CODEWITHSUNDEEP
java
Blangzo
Blangzo

Reputation: 13

I'm a beginner and I have no idea why I can't catch NumberFormatException

In my simple program(below) that just checks if a number is odd or even, I think I have tested everything that the user(me) could possible do, and everything works unless the user inputs a string or double instead of an integer for numStr. I attempted to try{...} and catch{...} but the catch isn't catching the exception that is thrown.

If for example I enter "s" instead of an integer an error shows up saying

java.lang.NumberFormatException: For input string: "s"

Could you please help me?

import javax.swing.JOptionPane;

public class HowToUseJOptionPane
{
   public static void main (String[] args)
   {
     String numStr, result;
     int num, again = 1;

     do
     {
        try 
        {
           numStr = JOptionPane.showInputDialog("Enter an integer: ");
        }
        catch (NumberFormatException nfe)
        {
           System.out.println("Requires an integer, try again");
           again = 0;
           numStr = null;
        }
        while (numStr != null)
        {
           num = Integer.parseInt(numStr); //gets number
           result = "That number is " + ((num % 2 == 0) ? "even" : "odd");
           JOptionPane.showMessageDialog(null, result); //shows result
           //JOptionPane.YES_OPTION =0, no = 1, cancel = 2
           again = JOptionPane.showConfirmDialog(null, "Do another?");
           numStr = null;
        }
     } while (again == 0);
   }
}

Upvotes: 0

Views: 107

Answers (3)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726589

You cannot catch the NumberFormatException because you are looking for it in a wrong place.

The first thing you do to debug a situation like that is to check which method would throw the exception of interest. In this case that would be the Integer.parseInt.

Once you identify the method, make sure the try/catch block includes the method call of interest. In your case, it does not, which means that your code would not catch the NumberFormatException exception.

Moving the entire try/catch into the while loop so that it encloses the call of Integer.parseInt method would fix this issue.

Upvotes: 1

cruftex
cruftex

Reputation: 5723

You need to put the try block where you do the parsing:

    try 
    {
       num = Integer.parseInt(numStr); //gets number
    }
    catch (NumberFormatException nfe)
    {
    . . .

Upvotes: 3

Henry Zhu
Henry Zhu

Reputation: 2618

The problem is that you're trying to parse the number outside the try catch block. Place it inside the try catch block.

try 
{
    numStr = JOptionPane.showInputDialog("Enter an integer: ");
    num = Integer.parseInt(numStr); //gets number
}

Upvotes: 0

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