将类型传递给通用Swift扩展，或者理想地推断它

Question

Say you have

 class Fancy:UIView

you want to find all sibling Fancy views. No problem ...

    for v:UIView in superview!.subviews
        {
        if let f = v as? Fancy
            { f.hungry = false }
        }

So, try an extension,

public extension UIView
    {
    internal func fancySiblings()->([Fancy])
        {
            return (self.superview!
                .subviews
                .filter { $0 != self }
                .flatMap { $0 as? Fancy }
                )
        }
    }

Awesome, you can now

    for f:Fancy in self.fancySiblings()
        { f.hungry = false }

Fantastic.

But,

How to generalize that extension to work with any UIView subtype?

Ideally, can the extension infer the type , even? As well as taking a type?

So, something like ...

public extension UIView
    {
    internal func siblings<T>( something T )->([T])
        {
            return (self.superview!
                .subviews
                .filter { $0 != self }
                .flatMap { $0 as? T }
                )
        }

and then you could call it something like this ...

    for f in self.siblings(Fancy)
    for p in self.siblings(Prancy)
    for b in self.siblings(UIButton)

How can you "tell" a generic extension the type to use, like that??

It seems you can "infer it backwards",

public extension UIView
    {
    internal func incredible<T>()->([T])
        {
        return (self.superview!
         .subviews
         .filter { $0 != self }
         .flatMap { $0 as? T }
         )
        }


    for f:Fancy in self.incredible()

    for p:Prancy in self.incredible()

Which is amazing but doesn't work the other way.

You can even...

    self.siblings().forEach{
        (f:Fancy) in
        d.hasRingOn = false
        }

So I would still like to know how to "pass in" a type something like for f in self.siblings(Fancy) and, ideally, even infer it also.

Answer 1

Simply use the .Type :

internal func siblings<T>( something : T.Type)->([T]) {
    ...
}

Afterwards for f in self.siblings(Fancy) should work exactly as expected.

Full working example:

class Fancy : UIView {}

public extension UIView {
    internal func siblings<T>( _ : T.Type)->([T]) {
        return (self.superview!
            .subviews
            .filter { $0 != self }
            .flatMap { $0 as? T }
        )
    }
}

let superView = UIView()
let view = UIView()
superView.addSubview(view)
superView.addSubview(UIView())
superView.addSubview(Fancy())

print(view.siblings(Fancy))

Correctly outputs the one Fancy view!

---

To address the requested addition for optionally using the explicit type parameter or take effect of the type inference of the compiler. You can create a second method in the same extension

internal func siblings<T>()->([T]) {
    return siblings(T)
}

That way providing a explicit type parameter calls method one, omitting it will require you to make it inferable and will call the second function which in terms calls the first one internally.

---

Or , you can use the far more swifty way and make the explicit type argument an optional with default nil . That, remarkably, will force the inference in case of omitting the type argument:

// power extension, it provides both infered or stated typing
internal func siblings<T>(_ : T.Type? = nil) -> ([T]) {
    return (self.superview!
        .subviews
        .filter { $0 != self }
        .flatMap { $0 as? T }
        )
}

That will enable you to call the method either via

for f in self.siblings(Fancy)

or even

for f : Fancy in self.siblings()

Both will work while still only defining one function.

Answer 2

Similar answer to what's been said previously, but a little more streamlined and without having to pass anything or iterate over the subviews more than once:

extension UIView {
    internal func siblings<T: UIView>() -> [T] {
        return superview?.subviews.flatMap {return ($0 == self) ? nil : ($0 as? T) } ?? []
    }
}

or my preference using optionals:

internal func siblings<T: UIView>() -> [T]? {
        return superview?.subviews.flatMap {return ($0 == self) ? nil : $0 as? T } 
}

Example Usage:

class ExampleView: UIView {

    func getMatchingSiblings(){
        let foundSiblings: [ExampleView] = siblings()
    }

    //or with the for loop in the question:
    for item: ExampleView in siblings() {

    }
}

When dealing with generics, you simply need one instance of the generic type in the signature of the method. So if you have either a parameter, or return type that uses the generic, you don't need to pass the type.