Simple exception handling in Java

Question

Can you please let me know if my code is correct? I'm studying for my test in two hours so I dont really have time to write an application to test it.

Question is: if I have a JLabel with a number as its label. simply a label that says 34 for example. I want to extract the number from the label. but i need to handle exceptions, i.e it's not a number, it can be a letter.

would my code below handle the exception correctly?

JLabel label = new JLabel("34");
int extracted;

this is what i would do

try{
    extracted = Integer.parseInt(extracted.getText());
    System.out.println("the number was: "+ extracted);
}
catch(IOException exception){
    System.out.println(label.getText() + " is not a number");
}

Craig Otis · Accepted Answer · 2009-04-09 16:02:09Z

8

Close, but catching an IOException won't work because that exception type is not thrown by the parseInt() method. Try catching a NumberFormatException instead:

try{
    extracted = Integer.parseInt(extracted.getText());
    System.out.println("the number was: "+ extracted);
} catch(NumberFormatException exception) {
    System.out.println(label.getText() + " is not a number");
}

answered Apr 9, 2009 at 16:02

Craig Otis

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Comments

Brian Agnew · Accepted Answer · 2009-04-09 15:58:34Z

6

I would check the documentation for Integer.parseInt()

Furthermore, I'd strongly recommend setting up a test project in whatever IDE you use so you can test this stuff yourself with a rapid turnaround! Even if it's a vim/javac+make script.

answered Apr 9, 2009 at 15:58

Brian Agnew

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Comments

Dan Lew · Accepted Answer · 2009-04-09 15:59:18Z

2

It's almost correct, except that you're catching the wrong exception; parseInt() throws a NumberFormatException.

answered Apr 9, 2009 at 15:59

Dan Lew

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Comments

cletus · Accepted Answer · 2009-04-09 15:59:39Z

1

Integer.parseInt() throws a NumberFormatException not an IOException.

answered Apr 9, 2009 at 15:59

cletus

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Comments

Ted Heich · Accepted Answer · 2009-04-09 16:08:34Z

1

NumberFormatException is a RunTimeException (unchecked), for compilation purposes, you don't really have to write it in the catch portion.

If what you are trying to do is determine if the user will type numbers in the JTextField (and not any other character), you should look at regex (Regular expressions), instead of catching this one using the try .. catch mechanism.

answered Apr 9, 2009 at 16:08

Ted Heich

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2 Comments

Bill K Over a year ago

Why in earth would you want to do that? Regex have GOT to be slower, and are certainly more fallible. You would lose internationalization and any other things the built-in implementation gives you. Although a test would be better than an exception, anything is better than regex

Ted Heich Over a year ago

Well yes, I agree, but there wasn't anything in the code sample and on the original question which pertains to internationalization. And I based recomm on regex just by second guessing his intentions, which is to test if the characters inputted were numerics. I respect your opinion on regex though.

Randolpho · Accepted Answer · 2009-04-09 15:59:26Z

0

You should catch NumberFormatException. Otherwise, fine.

answered Apr 9, 2009 at 15:59

Randolpho

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Simon Nickerson · Accepted Answer · 2009-04-09 16:01:20Z

0

The javadoc for Integer.parseInt() states that it can throw a NumberFormatException, not an IOException.

The code you have written will not compile, because IOException is a checked exception which cannot be thrown by any of the code in the try block.

answered Apr 9, 2009 at 16:01

Simon Nickerson

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Arthur Thomas · Accepted Answer · 2009-04-09 16:02:05Z

0

http://java.sun.com/javase/6/docs/api/java/lang/Integer.html#parseInt(java.lang.String)

you should use a NumberFormatException

http://java.sun.com/javase/6/docs/api/java/lang/NumberFormatException.html

answered Apr 9, 2009 at 16:02

Arthur Thomas

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