线程“main”中的异常，这是什么意思？

Question

I'm working on a homework problem for class. Where you have to calculate the distance between two points. The code is basically done, but I have one question. When I enter q to end the loop. I get a message back.

Exception in thread "main" java.lang.NumberFormatException: For input string: "q"

at.sun.misc.FloatingDecimal.readJavaFormatString(Unkown Source)

at java.lang.Double.parseDouble(Unkown Source)

atDistance.main(Distance.java:11)

import java.util.Scanner;
public class Distance {
public static void main(String[] args){
    Scanner input = new Scanner(System.in);
    while (true){
        System.out.print("Enter coordinate for x1: ");
        String x1String = input.next();
        if (x1String == "q")
            break;
            double x1 = Double.parseDouble(x1String);

        System.out.print("Enter coordinate for y1: ");
        String y1String = input.next();
        if (y1String == "q")
            break;
            double y1 = Double.parseDouble(y1String);

        System.out.print("Enter coordinate for x2: ");
        String x2String = input.next();
        if (x2String == "q")
            break;
            double x2 = Double.parseDouble(x2String);

        System.out.print("Enter coordinate for y2: ");
        String y2String = input.next();
        if (y2String == "q")
            break;
            double y2 = Double.parseDouble(y2String);

        double distance = (Math.pow(x2 - x1,2)) + (Math.pow(y2 - y1,2));
        distance = Math.sqrt(distance);
        System.out.printf("The distance is %5.2f",distance);
        System.out.println("");
    }
}//main
}//Distance

That is the code I have written. Any help is appreciated.

Answer 1

It means 'q' is not a number. To compare strings you have to use equals , == just compares references.

Answer 2

You don't do a string compare with ==. The "==" comparison checks to see if they are the exact same objects, not if the strings contain the same characters. Try x1String.equals("q") instead.

What's happening now is that the "==" will say "these aren't the same object" and then it will attempt to parse the "q" as a double in the next line, which is throwing the exception.

Answer 3

This old chestnut...

You can not compare Strings (safely) using == . Use .equals() instead:

if (x1String.equals("q")) // change every == to .equals()

Java != Javascript

Answer 4

x1String == "q" simply compares references. In order to compare the input string with the string "q" , you need to use compareTo or equals :

if (x1String.compareTo("q") == 0)
// or
if (x1String.equals("q"))

Answer 5

As others said, you should be using a function such as equals to compare the contents of the objects ( x1String and "q" ) rather than comparing the actual objects/references, but you should also be checking if the string is numeric and/or catch ing any exceptions that are thrown by parseDouble .

What if someone inputs "asdfg" ?