Swift unique property in multiple instances

Question

I know that theoretically it's possible to create multiple instances of the same class with a property that would have a different value for each instance.

The thing is, I can't make it happen.

Each time I'm creating a new instance, it gets the property's value of the other instances, and when I'm changing one value for an instance, it changes the other's too.

So my guess is that I'm doing something wrong (obviously), like accessing the class property value instead of the instance property value... Here's the code.

class CustomUIImageView: UIImageView {
    var someParameter: Bool = false // This is the property I want to be different in each version of the instance. 
}


class ClassSiege: UIViewController, UIGestureRecognizerDelegate {    

    var myView: CustomUIImageView! //the instance declaration. 

    // I use this gesture recognizer to find out the value of the instance I'm tapping on.
    func handleTap (sender: UITapGestureRecognizer) {
        print("value of someParameter \(self.myView.someParameter)")
    }

    func handlePan(recognizer: UIPanGestureRecognizer) {
        let iv: UIView! = recognizer.view
        let translation = recognizer.translationInView(self.view)
        iv.center.x += translation.x
        iv.center.y += translation.y
        recognizer.setTranslation(CGPointZero, inView: self.view)

        var centerBoardX = BlackBoard.center.x // 'Blackboard' is a fixed image on the screen. 
        var centerBoardY = BlackBoard.center.y
        var centerRondX = iv.center.x
        var centerRondY = iv.center.y

        if centerRondY - centerBoardY < 100 { 
            self.myView.someParameter = true // If the distance between myView and the blackboard is under 100 I want the instance's property to become true. 
        } else {
            self.myView.someParameter = false // On the other hand, if the distance is greater than 100, I want it to be false. 
        }
    }

    // When the user pushes a button, it triggers this method that creates a new instance of myView and add it to the screen. 
   @IBAction func showContent(sender: AnyObject) {
       // some code...
       // Here I'm creating the instance of the view and I give it the gesture recognizer parameters. I don't think that relevant to the issue, so I'm not adding the code.
    }
}

So clearly that's not the good way to do it, but what's wrong, and how can it be solved?

Answer 1

I've found the solution, and if you've been facing the same issu, here's how to deal with it. The secret is to downcast the recognizer.view to take the parameter of the subclass CustomUIImageView.

here's how :

func handleTap (sender: UITapGestureRecognizer) {

println("value of someParameter \(self.myView.someParameter)") //I use this gesture recognizer to find out the value of the instance I'm tapping on. 

}

func handlePan(recognizer:UIPanGestureRecognizer) {
let iv : UIView! = recognizer.view
let translation = recognizer.translationInView(self.view)
iv.center.x += translation.x
iv.center.y += translation.y
recognizer.setTranslation(CGPointZero, inView: self.view)

var centerBoardX = BlackBoard.center.x //blackboard is a fixed image on the screen. 
var centerBoardY = BlackBoard.center.y
var centerRondX = iv.center.x
var centerRondY = iv.center.y

var myParameter = recognizer.view as CustomUIImageView //<- this is the key point. Downcasting let you access the custom subclass parameters of the object that is currently moved

if centerRondY - centerBoardY < 100 { 

  myParameter.someParameter = true //so now I'm really changing the parameter's value inside the object rather than changing a global var like I did before. 



} else {
    myParameter.someParameter = false 

    }

}

//when user pushes a button, it triggers this func that creates a new instance of myView and add it to the screen. 
@IBAction func showContent(sender: AnyObject) {

some code...
 //here I'm creating the instance of the view and I give it the gesture recognizer parameters. I don't think that relevant to the issue, so I'm not adding the code. 

 }