CODEWITHSUNDEEP
javacommand-line-arguments
SyM
SyM

Reputation: 35

How to print when NumberFormatException occurs?

How do I print something when a command line argument is not an integer and a NumberFormatException occurs?

My program takes 3 command line arguments and prints certain text depending on what they are.

Here's the code:

    public class CommandLine {

  public static void main(String[] args) {

      if(args.length !=3){
          System.out.println("Error. Must give 3 values");
        }
     int x = Integer.parseInt(args[0]);
     int y = Integer.parseInt(args[1]);
     int z = Integer.parseInt(args[2]);

     if((x%2)==0 && (y%2)==0 &&(z%2)==0)
     System.out.println("Even");

else
    System.out.println("odd");

  }

}

Upvotes: 0

Views: 759

Answers (3)

Ketan G
Ketan G

Reputation: 517

Try this 

     public class CommandLine {

      public static void main(String[] args) {

          if(args.length !=3){
              System.out.println("Error. Must give 3 values");
            }
         try{
         int x = Integer.parseInt(args[0]);
         int y = Integer.parseInt(args[1]);
         int z = Integer.parseInt(args[2]);

         if((x%2)==0 && (y%2)==0 &&(z%2)==0)
         System.out.println("Even");

    else
        System.out.println("odd");


    }catch(Exception e){
     System.out.println("Exception Caught !");
    }
    }
}

Upvotes: 1

SatyaTNV
SatyaTNV

Reputation: 4135

       if(args.length !=3){
          System.out.println("Error. Must give 3 values");
        }
        else//if the above condition if true so skip these statements
        {
    try
    {
     int x = Integer.parseInt(args[0]);
     int y = Integer.parseInt(args[1]);
     int z = Integer.parseInt(args[2]);

     if((x%2)==0 && (y%2)==0 &&(z%2)==0)
     System.out.println("Even");

    else
    System.out.println("odd");
    }
    catch(NumberFormatException ne)
    {
      System.out.println("Plz! pass only integer values");//catching number format exception
    }
    }

Upvotes: 1

Karthik
Karthik

Reputation: 5040

You can catch that exception and print :

int x=y=z=Integer.MIN_VALUE;
try{
   x = Integer.parseInt(args[0]);
   y = Integer.parseInt(args[1]);
   z = Integer.parseInt(args[2]);
}catch (NumberFormatException e) {
   System.out.println("x:" +x + " y:" +y +" z:" +z); 
   e.printStackTrace();
}

The first value which is still Integer.MIN_VALUE caused your exception(unless your number is Integer.MIN_VALUE)

Upvotes: 3

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