What is going on in this code

[Vikas J Yadav]

package com.infy.cert;

public class Chapter2_Array {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		SubClass sc=new SubClass();
	}
}
class SuperClass{
	public SuperClass(){
		System.out.println("Superclass no-arg");
	}
	public SuperClass(int n){
		System.out.println("Superclass arg");
	}
	public static int method(){
		System.out.println("sup static");
		return 4;
	}
}
class SubClass extends SuperClass{
	public SubClass(int n){
		System.out.println("Subclass arg");
	}
	public SubClass(){
		super(method());
		System.out.println("Subclass no-arg");
	}
	public static int method(){
		System.out.println("sub static");
		return 5;
	}
}

[Added code tags - see UseCodeTags for details]

Produced Output as
sub static
Superclass arg
Subclass no-arg

I have written a code to call SuperClass satic method; I know that SuperClass static methods cannot be overriden but what happens with this line super(method()); of code

[Jack McLaren]

Vikas J Yadav wrote:

what happens with this line super(method()); of code

It will call super class constructor with int parameter.

[Matthew Brown]

When you call a static method without qualification it's going to look in the current class. If you want to call the superclass version here you need to qualify it: SuperClass.method().

[Gajendra Kangokar]

please use code tags if you want to post any code snippet.
and in your code method() has a return type of int ,so super(method()) passes 5 value to superclass constructor.

Regards
Gajendra

[Vikas J Yadav]

Hey Thanks all for quick reply. It solved my doubt

Thanks Gajendra for your suggesion. I will make sure that i wll post the code with proper format

-Vikas