WIDENING and VAR-ARG

[Bob Wheeler]

Source: ExamLab from Deveka (practise exam) - Exercise 9

  
public class VarArg{

    public static void main(String argv[]){
         doNumber(2);
         doNumber(2L);
         doNumber(new int[]{1,2,3,4}); // i expected WIDENING before var-arg
         doNumber(new Integer(2));
         doNumber(new Integer[2]);
    }

    public static void doNumber(int dn){
         System.out.print("a");
    }

    public static void doNumber(int... dn){
         System.out.print("b");
    }

    public static void doNumber(Integer dn){
         System.out.print("c");
    }

    public static void doNumber(Integer... dn){ // but instead this is called
         System.out.print("d");
    }

    public static void doNumber(Object dn){
         System.out.print("e");
    }

}
  

Output: aebcd

I was surprised that the call with the int array goes to the var-arg method and not the object method. I thought var-arg is always used last???
But not here? I mean, an array is an object and that int array could be widened to an object. And because widening is before var-arg ...
Where is my error? (same for the Integer array)

Thanks
cheers
Bob

[Ankit Garg]

First of I think you pointed out the wrong method that's called. Try this

  
public class VarArg{

    public static void main(String argv[]){
         doNumber(2);
         doNumber(2L);
         doNumber(new int[]{1,2,3,4}); // i expected WIDENING before var-arg
         doNumber(new Integer(2));
         doNumber(new Integer[2]);
    }

    public static void doNumber(int dn){
         System.out.print("a");
    }

    /*public static void doNumber(int... dn){
         System.out.print("b");
    }*/

    public static void doNumber(Integer dn){
         System.out.print("c");
    }

    public static void doNumber(Integer... dn){ // but instead this is called
         System.out.print("d");
    }

    public static void doNumber(Object dn){
         System.out.print("e");
    }

}
  

Now the method with Object argument will be called. when you create a var-arg method, then it is converted into a normal array accepting method (I think you must be knowing that already). Here var-arg is chosen because it is a direct match to the int... method. So that's why the compiler didn't go for widening...

[Bob Wheeler]

Ankit Garg wrote:...
Now the method with Object argument will be called. when you create a var-arg method, then it is converted into a normal array accepting method (I think you must be knowing that already). Here var-arg is chosen because it is a direct match to the int... method. So that's why the compiler didn't go for widening...

Yeah, I SHOULD have known that
But at least I know it now.

Thanks a lot
Bob