Pass in a type to a generic Swift extension, or ideally infer it

Question

Say you have

 class Fancy:UIView

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

    for v:UIView in sup         


        

Answer 1

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] {
        return superview?.subviews.flatMap {return ($0 == self) ? nil : ($0 as? T) } ?? []
    }
}

or my preference using optionals:

internal func siblings() -> [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.