Static variable gets unexpected value in method. Qestion of initialization?

[Sven Sylta]

Can someone please explain what is happening in this piece of code? The first three should print "22" and the last two should print "44" but it doesn't. Is this about initialization? I've added a couple of comments.
Thank you.

public class Q {
	
	private static int x = 11;
	private int y = 33;
	
	public static void main(String[] args) {
		
		Q q = new Q();
		q.call(5);
		
	}
	
	public void call(int x) {  
		Q q = new Q();
		x = 22;
		y = 44;
		System.out.println("Output: " + Q.x);//This should print 22 because of x= 22. The x variable changes and should be the same everywhere. Same below
		System.out.println("Output: " + x);    
		System.out.println("Output: " + q.x);      
		System.out.println("Output: " + q.y);// This should print 44 because of y = 44. The y variable changes and should be the same everywhere. Same below.
		System.out.println("Output: " + this.y);    
	}
}

[riteshhh kumar singh]

In line 17 you have used Q.x where Q is class that you have created and x is static variable which can be accessed by using classname.variablename,so it wil print 11 because the value assigned to the instance variable x is 11 in the class Q,and in line 20, it will print 33 because it is called by using the object of the class which you have created because whenever you create a object, memory is allocated for that particular object,so if if you to print 44 instead of 33 you have to change the value of the instance variable y in that object,and then you can use q.y to print 44.

[manish ghildiyal]

I ran theprogram and got whatI expected....

Output: 11
Output: 22
Output: 11
Output: 33
Output: 44

You are confusing x variable input to call() with class variable x.
Comment the line....x = 22;
...and run it, you would get the idea.

Manish

[Sven Sylta]

Thanks for clearing this up. Very enlightening.