Manual GestureRecognizer handler in Swift can't have arguments? - swift

Question

If I create a GestureRecognizer handler through InterfaceBuilder, I get a method that looks like this:
#IBAction func Tap(tap: UITapGestureRecognizer) {
var touchLocation = tap.locationInView(self.view)
imageView.transform.tx = touchLocation.x
imageView.transform.ty = touchLocation.y
}
//works great
If I manually create a GestureRecognizer, I have to create a function with no arguments, like this:
let tap = UITapGestureRecognizer()
override func viewDidLoad() {
super.viewDidLoad()
let imageSize = CGSize(width: 100, height: 100)
var iView = UIImageView(frame: CGRect(origin: CGPoint(x: 100, y: 100), size: imageSize))
self.view.addSubview(iView)
let image2 = UIImage(named: "square.png");
iView.image = image2
iView.userInteractionEnabled = true
iView.addGestureRecognizer(tap)
tap.addTarget(self, action: "Tap")
}
func Tap( //why can't I put an argument in here?// ){
let tapAlert = UIAlertController(title: "Tap Pressed", message: "You just tapped", preferredStyle: .Alert)
tapAlert.addAction(UIAlertAction(title: "OK", style: .Destructive, handler: nil))
self.presentViewController(tapAlert, animated: true, completion: nil)
}
//works but I've lost access to the sending object
If I add an argument to the Tap handler
func Tap(sender: UITapGestureRecognizer){
}
//crashes: [tapHandler]: unrecognized selector sent to instance
, it crashes on the Tap. I'd prefer to have a method with access to the sender object. Is there a way to manually create a GestureRecognizer handler and have access to the sender object, the same way you can when you do it through the Interface Builder?

Answer 1

Selector names are something that Swift inherits from Objective C. In Objective C a selector with single argument has name selector: as opposite to just selector, which means - no arguments. Change the name of the action in addTarget from Tap to Tap: and it should work then.