If statement doesn't work, the program enters directly the "else" statement

Question

I am trying to write a program which detects "idle" status but I don't see the problem in my code. Could someone help me please with an useful tip? Here's my code:

package idlestatus;

import java.awt.MouseInfo;

public class Idlestatus {

    public static void main(String[] args) throws InterruptedException {
        Integer firstPointX = MouseInfo.getPointerInfo().getLocation().x;
        Integer firstPointY = MouseInfo.getPointerInfo().getLocation().y;
        Integer afterPointX;
        Integer afterPointY;
        while (true) {
            Thread.sleep(10000);
            afterPointX = MouseInfo.getPointerInfo().getLocation().x;
            afterPointY = MouseInfo.getPointerInfo().getLocation().y;
            if (firstPointX == afterPointX && firstPointY == afterPointY) {
                System.out.println("Idle status");
            } else {
                System.out.println("(" + firstPointX + ", " + firstPointY + ")");
            }
            firstPointX = afterPointX;
            firstPointY = afterPointY;

        }

    }
}

or use `int` not `Integer`. – Hovercraft Full Of Eels Jun 04 '17 at 12:27 — Hovercraft Full Of Eels, Jun 04 '17 at 12:27
Well yeah...It solved, thanks sir ! – Dániel István Jun 04 '17 at 12:28 — Dániel István, Jun 04 '17 at 12:28

Kaushal28 · Accepted Answer · 2017-06-04T12:34:38.750

0

If is working but your condition is always getting false, because you are using Integer instead of primitive int. Note that when you use Object, compare them with .equals() method instead of ==.

Therefore :

if (firstPointX.equals(afterPointX) && firstPointY.equals(afterPointY)) {
    //your code...
}

See this for difference between == and Object.equals() method.

And as mentioned in comments, you can always use int for such purposes, instead of Integer.

See this for difference between Integer and int.

edited Jun 04 '17 at 12:34

answered Jun 04 '17 at 12:28

Kaushal28

4,823
4
30
57

It was enough to use .equals instead of ==, also int for such purposes might be better. Thank you very much ! :) – Dániel István Jun 04 '17 at 12:32
Just did it ! :) – Dániel István Jun 04 '17 at 12:46

Blasanka · Answer 2 · 2017-06-04T12:48:17.477

You are comparing two objects memory addresses , which is Integer object(wrapper class).

if (firstPointX == afterPointX && firstPointY == afterPointY)

What you want to do is compare values in these two objects. To do that you need to use like below:

if (firstPointX.equals(afterPointX) && firstPointY.equals(afterPointY))

Wrapper / covering classes:

There is a wrapper class for each primitive data type.
Primitive types are use for performance reasons(Which better for your program).
Cannot create objects using primitive types.
Allows a way to create objects and manipulate basic types(i.e. converting types).

Exsample:

Integer - int
Double - double

If statement doesn't work, the program enters directly the "else" statement

2 Answers2