Why does this method call compile with only one argument ?

[Harry Henriques]

public class Junk {

     public static void main(String argv[]){
          // why does this compile ?
          Object ob3=getFormattedObject(null); 
     }

     public static Object getFormattedObject(Object ob, String... s){

             return new Object();

     }

}    

[Sebastian Janisch]

Because String... is a var-arg which is a convenience for passing an array.

So if you do

getFormattedObject(Object ob, "String1", "String2");

The compiler will turn it into

getFormattedObject(someObject, new String[] {"String1", "String2"});

which is a valid String array.

If you omit this argument completely as in your example, the compiler does this:

getFormattedObject(someObject, new String[0]);

Which is a 0 sized String array and perfectly legal.

[Salil Vverma]

Hey Harry,

... operator represents 0 or more arguments . In this case the function can be called in various formats.. I believe, the below code would clear your doubt.

   
public class Gened {

	public static void main(String argv[]) {

		doNumber(new Integer(21), new Integer(22));
		doNumber(20, 30, 40);
		doNumber(new int[] { 1, 2, 3, 4 });
		doNumber();
	}

	public static void doNumber(int... dn) {
		System.out.println("The array elements are -");
		for (int i : dn) {
			System.out.print(i + " ");
		}
		System.out.println();
	}

}

 

The output would be

The array elements are -
21 22
The array elements are -
20 30 40
The array elements are -
1 2 3 4
The array elements are -