overloading example from examlab

[S Ali]

   public class Fex2{
    public static void main(String args[]){
         byte b1=12;
         Byte B1=new Byte(b1);
         short s1=6;
         Short S1=new Short(s1);
         int i1=7;
         Integer I1=new Integer(i1);
         Long l1=new Long(45L);
         Number nm=new Short("4");

         capture(b1);
         capture(B1);
         capture(s1);
         capture(s1+=20000);
         capture(S1);
         capture(nm);
         capture(I1.intValue());
         capture(Short.valueOf("110110",2));
         capture(Short.parseShort("110110",2));
         capture();
    }

    static void capture(short x){
         System.out.print("A");
    }
    static void capture(int x){
         System.out.print("B");
    }
    static void capture(long x){
         System.out.print("C");
    }
    static void capture(Short x){
         System.out.print("D");
    }
    static void capture(Integer x){
         System.out.print("E");
    }
    static void capture(Long... x){
         System.out.print("F");
    }
    static void capture(Number x){
         System.out.print("G");
    }
} 

My doubt is how come capture(s1+=20000) evaluate to short !! isn't 20000 considered an integer literal therefore producing the output as an int?

[S Ali]

ok then s1+=20000 is different from s1 = s1 + 20000

[Dejan Miler]

capture(s1+=20000); 

First thing that is happens here is that JVM pass reference value s1 which is 6 to the method and when that method execution
is over then JVM add value of 20000 to the value of s1.

Dejan

[Chandana Garlapati]

What about capture();?? why it prints "F"??
I dont know why it is considering as varargs i.e; Long... x in this program..

can anyone help me??

[Siva Masilamani]

var...arg means zero or more argument.

In the above program the only function that can take zero argument is capture and hence the output F.

But if the function call has arguments then compiler will choose var...arg method as the last option to select from.

[Chandana Garlapati]

Thanks Siva Masilamani

[Francisco Montes]

Dejan,

I´m sorry but it does not work the way you mentioned. Try this:

public class Test {
    public static void main(String ... args) {
        short s = 6;
        capture(s+=20000);
        System.out.println("Value after calling capture: "+s);
    }

    public static void capture(short s) {
        System.out.println("Value received in capture method: "+s);
    }
}

This is the output:
Value received in capture method: 20006
Value after calling capture: 20006

So in fact, the expression is completely evaluated before passing its result into the method.

[Punit Singh]

compiler converts this

 
capture(s+=20000);
   

to this

 
        s += 20000;
        capture(s);  

and s+=20000; works differently then s=s+2000; as
for s+=2000; implicit conversion is done from int to short.
and for s=s+2000; we need to do explicit conversion.