Try and Catch using Integer.parseInt() syntax error [Java]

Question

I am trying to make a simple method which test to see if a provide String contains only numbers, to do this I am trying to use try and catch (just learnt it and I would like to practise putting it to use) where I try to parseInt() the given String and if there's an error (not a number) then it will catch it and return false;

    public boolean checkNumber(String s){
    try(Integer.parseInt(s)){
        return true;
    }
    catch(Exception E){
        return false;
    }
}

It says I have a misplaced constructor.

You actually should be using regex expressions... But that is outside the scope of your question. — skiwi
– skiwi, Commented Mar 4, 2014 at 16:01
To elaborate on skiwis comment, a string representation of a number falling ouside the range of an integer will throw an exception when parsed by Integer.parseInt — William Morrison
– William Morrison, Commented Mar 4, 2014 at 16:04
The syntax you are using is for "try with resources". You might want to look it up, can be useful in other circumstances. — Drunix
– Drunix, Commented Mar 4, 2014 at 16:06
@Khaelid On everything that is not a number, an Exception will be thrown and caught, which are generally a big order of magnitude slower than simple checks. Exceptions are for exceptional situations. — skiwi
– skiwi, Commented Mar 4, 2014 at 16:09

David · Accepted Answer · 2014-03-04 16:02:30Z

6

Catch the correct exception and move the try check into the block rather than in brackets.

 try {
       Integer.parseInt(s);
       return true;
    }
    catch(NumberFormatException e){
        return false;
    }

answered Mar 4, 2014 at 16:02

David

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1 Comment

David Over a year ago

This is true but it'd be defensive programming to enforce a null check, it could be expected for a NPE to throw if s is passed in as null as just blindly checking for it could hide other errors.

Khaelex · Accepted Answer · 2014-03-04 16:03:29Z

1

That's not the correct syntax for try. Use

try
{
    Integer.parseInt(s);
    return true;
}
catch (Exception ex)
{
    return false;
}

answered Mar 4, 2014 at 16:03

Khaelex

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Comments

poitevinpm · Accepted Answer · 2014-03-04 16:05:28Z

1

syntax error

try {
  Integer.parseInt(s);
} catch(Exception e) {
  return false;
}
return true;

answered Mar 4, 2014 at 16:05

poitevinpm

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