Question

我正在编写一个像魔术8球一样的程序。它询问一个问题，然后打印出20个我随机输入的随机响应。它没有显示编码中的任何错误，但是当我运行它时，程序完美地工作，直到它需要打印随机响应。这是一项家庭作业，所以我不是在寻找有人来纠正它，但我正在努力弄清楚如何让它发挥作用。谢谢，我感谢你的帮助。

import java.util.Scanner;


public class MagicBall
{

    public static void main(String[] args)
    {

        Scanner input = new Scanner(System.in);
        System.out.println("Please type in a question: ");
        double question = input.nextDouble();


        int number = (int)(Math.random() * 20);

        if (number == 0)
            System.out.println("As I see it, yes");

        if (number == 1)
            System.out.println("It is certain");

        if (number == 2)
            System.out.println("It is decidedly so");

        if (number == 3)
            System.out.println("Most likely");

        if (number == 4)
           System.out.println("Outlook good");

        if (number == 5)
            System.out.println("Signs point to yes");

        if (number == 6)
            System.out.println("Without a doubt");

        if (number == 7)
            System.out.println("Yes");

        if (number == 8)
            System.out.println("Yes – definitely");

        if (number == 9)
            System.out.println("You may rely on it");

        if (number == 10)
            System.out.println("Reply hazy, try again");

        if(number == 11)
            System.out.println("Ask again later");

        if (number == 12)
            System.out.println("Better not tell you now");

        if (number == 13)
            System.out.println("Cannot predict now");

        if (number == 14)
            System.out.println("Concentrate and ask again");

        if (number == 15)
            System.out.println("Don't count on it");

        if (number == 16)
            System.out.println("My reply is no");

        if (number == 17)
            System.out.println("My sources say no");

        if (number == 18)
            System.out.println("Outlook not so good");

        if (number == 19)
            System.out.println("Very doubtful");
    }
}

Answer 1

double question = input.nextDouble();

是什么让你认为问题是浮点数？

Answer 2

Apache commons非常有帮助：

RandomUtils.nextInt(maxVal);

Answer 3

您的代码适用于我，但它会强制您输入双精度（例如123.456）。你应该改变：

double question = input.nextDouble();

为：

String question = input.nextLine();

此外，您可能希望使用switch语句而不是所有ifs。

Answer 4

而不是int number = (int)(Math.random() * 20);

替换为int number = new Random().nextInt(20);

请参阅http://download.oracle.com/javase/1.4.2/docs/api/java/util/Random.html#nextInt(int)

Java中的随机数生成器

4 个答案: