How to return `[Self]` from a Swift Protocol?

Question

I am having a Protocol implementation as follows.

protocol DatabaseInjectable {

    static func deriveObjectFromDBRow(row: [String]) -> Self? // Method - 1

    static func collectAllObjectsForDatabaseAction(action: (Database) -> Void) -> [Self]? // Method - 2
}

Where I am successful with the correspondent implementation of Method - 1 like this:

static func deriveObjectFromDBRow(row: [String]) -> Self? {

    ...
}

But I could not implement the Method - 2 like this:

static func collectAllObjectsForDatabaseAction(action: (WWDatabase) -> Void) -> [Self]? {

    ...
}

There I am getting an error like this:

'Self' is only available in a protocol or as the result of a method in a class;

Any help to return the array form of Self (the class it self) would be nice.

Despite having answered several *similar* questions myself (none of which I think answer your *specific* question) I still find the treatment of `Self`unintuitive. Maybe someone can write a definitive guide to this... — Grimxn, Jan 22 '16 at 09:20

phimage · Answer 1 · 2016-01-22T09:46:51.460

1

If you can set your class final you can replace Self by the class name

final class SampleClass: DatabaseInjectable {
    init() {
    }
    static func deriveObjectFromDBRow(row: [String]) -> SampleClass? {
        return SampleClass()
    }

    static func collectAllObjectsForDatabaseAction(action: (Database) -> Void) -> [SampleClass]? {
        let array = [SampleClass]()
        return array
    }
}

edited Jan 22 '16 at 09:46

answered Jan 22 '16 at 09:19

phimage

274
1
6

It could be a solution but I need to subclass it. Therefore I can't set it as a ˚Final˚ class. – Goppinath Jan 22 '16 at 09:52

score 0 · Answer 2 · edited May 23 '17 at 12:23

0

There's a very well written answer here, but in short, you can define your protocol as such:


protocol DatabaseInjectable {

    static func deriveObjectFromDBRow(row: [String]) -> DatabaseInjectable? // Method - 1

    static func collectAllObjectsForDatabaseAction(action: (Database) -> Void) -> [DatabaseInjectable]? // Method - 2
}

edited May 23 '17 at 12:23

Community

1
1

answered Jan 22 '16 at 09:30

Ramy Kfoury

937
5
8

Thanks, but I need the specific class type as a return type. Same as `instancetype` in Objective-C – Goppinath Jan 22 '16 at 09:40

dfrib · Answer 3 · 2016-01-22T12:10:45.003

You can make use of a typealias in your protocol DatabaseInjectable, and use this as an alias to Self type in the classes conforming to your protocol.

class Database {
    var desc : String = "Default"
}

protocol DatabaseInjectable {
    typealias MySelf

    static func deriveObjectFromDBRow(row: [String]) -> MySelf?

    static func collectAllObjectsForDatabaseAction(action: (Database) -> Void) -> [MySelf]?
}

class MyClass : DatabaseInjectable {
    typealias MySelf = MyClass

    static func deriveObjectFromDBRow(row: [String]) -> MySelf? {
        return MyClass()
    }

    static func collectAllObjectsForDatabaseAction(action: (Database) -> Void) -> [MySelf]? {
        return [MyClass(), MyClass()]
    }
}

/* example */
let closure : (Database) -> () = { print($0.desc) }
var arr : [MyClass]? = MyClass.collectAllObjectsForDatabaseAction(closure)
/* [MyClass, MyClass] */

One drawback here is, however, that you could set e.g. typealias MySelf = Int (in the class) and have your functions return an integer/array of integers (rather than self/[Self]), and still conform to you protocol. Possibly this is a deal-breaker.

How to return `[Self]` from a Swift Protocol?

3 Answers3