Java 用户界面改进

Question

我一直在努力向用户更好地展示这一点。我正在尝试设置，如果他输入错误的输入，那么他必须重新开始，直到他输入 0 到 9000 之间的正确数字集。它也是乘法迭代和递归。但是我不确定为什么当他出错时我设置的消息没有显示。知道原因和改进想法，有代码示例吗？

import java.util.InputMismatchException;
import java.util.Scanner;

public class Multiplication {

public static int multIterative(int a, int b) {

    int result = 0;

    while (b > 0) {

        result += a;
        b--;
    }
    return result;
}

public static int multRecursive(int a, int b) {

    if (a == 0 || b == 0) {

        return 0;
    }

    return a + multRecursive(a, b - 1);

}

public static void main(String[] args) {

    int a = 0;
    int b = 0;

    Scanner userInput = new Scanner(System.in);

    do {

        System.out.print("Please enter first Integer: ");
        System.out.print("Please enter second Integer: ");

        try {

            a = userInput.nextInt();
            b = userInput.nextInt();

        } catch (InputMismatchException e) {

            System.out.println("Must enter an integer!");
            userInput.next();

        } catch (WhosebugError e) {

            System.out.println("Thats too much");
            userInput.next();

        }

    } while (a >= 9000 || b >= 9000);

    System.out.println("The Multiplication Iteration would be: "
            + multIterative(a, b));

    System.out.println("The Multiplication Recursion would be: "
            + multRecursive(a, b));

}
}

Answer 1

如果用户输入的值大于 9000，则不会抛出任何错误，因此您的 "Thats too much" 消息将永远不会显示，要显示该消息，您必须执行如下检查：

if(a >= 9000 || b >= 9000){
   System.out.println("Thats too much");
}

Answer 2

我建议改用 Integer.parseInt(...) 并检查 NumberFormatException。

以下临时工作：

do
        try {
            System.out.print("Please enter first Integer: ");
            a = Integer.parseInt(userInput.next());
            System.out.print("Please enter second Integer: ");
            b = Integer.parseInt(userInput.next());
            if (a >= 9000 || b >= 9000)
                throw new WhosebugError();
            break;

        } catch (NumberFormatException e) {

            System.out.println("Must enter an integer!");

        } catch (WhosebugError e) {

            System.out.println("Thats too much");

        }
    while (true);

Java 用户界面改进

Java User Interface Improvement

java

error-handling

recursion

user-interface

user-input