varargs overloaded method

[Nidhi Sar]

This works fine:

   public class Ex{
	static void doStuff(int i){System.out.println("int");}
	static void doStuff(Integer iI){System.out.println("Integer");}

	public static void main(String[] args){
		doStuff(6);
	}
}
 

The output is int

However, this fails to compile (ambiguous method call)

   public class Ex{
	static void doStuff(int... i){System.out.println("int");}
	static void doStuff(Integer... iI){System.out.println("Integer");}

	public static void main(String[] args){
		doStuff(6);  //This line doesn't compile - ambiguous call
	}
}
 

Why the difference between direct call and var-args call??

Thanks.

[Pramod P Deore]

here you have overloaded methods in Ex class one which takes var args then you cannot pass a single int or Integer variable to call the method because of this compiler gives you ambiguous error.
Here you have to pass int[] or Integer[] to call particular method.

[Rajiv Chopra]

Nidhi Sar wrote:Why the difference between direct call and var-args call??

Boxing + Varargs and Widening + varargs you can do only manually exclusive manner not togather.

[Ankit Garg]

There has been a long discussion on this these days here...

[Nidhi Sar]

There has been a long discussion on this these days here...

Thanks for pointing that out. Yes, I have been following that great discussion. In fact, this question is an offshoot of one of the posts in that discussion. Started a new topic, since it seemed a little OT to the original post.

I'm sorry, but can someone please explain this more.

In example 1, there is no widening involved, just boxing... if needed. Since a method call that takes a primitive variable(int) is present, the boxing is not needed either.
In example 2, there are two options - varargs only/ boxing + varargs. If in example 1, the method knows to go to the primitive method call, and not do the boxing needlessly, then why not the same in example 2. Yes, there is varargs also, but again the choice is between primitive and boxing so why not go just to primitive.

The question seems clear enough in my head, but not sure if I am being able to make sense!!!

[Rajiv Chopra]

Nidhi Sar wrote:

There has been a long discussion on this these days here...

Thanks for pointing that out. Yes, I have been following that great discussion. In fact, this question is an offshoot of one of the posts in that discussion. Started a new topic, since it seemed a little OT to the original post.

I'm sorry, but can someone please explain this more.

In example 1, there is no widening involved, just boxing... if needed. Since a method call that takes a primitive variable(int) is present, the boxing is not needed either.
In example 2, there are two options - varargs only/ boxing + varargs. If in example 1, the method knows to go to the primitive method call, and not do the boxing needlessly, then why not the same in example 2. Yes, there is varargs also, but again the choice is between primitive and boxing so why not go just to primitive.

The question seems clear enough in my head, but not sure if I am being able to make sense!!!

Hi you are passing int to an Array argument not to int argument. To create int to Array means you are boxing int in an Array Type Object so it is Boxing + Varargs. Correct me...