发生错误时的无限循环

Question

When this code is run with a string as the input so that an error occurs it causes an infinite loop of the error message. I have tried inserting break; after the error message which does stop the loop but also stops the program. I would like it to loop back to the input request after an error occurs.

import java.util.Scanner;

public class CubeUser                           
{                                               

    public static void main(String argv[])      
    {
        Scanner in = new Scanner(System.in);
        boolean error = true;
        System.out.print("Please input the length of the cube:");                               
        while(error == true)
        {
            if (in.hasNextDouble())
            {
                double length = in.nextDouble();
                Cube cube1 = new Cube(length);                                                          
                System.out.println("The surface area of cube1 is " + cube1.calculateSurfArea() );       
                System.out.println("The volume of cube1 is " + cube1.calculateVolume() );
                error = false;
            }
            else
            {
                System.out.println("Please enter a numerical value for the cube's length.");
            }
        }
        in.close(); 

    }  
}   

Answer 1

Move your scanner's cursor in case of error, otherwise it will keep reading the same value .

else {
     System.out.println("Please enter a numerical value for the cube's length.");
     in.next();
}

Off note: Use if(error) instead of (error == true) . Latter is bit frowned upon.

Answer 2

if (in.hasNextDouble())

This will fire the first time, when the user types it in. When there is an error though, it doesn't give the user a chance to type in a double value, hence the infinite loop.

Restructure your loop like so:

String input;
while((input = in.nextDouble()) != null)
{
    // Force the user to type a value.
    // The rest of your code here.
}

Answer 3

您必须阅读，输入，如果它不是一个双例如使用in.next()否则， in.nextDouble()当然会在接下来的迭代中真实的还有在“排队”非双值。