Question

我想问几个问题，请参阅我在代码中添加的评论部分，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

阅读java手册后我还不熟悉nextLine（）参数，你介意解释一下吗？我想做的就是＆＃34; Clear Scanner Buffer＆＃34;因此它不会为println循环并要求用户再次输入A，这是一种正确的方法吗？

在input.nextLine();之后使用input.nextInt();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，将内容保留在缓冲区中如果input.nextInt();没有先清除<{1}}，会导致Exception继续投放

由于这与上述情况相似，是否可以重用try-catch块来处理b（甚至更多的输入，如c d e ...）异常？

你可以，但如果输入b错误会怎样？你要求用户重新输入输入吗？如果您有100个输入并且它们最后一个输入错误会发生什么？您实际上最好还是编写一个方法来执行此操作，即提示用户输入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;
}

为何选择＆amp; b找不到？

因为它们尚未初始化且编译器无法确定它们是否具有有效值...

尝试更改它的更多内容。

int a = 0;
int b = 0;

Answer 2

是的，没关系。并将使用非整数输入。
是。如果我们将其提取到方法中。
因为编译器认为它们可能未初始化。

让我们简化并提取一种方法，

private static int readInt(String name, Scanner input) {
    while (true) {
        try {
            System.out.printf("Input %s: ", name);
            return input.nextInt();
        } catch (InputMismatchException ex) {
            System.out.printf("Input %s is Wrong%n", input.nextLine());
        }
    }
}

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

    int a = readInt("A", input);
    int b = readInt("B", input);
    System.out.println("The sum is " + (a + b));
}

Answer 3

我已对该问题进行评论。

package test;

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

public class Test {

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

    boolean accept_a = false;
    boolean accept_b = false;
    int a=0;
    int b=0;

    System.out.print("Input A: ");

    while (accept_a == false) {
        try {

            a = input.nextInt();  // it looks for integer token otherwise exception     

            accept_a = true;
        } catch (InputMismatchException ex) {
            System.out.println("Input is Wrong");
            input.next();    // Move to next other wise exception // you can use hasNextInt()         

        }
    }

    System.out.print("Input B: ");
    while (accept_b == false) {
        try {

            b = input.nextInt();
            accept_b = true;
        } catch (InputMismatchException ex) {     

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

    }
    System.out.println("The sum is " + (a + b)); // complier doesn't know wheather they have initialised or not because of try-catch blocks. so explicitly initialised them.

}

}

Answer 4

Check out this "nextLine() after nextInt()"

并将变量a和b初始化为零

nextInt（）方法不会读取最后一个换行符。

关于Java中的一些简单异常处理

4 个答案: