Confusion about autoboxing

[Abdullah Mamun]

1 public class Test1 {
2    static void printf(Number a, Number b) {System.out.println("Method 1");}
3    static void printf(Integer a, Integer b) {System.out.println("Method 2");}
4    
5    public static void main(String... args) {
6        byte s = 127;
7        printf(s,s);
8        printf(1,1);
9    }
10}

I was expecting, line 7 will print "Method 2" but the actual output is "Method 1".
I think its because of "byte s" is wrapping to Byte object and since there is no such method with this type it is calling the method with Number type parameter.
I want to know if there is any other explanation.
[ July 06, 2007: Message edited by: Abdullah Al Mamun ]

[Manfred Klug]

Originally posted by Abdullah Al Mamun:
I think its because of "byte s" is wrapping to Byte object and since there is no such method with this type it is calling the method with Number type parameter.

Your assumption is correct.

[Louis Moloney]

byte will never box to Integer as the Number subclasses are peers
ie Integer does not subclass Byte
eg

    
         Number
            |
       ------------------------
       |            |         |
      Byte     Integer     Double etc
 

[ July 06, 2007: Message edited by: Louis Moloney ]

[Veeraraagavan Rajachandrasekhar]

Your assumption is right.....If you want a more detailed answer read the next few lines.

byte is autoboxed to Byte.
In the code there is no method with a Byte,Byte argument.
A "Byte" cannot be autoboxed to an "Integer" since they(these Wrapper Classes) are different from each other. Remember that the following classes extend "Number" class:
AtomicInteger, AtomicLong, BigDecimal, BigInteger, Byte, Double, Float, Integer, Long, Short. Therefore, a "Byte" is a "Number" and hence the output "Method1".
[ July 06, 2007: Message edited by: Veeraraagavan Rajachandrasekhar ]