java - 限制 float 参数的允许值

Question

我是 Java 新手，正在开发管理日程的程序。

是否可以限制作为函数参数的float的值？

由于这是一个时间表，因此小时值必须介于 0 和 24 之间。

Answer 1

一个简单的异常处理示例是:

     class ValidateDouble {

        public static void main(String[] args) {

        // Your initial value
        double value = 0;
        System.out.println("Your initial value is: " + value);

        // Not valid argument, should print 0 as returned from method
        value = inputDouble(-1);
        System.out.println("After entering -1, your value is: " + value);

        // Not valid argument, should print 0 as returned from method
        value = inputDouble(-1.5);
        System.out.println("After entering -1.5, your value is: " + value);

        // Not valid argument, should print 0 as returned from method
        value = inputDouble(25);
        System.out.println("After entering 25, your value is: " + value);

        // Valid argument, should print the returned argument value
        value = inputDouble(2);
        System.out.println("\nAfter entering 2, your value is: " + value);

        // Valid argument, should print the returned argument value 
        value = inputDouble(2.54);
        System.out.println("\nAfter entering 2.54, your value is: " + value);

    }

    private static double inputDouble (double number) {

        // Program will try to execute this
        try {

            // If number is out of range, throw exception if not, return the argument.
            if (number < 0 || number > 24) {
                throw new IllegalArgumentException("\nNot a valid argument...");
            } else {
                return number;
            }

        } catch (IllegalArgumentException exception) {

            // Print message thrown in the exception and return 0.
            System.out.println(exception.getMessage());
            return 0;
        }

    }

}

我知道这可能不是您想要的，我已经根据您的问题发布了一个示例，如何验证方法参数。有很多方法可以验证用户输入，但只要您询问验证方法参数，我认为这可以帮助您实现目标。干杯。

您将无法在方法原型(prototype)中执行此操作。