Question 9





The following code will give

1:     Byte b1 = new Byte("127");
2:
3:     if(b1.toString() == b1.toString())
4:        System.out.println("True");
5:     else
6:        System.out.println("False");

A) Compilation error, toString() is not avialable for Byte.
B) Prints "True".
C) Prints "False".