Best practice to implement a failable initializer in Swift

That doesn't make sense. Why should I initialize those properties when I plan to return nil?

According to Chris Lattner this is a bug. Here is what he says:

This is an implementation limitation in the swift 1.1 compiler, documented in the release notes. The compiler is currently unable to destroy partially initialized classes in all cases, so it disallows formation of a situation where it would have to. We consider this a bug to be fixed in future releases, not a feature.

Source

EDIT:

So swift is now open source and according to this changelog it is fixed now in snapshots of swift 2.2

Designated class initializers declared as failable or throwing may now return nil or throw an error, respectively, before the object has been fully initialized.

Update: From the Swift 2.2 Change Log (released March 21, 2016):

Designated class initializers declared as failable or throwing may now return nil or throw an error, respectively, before the object has been fully initialized.

For Swift 2.1 and earlier:

According to Apple's documentation (and your compiler error), a class must initialize all its stored properties before returning nil from a failable initializer:

For classes, however, a failable initializer can trigger an initialization failure only after all stored properties introduced by that class have been set to an initial value and any initializer delegation has taken place.

Note: It actually works fine for structures and enumerations, just not classes.

The suggested way to handle stored properties that can't be initialized before the initializer fails is to declare them as implicitly unwrapped optionals.

Example from the docs:

class Product {
    let name: String!
    init?(name: String) {
        if name.isEmpty { return nil }
        self.name = name
    }
}

In the example above, the name property of the Product class is defined as having an implicitly unwrapped optional string type (String!). Because it is of an optional type, this means that the name property has a default value of nil before it is assigned a specific value during initialization. This default value of nil in turn means that all of the properties introduced by the Product class have a valid initial value. As a result, the failable initializer for Product can trigger an initialization failure at the start of the initializer if it is passed an empty string, before assigning a specific value to the name property within the initializer.

In your case, however, simply defining userName as a String! does not fix the compile error because you still need to worry about initializing the properties on your base class, NSObject. Luckily, with userName defined as a String!, you can actually call super.init() before you return nil which will init your NSObject base class and fix the compile error.

class User: NSObject {

    let userName: String!
    let isSuperUser: Bool = false
    let someDetails: [String]?

    init?(dictionary: NSDictionary) {
        super.init()

        if let value = dictionary["user_name"] as? String {
            self.userName = value
        }
        else {
            return nil
        }

        if let value: Bool = dictionary["super_user"] as? Bool {
            self.isSuperUser = value
        }

        self.someDetails = dictionary["some_details"] as? Array
    }
}

I accept that Mike S's answer is Apple's recommendation, but I don't think it's best practice. The whole point of a strong type system is to move runtime errors to compile time. This "solution" defeats that purpose. IMHO, better would be to go ahead and initialize the username to "" and then check it after the super.init(). If blank userNames are allowed, then set a flag.

class User: NSObject {
    let userName: String = ""
    let isSuperUser: Bool = false
    let someDetails: [String]?

    init?(dictionary: [String: AnyObject]) {
        if let user_name = dictionary["user_name"] as? String {
            userName = user_name
        }

        if let value: Bool = dictionary["super_user"] as? Bool {
            isSuperUser = value
        }

        someDetails = dictionary["some_details"] as? Array

        super.init()

        if userName.isEmpty {
            return nil
        }
    }
}

Another way to circumvent the limitation is to work with a class-functions to do the initialisation. You might even want to move that function to an extension:

class User: NSObject {

    let username: String
    let isSuperUser: Bool
    let someDetails: [String]?

    init(userName: String, isSuperUser: Bool, someDetails: [String]?) {

         self.userName = userName
         self.isSuperUser = isSuperUser
         self.someDetails = someDetails

         super.init()
    }
}

extension User {

    class func fromDictionary(dictionary: NSDictionary) -> User? {

        if let username: String = dictionary["user_name"] as? String {

            let isSuperUser = (dictionary["super_user"] as? Bool) ?? false
            let someDetails = dictionary["some_details"] as? [String]

            return User(username: username, isSuperUser: isSuperUser, someDetails: someDetails)
        }

        return nil
    }
}

Using it would become:

if let user = User.fromDictionary(someDict) {

     // Party hard
}

Although Swift 2.2 has been released and you no longer have to fully initialize the object before failing the initializer, you need to hold your horses until https://bugs.swift.org/browse/SR-704 is fixed.

Best practice to implement a failable initializer in Swift

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