/**
* Summary description for Class1.
*/
public class Class1
{
   private final double one = 1.0000000000000000000000;
   private final double two = 100000000000000.00000000000000;
   private final double three = 1.0000000000000000000000000000;
   private final double four = 0.123456789;
   private final double five = 0.123456789;
   private final double six = 0.000000000123456789;
   private final double seven = 0.000000000000000000123456789;
   private final double eight = -7.9228162514264337593543950335;

   public Class1()
   {
       ShowDecimalGetBits( one );
       ShowDecimalGetBits( two );
       ShowDecimalGetBits( three );
       ShowDecimalGetBits( four );
       ShowDecimalGetBits( five );
       ShowDecimalGetBits( six );
       ShowDecimalGetBits( seven );
       ShowDecimalGetBits( eight );
    }

   public void ShowDecimalGetBits( double numb )
   {
       System.Console.WriteLine("[out] ->"+numb );
   }

   /** @attribute System.STAThread() */
   public static void main(String[] args)
   {
       new Class1();
   }
}

There is no equivalent to C#'s M in J#. You should be able to just be able
to specify the decimal constant.

Thanks,

Michael Green
Microsoft Developer Support

--------------------

For Example,
       ShowDecimalGetBits( 10000000000000000000000000000M );
       ShowDecimalGetBits( 100000000000000.00000000000000M );
       ShowDecimalGetBits( 1.0000000000000000000000000000M );
       ShowDecimalGetBits( 0.123456789M );
       ShowDecimalGetBits( 0.000000000123456789M );
       ShowDecimalGetBits( 0.000000000000000000123456789M );
       ShowDecimalGetBits( -7.9228162514264337593543950335M );
where ShowDecimalGetBits is a method which takes argumant as decimal.
What is the equivalent for M in J#?If we write M in J#, it gives error.

Thanks