Question

无法弄清楚为什么我的代码只在需要字符串时才返回Int，并且帮助会很棒。代码如下。我尝试将变量声明为String而没有运气。

我想要返回3个随机字符串：cherry，grape，bell或x

import java.util.Scanner;
import java.util.Random;

public class slot {
    public static void main(String[] args)
    {
        String answer = "y";
        int cherry;
        int grape;
        int bell;
        int x;



        Random generator = new Random(); // random generator
        Scanner scan = new Scanner (System.in); // scanner class

        System.out.println("Would you like to play the slot machine?(y/n): ");
        answer = scan.nextLine();

        while(answer.equalsIgnoreCase("y"))
        {
             cherry = generator.nextInt(5); // generates a random number
            grape = generator.nextInt(5);
            bell = generator.nextInt(5);

            System.out.println("The three numbers of the slot machine are: " + cherry +grape +bell);

            if(cherry == grape && grape == bell)
               {
                System.out.println("JACKPOT! All three of the same");
               }

            if(cherry == grape || cherry == bell || grape == bell )
               {
                System.out.println("Close, you got two of the same!!");
               }
            else
               {
                System.out.println("Not a winner");
               }

            System.out.print("Try again?(y/n): ");
            answer = scan.nextLine();
            System.out.println();
        }


        System.out.println("Bye!!");

    }

}

Answer 1

我会这样表达：

// The different results each "wheel" / "column" on the slot machine.
String[] results = { "cherry", "bell", "grape", "x" };

// Create a random result for each wheel.
String wheel1 = results[generator.nextInt(results.length)];
String wheel2 = results[generator.nextInt(results.length)];
String wheel3 = results[generator.nextInt(results.length)];

然后继续您的if语句。（但是对于第二和第三个陈述，请else if。）

if (wheel1 == wheel2 && wheel2 == wheel3) {
    // jackpot
} else if (wheel1 == wheel2 || wheel2 == wheel3 || wheel1 == wheel3) {
    // two equal
} else {
    // all three different.
}

如果您想深入了解该语言，我建议您查看enum s。

（请注意，使用==比较字符串在9个案例中有10个是坏主意。但是，我们不需要费心去比较字符串内容但可以逃脱通过比较参考值。）

Answer 2

将字符串声明为int并不会使它们成为整数值。你需要做的是创建一个包含你正在使用的单词的数组。然后生成一个随机int值，该值在数组范围内。

然后你将从数组中选择你所拥有的随机int所指定位置的单词。

编辑：对不起，我没有看完你的整个问题。你能告诉我们印刷什么吗？

Answer 3

您正在打印int的值。试试这个：你生成一个数字，根据这个数字你选择你的字符串。

Random generator = new Random();
    int a = generator.nextInt(5); 
    int b = generator.nextInt(5);  
    int c = generator.nextInt(5); 

    String roll1 = null;
    switch(b){
    case 1: roll1 = "cherry";
            break;
    case 2: roll1 = "grape";
            break;
    case 3: roll1 = "bell";
            break;
    default: roll1 = "xxx";
             break;
    }
    //repeat for b and c with roll2 and roll3
    System.out.println(roll1);

需要返回字符串但只能获得带随机数生成器的int

3 个答案: