What's .self, .Type and .Protocol? Understanding Swift Metatypes

Ah metatypes. That's another one for the list of things I use everyday but couldn't explain in an interview if my life depended on it.

Metatypes are pretty useful in Swift, and you have certainly used it in multiple occasions. Unfortunately they look pretty weird in code, which can cause some confusion when trying to understand what they really are.

I for one know how these weird suffixes can confuse you, but don't worry, they are actually pretty straightforward once you get to know the differences between each of them. But before getting into that, let's take a step back:

What's a Metatype?

If you take a look at Apple's docs, you'll see that a metatype is defined as being the type of a type. Wait, isn't String a type? What could possibly be type of a String that's already a type? SuperString??

It sounds weird in theory, but that's because we got used to Swift's syntax which specifically hides some of these details from us in order to make the language easy to use. To understand metatypes, try to stop seeing things as types and start seeing them more as instances and classes (the usage keyword, not the object!) instead.

Considering the following snippet: How would you define SwiftRocks() and : SwiftRocks?

struct SwiftRocks {
    static let author = "Bruno Rocha"
    func postArticle(name: String) {}

let blog: SwiftRocks = SwiftRocks()

You can say that SwiftRocks() is an object and SwiftRocks is its type, but instead, try seeing SwiftRocks() as an instance, and : SwiftRocks itself as the representation of the type of an instance. After all, you can call the instance method postArticle() from blog, but you can't access the class property author.

Now, how can we access author? The most common way would be through SwiftRocks.author which will directly return you a String, but I will ask you to forget about that one for a moment. Is there another way?

I know that Bruno! You can call type(of: blog).author!

Yup! That is also correct, as type(of) transforms something an object into something that allows you to access all class properties. But have you ever tried to call just type(of: blog) to see what would happen?

let something = type(of: blog) // SwiftRocks.Type

One of the weird suffixes! The type of SwiftRocks is SwiftRocks.Type, which means that SwiftRocks.Type is SwiftRocks's metatype.

By using Xcode's code completion on the something property, you'll see that a reference to a metatype allows you to use all of that type's class properties and methods, including init():

let author: String = something.author
let instance: SwiftRocks = something.init()

That's very useful when you want a method to instantiate objects for you (like how UITableView cell reuse and Decodable work), access class properties or just overall do actions based on the type of an object. Doing so in a generic way is easy as you can pass metatypes as arguments:

func createWidget<T: Widget>(ofType: T.Type) -> T {
    let widget = T.init()
    return widget

Metatypes can also be used in equality checks, which I personally find handy when designing factories:

func create<T: BlogPost>(blogType: T.Type) -> T {
    switch blogType {
    case is TutorialBlogPost.Type:
        return blogType.init(subject: currentSubject)
    case is ArticleBlogPost.Type:
        return blogType.init(subject: getLatestFeatures().random())
    case is TipBlogPost.Type:
        return blogType.init(subject: getKnowledge().random())
        fatalError("Unknown blog kind!")

You can define the metatype of any type, including classes, structs, enums and protocols as being the name of that type followed by .Type. In short, while SwiftRocks refers to the type of an instance (which only lets you use instance properties), the metatype SwiftRocks.Type refers to the type of class itself, which lets you use the SwiftRocks's class properties. "type of a type" makes a lot more sense now, right?

type(of:) Dynamic Metatypes vs .self Static Metatypes

So type(of) returns the metatype of an object, but what happens if I don't have an object? Xcode gives me a compiler error if I try to call create(blogType: TutorialBlogPost.Type)!

To make it short, the reason you can't do that is the same reason why you can't call myArray.append(String): String is the name of the type, not the value! To get a metatype as a value, you need to type the name of that type followed by .self.

If that sounds confusing, you can see it like this: Just like String is the type and "Hello World" is the value of an instance, String.Type is the type and String.self is the value of a metatype.

let intMetatype: Int.Type = Int.self
let widget = createWidget(ofType: MyWidget.self)
tableView.register(MyTableViewCell.self, forReuseIdentifier: "myCell")

.self is what Apple calls a static metatype - a fancy word for what is the compile time type of an object. You use that more than you expect - remember when I told you to ignore SwiftRocks.author? The reason was because writing that is the same as writing SwiftRocks.self.author.

Static metatypes are everywhere in Swift, and you implicitly use them every time you access a type's class property directly. You might find interesting that the AnyClass type used by a table's register(cellClass:) is just an alias for AnyObject.Type:

public typealias AnyClass = AnyObject.Type

On the other hand, type(of) will return a dynamic metatype, which is the metatype of the object's real, runtime type.

let myNum: Any = 1 // Compile time type of myNum is Any, but the runtime type is Int.
type(of: myNum) // Int.type

The actual contents of type(of:) and its Metatype return type are compiler magic (a subject for another article), but here's the method's signature:

func type<T, Metatype>(of value: T) -> Metatype {}

In short, if the subclass of an object matters, you should use type(of) in order to have access to that subclass's metatype. Otherwise, you can simply access the static metatype directly through (name of the desired type).self.

An interesting property of metatypes is that they are recursive, which means you can have meta-metatypes like SwiftRocks.Type.Type, but thankfully for our sanity, you can't do much with these as it's currently impossible to write extensions for metatypes.

Protocol Metatypes

Although everything said before applies to protocols, they have an important difference. The following code will not compile:

protocol MyProtocol {}
let metatype: MyProtocol.Type = MyProtocol.self // Cannot convert value of...

The reason for that is that in the context of protocols, MyProtocol.Type doesn't refer to the protocol's own metatype, but the metatype of whatever type is inheriting that protocol. Apple calls this an existential metatype.

protocol MyProtocol {}
struct MyType: MyProtocol {}
let metatype: MyProtocol.Type = MyType.self // Now works!

In this case, metatype only has access to MyProtocol class properties and methods, but MyType's implementations will be called. To get the concrete metatype of the protocol type itself, you can use the .Protocol suffix. That's basically the same as using .Type on other types.

let protMetatype: MyProtocol.Protocol = MyProtocol.self

Because we're referring to the uninherited protocol itself, there's nothing you can really do with protMetatype besides simple equality checks like protMetatype is MyProtocol.Protocol. If I had to make a guess, I would say that a protocol's concrete metatype's purpose is more about making protocols work in the compiler side of things, which is likely why we never see it in iOS projects.

Conclusion: More uses for Metatypes

Representing a type through a metatype can help you build very intelligent and type-safe generic systems. Here's an example of how we use them in deep link handlers to prevent having to deal with strings directly:

public protocol DeepLinkHandler: class {
    var handledDeepLinks: [DeepLink.Type] { get }
    func canHandle(deepLink: DeepLink) -> Bool
    func handle(deepLink: DeepLink)

public extension DeepLinkHandler {
    func canHandle(deepLink: DeepLink) -> Bool {
        let deepLinkType = type(of: deepLink)
        //Unfortunately, metatypes can't be added to Sets as they don't conform to Hashable!
        return handledDeepLinks.contains { $0.identifier == deepLinkType.identifier }


class MyClass: DeepLinkHandler {
    var handledDeepLinks: [DeepLinks.Type] {
        return [HomeDeepLink.self, PurchaseDeepLink.self]

    func handle(deepLink: DeepLink) {
        switch deepLink {
        case let deepLink as HomeDeepLink:
        case let deepLink as PurchaseDeepLink:

And as a more recent example, here's how we use metatypes to represent and retrieve information about A/B tests (called "Experiments"):

if ExperimentManager.get(HomeExperiment.self)?.showNewHomeScreen == true {
    //Show new home
} else {
    //Show old home

// Experiment Manager

public static func get<T: Experiment>(_ experiment: T.Type) -> T? {
    return shared.experimentDictionary[experiment.identifier] as? T

public static func activate(_ experiment: Experiment) {
    shared.experimentDictionary[type(of: experiment).identifier] = experiment

Follow me on my Twitter - @rockthebruno, and let me know of any suggestions and corrections you want to share.

References and Good reads

Apple Docs: Types
Apple Docs: type(of:)

Contact Info

Bruno Rocha
Senior iOS Developer


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