Swift generic function not working - swift

Question

So, I made a generic function in a structure which has some static methods for helping to create customized UIButtons and so on. So I did this code:
static func createAlertPicker<T: UIViewController where T: UIPickerViewDelegate, T: UIPickerViewDataSource>(#title: String, inout forPicker picker: UIPickerView, viewController: T) -> UIAlertController {
let alert = UIAlertController(title: title, message: "\n\n\n\n\n\n\n\n\n\n", preferredStyle: .Alert)
alert.view.tintColor = data.backgroundColor
picker = createPickerViewWithFrame(CGRectZero, delegate: viewController, dataSource: viewController, backgroundColor: UIColor.clearColor(), addToView: alert.view)
picker.frame = CGRect(x: 2, y: 70, width: 266, height: 162)
return alert
}
I don't get an error doing this but when calling this method in a ViewController like so:
let alert = CreatorClass.createAlertPicker(title: "select sortage", forPicker: &self.pickerView!, viewController: self)
I get a crazy error telling me Type 'UIViewController' does not conform to protocol 'UIPickerViewDelegate'. Although it is implemented, that's the ViewController's declaration:
class PlayerDetails:UIViewController, UITableViewDataSource, UITableViewDelegate, UIPickerViewDelegate, UIPickerViewDataSource
By the way, this problem only occurs in one ViewController, I'm calling it several times. Maybe it should be mentioned that this line of code (let alert = ...) is not even compiled in the first place.
I really don't understand this. Thank for any help ! :]

Answer 1

Yeah that error is a total lie.
The problem is with the middle parameter: &self.pickerView!.
self.pickerView is an optional. Your optional contains a reference, but unwrapping that reference passes back you back a fresh copy of the reference by value. You don’t get access to the original reference inside the optional. So when you call !, you get an immutable value. You can’t change it or assign to it, and that means you can’t pass it as an inout parameter.
Here’s a simpler example:
let i: Int? = 5
func f(inout i: Int) { i = 6 }
f(&i!) // error: 'Int' is not convertible to '#lvalue inout $T3’
This is the compiler saving you from a potentially very confusing runtime bug – if the value were passed in and changed, it would make no difference to the value you actually intended to change. Only the temporary copy would have been changed.
It might be a bit confusing because classes are reference types so you’re not used to thinking about them in value terms. But references themselves are values. What you are getting out of the unwrap is a copy of the reference, not a copy of the thing referred to.
If you change your call to something like the following, it should work:
if var picker = self.pickerView {
let alert = CreatorClass.createAlertPicker(title: "select sortage", forPicker: &picker, viewController: self)
// don’t forget to assign the value back...
self.pickerView = picker
}
This version also has the benefit of not exploding in flames if you’ve ever forgotten to set forPicker to be a value before you force unwrap it.
But if all you are using the inout for is to return a new picker (doesn’t look in your createAlertPicker like you use the value passed in, only assign to it), then why not ditch the inout and make the function return a pair of values:
static func createAlertPicker
<T: UIViewController where T: UIPickerViewDelegate>
(#title: String, viewController: T)
// return a tuple
-> (UIPickerView,UIAlertController) {
// etc…
var picker = createPickerViewWithFrame(CGRectZero, delegate: viewController, dataSource: viewController, backgroundColor: UIColor.clearColor(), addToView: alert.view)
picker.frame = CGRect(x: 2, y: 70, width: 266, height: 162)
return (picker, alert)
}
let (picker, alert) = CreatorClass.createAlertPicker(title: "select sortage”, viewController: self)
self.pickerView = picker