java - 关于Java中的一些简单的异常处理

Question

有几个问题想请教，请引用我在代码中添加的注释部分，谢谢。

package test;

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

public class Test {
    /* Task:
     prompt user to read two integers and display the sum. prompt user to read the number again if the input is incorrect */

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        boolean accept_a = false;
        boolean accept_b = false;
        int a;
        int b;

        while (accept_a == false) {
            try {
                System.out.print("Input A: ");
                a = input.nextInt();            /* 1. Let's enter "abc" to trigger the exception handling part first*/

                accept_a = true;
            } catch (InputMismatchException ex) {
                System.out.println("Input is Wrong");
                input.nextLine();                 /* 2. I am still not familiar with nextLine() parameter after reading the java manual, would you mind to explain more? All I want to do is "Clear Scanner Buffer" so it wont loop for the println and ask user to input A again, is it a correct way to do it? */

            }
        }

        while (accept_b == false) {
            try {
                System.out.print("Input B: ");
                b = input.nextInt();
                accept_b = true;
            } catch (InputMismatchException ex) {              /*3. Since this is similar to the above situation, is it possible to reuse the try-catch block to handling b (or even more input like c d e...) exception? */

                System.out.println("Input is Wrong");
                input.nextLine();
            }

        }
        System.out.println("The sum is " + (a + b)); /* 4. Why a & b is not found?*/

    }
}

Answer 1

2. I am still not familiar with nextLine() parameter after reading the java manual, would you mind to explain more? All I want to do is "Clear Scanner Buffer" so it wont loop for the println and ask user to input A again, is it a correct way to do it?

在 input.nextInt(); 之后使用 input.nextLine(); 是为了清除输入流中的剩余内容，因为(至少)新行字符仍在缓冲区中，如果没有首先清除，则将内容保留在缓冲区中将导致 input.nextInt(); 继续抛出 Exception

3. Since this is similar to the above situation, is it possible to reuse the try-catch block to handling b (or even more input like c d e...) exception?

可以，但是如果输入 b 错误怎么办？是否要求用户重新输入输入a？如果你有 100 个输入，而最后一个输入错误，会发生什么？实际上，你最好编写一个方法来执行此操作，即提示用户输入一个值并返回该值的方法

例如...

public int promptForIntValue(String prompt) {
    int value = -1;
    boolean accepted = false;
    do {
        try {
            System.out.print(prompt);
            value = input.nextInt();
            accepted = true;
        } catch (InputMismatchException ex) {
            System.out.println("Input is Wrong");
            input.nextLine();
        }
    } while (!accepted);
    return value;
}

4. Why a & b is not found?

因为它们尚未初始化，编译器无法确定它们是否具有有效值...

尝试将其更改为类似的内容。

int a = 0;
int b = 0;