如何使用 Scanner 处理由无效输入(InputMismatchException)引起的无限循环 [英] How to handle infinite loop caused by invalid input (InputMismatchException) using Scanner

问题描述

所以，我被这段代码困住了:

So, I'm getting stuck with this piece of code:

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

public class ConsoleReader {

    Scanner reader;

    public ConsoleReader() {
        reader = new Scanner(System.in);
        //reader.useDelimiter(System.getProperty("line.separator"));
    }

    public int readInt(String msg) {
        int num = 0;
        boolean loop = true;

        while (loop) {
            try {
                System.out.println(msg);
                num = reader.nextInt();

                loop = false;
            } catch (InputMismatchException e) {
                System.out.println("Invalid value!");
            } 
        }
        return num;
    }
}

这是我的输出:

插入一个整数:
无效值！
插入一个整数:
无效值！
...

Insert a integer number:
Invalid value!
Insert a integer number:
Invalid value!
...

推荐答案

根据 javadoc 用于扫描仪:

当扫描仪抛出一个InputMismatchException，扫描器不会传递导致的令牌例外，所以它可能是通过其他一些检索或跳过方法.

When a scanner throws an InputMismatchException, the scanner will not pass the token that caused the exception, so that it may be retrieved or skipped via some other method.

这意味着如果下一个标记不是 int，它会抛出 InputMismatchException，但标记会留在那里.所以在循环的下一次迭代中，reader.nextInt() 再次读取相同的标记并再次抛出异常.你需要的是用完它.在你的 catch 中添加一个 reader.next() 来消费令牌，令牌无效，需要丢弃.

That means that if the next token is not an int, it throws the InputMismatchException, but the token stays there. So on the next iteration of the loop, reader.nextInt() reads the same token again and throws the exception again. What you need is to use it up. Add a reader.next() inside your catch to consume the token, which is invalid and needs to be discarded.

...
} catch (InputMismatchException e) {
    System.out.println("Invalid value!");
    reader.next(); // this consumes the invalid token
} 

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