parseInt Question.......

[sonir shah]

public class Parser
{
   public static void main( String[] args)
   {
       try 
       {
           int i = 0;
           i =  Integer.parseInt( args[0] );
       }
       catch(NumberFormatException e)
       {
          System.out.println("Problem in " + i );
       }
   }
}

I tried running the problem and it showed a compile time error saying that cannot resolve symbol 'i' in the statement"System.out.println("Problem in "+i)
Kindly explain.
Sonir

[Valentin Crettaz]

int i is declared in the try block.
System.out.println is trying to print i in the catch block but i is not in the scope.
Remember that try-catch block are two separate blocks and that they cannot share information. If you want some variable available in both blocks you have to declare it outside of them, i.e. before the try block.
HIH

[sonir shah]

Ooppssss..
I just forgot about the matter.It just went out of my mind..
Neways..Thanks val.
Sonir