how to avoid NumberFormatException

[Rajesh Kesarwani]

As we know, while parsing a string in to int/floal by using the method Integer.parseInt(String s)/Float.parseFloat(String s), it will return int/float value, if the String contains any numeric value like "101" or "101.11". but if it is not numeric like "a12", then it will throw NumberFormatException.

Please let me know, is there way to avoid this Exception except handling the exception means try catch block.

Thanks in advance.

[Paul Sturrock]

Personally I check the string first with a regular expression. But you could also split the sting in to a character array and check each character is acceptable in an number.

[Chris Beckey]

You could just let it happen and then catch the exception, i.e.

public static void main(String[] argv)
{
int i=42;
try{i=Integer.parseInt("X88");}
catch(NumberFormatException nfX)
{i=0;} // set a default value here?

System.out.println("i=" + i);
}

Personally I find that easier than doing a validity check beforehand.
What to do if the string is not a valid numeric is more an issue and that's app specific.

[Mahesh Shiwani]

To avoid this exception you have to feed a valid integer or float values to parsers... this is the idea only... as you can not peal coconuts using potato pealers.. so to avoid breakage (of pealer) use potato only... not the coconut.. otherwise.. just check before pealing.. whether you are trying with potato only or not.... simple..

[Chetan Raju]

Why dont you want to make use of try catch ? Java provides very robust exception handling mechanism.

[Jim Yingst]

JDK 5 provided another nice alternative:

        String input = "123";
 
        Scanner sc = new Scanner(input);
        if (sc.hasNextInt())
            System.out.println("Hey, look, an int: " + sc.nextInt());
        else
            System.out.println("That there's a sorry excuse for an int: " + sc.next());