wrapper class?

[srikanth reddy]

class Q3 
{
   public static void main(String[] args)
   {
      Byte b1 = new Byte("127");
      if(b1.toString() == b1.toString())
         System.out.println("true");
      else
         System.out.println("b1.toString()");
   }
}

here the Output iam getting is "b1.toString()"...here we have only one reference value ..what are we doing here by calling toString method...
can anyone help me???

thanks

srikanth
[ September 28, 2005: Message edited by: Michael Ernest ]

[Lalitha Gottumukkula]

I think if(b1.toString() == b1.toString()) is trying to compare two different String objects, not their values. b1.toString() returns a string equivalent of object b1.

Try to use if(b1.toString().equals(b1.toString()). This will compare values in the objects.

[Michael Ernest]

Maybe this will help:

public class Test
{
   public static void main(String[] args)
   {
      Byte b1 = new Byte("127");
      String s1 = b1.toString();
      String s2 = b1.toString();
 
      if (s1 == s2)
         System.out.println("toString() returns the same reference");
      else
         System.out.println("toString() returns different references");
 
      if (s1.equals(s2))
         System.out.println("s1 and s2 have the same value.");
      else
         System.out.println("s1 and s2 have different values.");
 
   }
}

Regards,

[agrah upadhyay]

 class Q3 
{
   public static void main(String[] args)
   {
      Byte b1 = new Byte("127");
      if(b1.toString() == b1.toString())
         System.out.println("true");
      else
         System.out.println("b1.toString()");
   }
}
 
here the Output iam getting is "b1.toString()"...here we have only one reference value ..what are we doing here by calling toString method...
can anyone help me???
 
 

Here on calling b.toString() in b1.toString() == b1.toString() we get two different String object because call to toString() is
done at run time.so == should result in false.Please correct me iff i am wrong
[ September 28, 2005: Message edited by: agrah upadhyay ]

[agrah upadhyay]

Haa Haa Haa
With above posting i have completed my 101 posts

[Arvind Giri]

yes! you are using only one reference b;
However keep in mind b.toString() returns a new String object, everytime it has been called.
That is why the actual checking here is between the reference returned by two b.toString(), which are different.