\nNumber of Guesses: " + count;
        success++;

我试图使用整数“ success”的值作为终止do / while循环的条件，但是即使我尝试增加该值，循环也会继续进行，就好像该值正在不断重置一样。如果是这样，我看不到哪里出了问题。同样，我对此很陌生，但是我会很感激。

import javax.swing.JOptionPane;

public class GuessingGame {

    public static void main(String[] args) {
        // generate a random number from 1 to 100
        int computerNumber = (int) (Math.random() * 100 + 1);
        // declare other variables
        int success = 0;
        int count = 0;
        // display the correct guess for testing purposes
        System.out.println("The correct guess would be " + computerNumber);
        // prompt user for a guess
        do {           
            count++;
            String response = JOptionPane.showInputDialog(null,
                    "Enter a guess between 1 and 100");
            int userAnswer = Integer.parseInt(response);
            // display result
            JOptionPane.showMessageDialog(null, determineGuess(userAnswer, computerNumber, success, count));
        } while (success == 0);

    }

    public static String determineGuess(int userAnswer, int computerNumber,int success, int count) {
        String message = null;        
        if (userAnswer <= 0 || userAnswer > 100) {
            message = "Invalid guess"
                    + "\nNumber of Guesses: " + count;
        } else if (userAnswer == computerNumber) {
            message = "Correct"
                    + "\nNumber of Guesses: " + count;
            success++;
        } else if (userAnswer > computerNumber) {
            message = "Incorrect, Too High"
                    + "\nNumber of Guesses: " + count;
        } else if (userAnswer < computerNumber) {
            message = "Incorrect, Too Low"
                    + "\nNumber of Guesses: " + count;
        }
        return message;
    }
}

Answer 1

您不更新成功值，并且每次循环运行时，成功值都会为0，从而导致无限循环。

int success = 0;
    int count = 0;
    // display the correct guess for testing purposes
    System.out.println("The correct guess would be " + computerNumber);
    // prompt user for a guess
    do {           
        count++;
        String response = JOptionPane.showInputDialog(null,
                "Enter a guess between 1 and 100");
        int userAnswer = Integer.parseInt(response);
        // display result
        JOptionPane.showMessageDialog(null, determineGuess(userAnswer, computerNumber, success, count));
success=1;
    } while (success == 0);

Answer 2

在Java中，一切都是按值传递的。

在这种情况下，已将原语（int）传递给方法，然后更改其值并希望其值反映在调用方法中。 Java不能那样工作

public class SuccessTest {
public static void main(String[] args) {
    int success = 0;
    updateSuccess(success);
    System.out.println(success); //will print 0
}

private static void updateSuccess(int success) {
    //changing value of success here will not reflect in main method 
    success=2;
    System.out.println(success);//will print 2
}
}

为了使这项工作声明成功为类级变量

private static int success = 0;

然后，如果您在 determineGuess 中更新成功的值，则无需将此成功传递给 determineGuess >方法，它将在主要方法中可用

Answer 3

success变量未在main方法中进行更新的原因是，它无权访问determineGuess方法的success变量。它们是在不同范围中的两个独立变量。

determineGuess接收success作为int方法参数。在Java中，诸如int，char和float之类的类型是按值传递的：这意味着在将值赋给方法或设置为变量时，该值实际上是复制的，这样，如果您修改复制的值，则原始值不会被修改。（实际上是all types are pass-by-value，但是对象的内容是通过引用提供的。）

有几种方法可以为success方法更新main变量，其中两种是：

在类中为success设置一个字段，这样该类中的所有方法都可以访问它。因为您正在main中进行所有操作，所以现在您需要success是静态的：private static int success = 0;。更好的方法是使所有内容均为非静态，并让main实例化一个GuessingGame对象，然后在其上调用一个run方法。
从determineGuess返回一个值，该值将使您知道答案适合的类别：成功，错误，过高或过低。然后，您将具有使用此输出选择要显示的消息的第二种方法。如果main看到输出成功，它将更新其success变量。这可能更好，但涉及更多。

对于这个简单的示例，我建议使用选项1。实际上，由于仅检查一次成功，因此success变量可以只是一个布尔值。这将使您的代码具有以下（经过修改）：

import javax.swing.JOptionPane;

public class GuessingGame {
    // Determines whether the user has successfully guessed the number
    private static boolean success = false;

    public static void main(String[] args) {
        // Generate a random integer between 1 and 100 inclusive
        int computerNumber = (int) (Math.random() * 100 + 1);
        // Count the number of guesses that the user makes, to report it to them
        int count = 0;
        // FIXME: only for testing purposes, remove this
        System.out.println("The correct guess would be " + computerNumber);
        do {           
            count++;
            String response = JOptionPane.showInputDialog(null, "Enter a guess between 1 and 100");
            int userAnswer = Integer.parseInt(response);
            JOptionPane.showMessageDialog(null, determineGuess(userAnswer, computerNumber, count));
        } while (!success);
    }

    public static String determineGuess(int userAnswer, int computerNumber, int count) {        
        if (userAnswer <= 0 || userAnswer > 100) {
            return "Invalid guess\nNumber of Guesses: " + count;
        } else if (userAnswer == computerNumber) {
            success = true;
            return "Correct\nNumber of Guesses: " + count;
        } else if (userAnswer > computerNumber) {
            return "Incorrect, Too High\nNumber of Guesses: " + count;
        } else if (userAnswer < computerNumber) {
            return "Incorrect, Too Low\nNumber of Guesses: " + count;
        }
        return null;
    }
}

如果要使用选项2，则您可能有一个enum GuessOutcome { INCORRECT, SUCCESS, TOO_LOW, TOO_HIGH }从determineGuess返回。然后，您将使用带有getOutcomeMessage(GuessOutcome outcome)的{{1}}方法来选择要显示的消息。如果为switch (outcome) { ... }，则为outcome == GuessOutcome.SUCCESS。在此版本中，success = true可以是success的局部变量。

Java中的无限do / while循环。请指教

3 个答案: