[dulvinrivindu]

public class Exampleprg {
	Exampleprg(String... person){
		for(String name: person){
			System.out.println("Hello " + name);
		}
	}
	public static void main(String args[]){
		Exampleprg("john","david", "suhel");
	}
}

It tells "Exception in thread "main" java.lang.Error: Unresolved compilation problem:
The method Exampleprg(String, String, String) is undefined for the type Exampleprg

at Exampleprg.main(Exampleprg.java:9)"

[sepp2k]

On line 2 you're defining a constructor of the class Exampleprg - you're not defining a method named Exampleprg.

Note that this is not related to varargs. You'd have the same problem if you defined Exampleprg to take three strings as arguments.

[ShawnaInMaine]

What I have learned about arrays is that they have to use square braces []

Something to this effect will work

static String[] names = {"John", "David", "Suhel"} ;//this declares the array of names
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {


System.out.println("Hello " + names[1]);//this will print hello and the first person name in array (which in example above is david because john is 
0 in array
    }
}

This post has been edited by macosxnerd101: 02 January 2013 - 01:03 PM
Reason for edit:: Please use code tags

[sepp2k]

ShawnaInMaine, on 01 January 2013 - 05:40 PM, said:

What I have learned about arrays is that they have to use square braces []

There is nothing wrong with OP's code other than what I explained in my earlier post.

Variadic methods are defined with TypeName ... as the type of the last parameter - not TypeName[]. If you use the latter, you end up defining a regular method taking an array as its argument - not a variadic method. Such a method would need to be called with an actual array as its argument and thus could not be called in the way the OP intends.

[dulvinrivindu]

sepp2k, on 01 January 2013 - 09:37 AM, said:

On line 2 you're defining a constructor of the class Exampleprg - you're not defining a method named Exampleprg.

Note that this is not related to varargs. You'd have the same problem if you defined Exampleprg to take three strings as arguments.

So how shud I recorrect this sir. I mean i m new to this varargs sstuf. I saw this in a book that im using.

[smohd]

Like sepp2k said, your problem is with how you are calling the constructor. That Exampleprg(String... person) is not a regular method, it is a constructor because it has the same name as the class with no return type.
To call a constructor you have to precede it with the keyword new which means we are instantiating an object and a constructor will be called automatically:

 new Exampleprg("john","david", "suhel");

[sepp2k]

smohd, on 02 January 2013 - 06:36 AM, said:

To call a constructor you have to precede it with the keyword new which means we are instantiating an object and a constructor will be called automatically:

 new Exampleprg("john","david", "suhel");

However, as it doesn't look like it was intended to be a constructor and it certainly shouldn't be one from a design point of view, it would make more sense to turn it into a (static) method and keep the method call unchanged (i.e. no new).

[pbl]

I guess this is what you try to do
Would be a lot clearer if you do not name a method the same as the name as the class

public class Exampleprg {

	static void foo(String... person){
		for(String name: person){
			System.out.println("Hello " + name);
		}
	}
	public static void main(String args[]){
		foo("john","david", "suhel");
	}
}