Try-Catch 异常处理未提供正确的响应

Question

我不确定为什么当我输入整数值以外的任何内容时，输出没有显示 Invalid Format!。

这就是我要实现的异常处理。另外，如何在 finally 子句中关闭 Scanner 而不会导致无限循环。

class ConsoleInput {

public static int getValidatedInteger(int i, int j) {
    Scanner scr = new Scanner(System.in);
    int numInt = 0;

    while (numInt < i || numInt > j) {
        try {
            System.out.print("Please input an integer between 4 and 19 inclusive: ");
            numInt = scr.nextInt();

            if (numInt < i || numInt > j) {
                throw new Exception("Invalid Range!");
            } else {
                throw new InputMismatchException("Invalid Format!");
            }

        } catch (Exception ex) {
            System.out.println(ex.getMessage());
            scr.next();

        }

    }
    scr.close();
    return numInt;
}

这是我试图获得的输出：

Answer 1

如果您输入除 int 以外的任何内容，将在以下行抛出错误：

 numInt = scr.nextInt();

然后陷入catch块，从而跳过打印语句。您需要检查是否有下一个整数：

if(!scr.hasNextInt()) {
   throw new InputMismatchException("Invalid Format!");
}

Also, how can I close the Scanner in a finally clause without causing an infinite loop.

您不需要在 finally 块中关闭 Scanner。事实上，你根本不应该关闭它。关闭 System.in 是不好的做法。通常，如果您没有打开资源，则不应关闭它。

Answer 2

以下代码适合您，这是您可以用来获得所需输出的另一种方法。在这段代码中，我处理了 catch 块中的 InputMismatch。

public static int getValidatedInteger(int i, int j) {
            Scanner scr = new Scanner(System.in);
            int numInt = 0;

            while (numInt < i || numInt > j) {
                try {
                    System.out.print("Please input an integer between 4 and 19 inclusive: ");
                    numInt = scr.nextInt();

                    if (numInt < i || numInt > j) {
                        System.out.println("Incorrect Range!");
                    }

                } catch (InputMismatchException ex) {
                    System.out.println("Incorrect format!");
                    scr.next();

                }

            }
            scr.close();
            return numInt;
        }

Answer 3

您需要在异常捕获块之前的单独捕获块中捕获 InputMismatchException 并添加 scr.next();如下所示：

public static int getValidatedInteger(int i, int j) {
    Scanner scr = new Scanner(System.in);
    int numInt = 0;

    while (numInt < i || numInt > j) {
        try {
            System.out.print("Please input an integer between 4 and 19 inclusive: ");
            numInt = scr.nextInt();

            if (numInt < i || numInt > j) {
                throw new Exception("Invalid Range!");
            }
        } catch (InputMismatchException ex) {
            System.out.println("Invalid Format!");
            scr.next();
        } catch (Exception ex) {
            System.out.println(ex.getMessage());
            scr.next();

        }

    }
    scr.close();
    return numInt;
}

Try-Catch 异常处理未提供正确的响应

Try-Catch Exception Handling does not provide the correct response

java

error-handling

exception-handling

exception

try-catch