Swift - what's the difference between metatype .Type and .self?

Question

What\'s the difference between metatype .Type and .self in Swift?

Do .self and .Type return a struct

Answer 1

Here is a quick example:

func printType<T>(of type: T.Type) {
    // or you could do "\(T.self)" directly and
    // replace `type` parameter with an underscore
    print("\(type)") 
} 

printType(of: Int.self) // this should print Swift.Int


func printInstanceDescription<T>(of instance: T) {
    print("\(instance)")
} 

printInstanceDescription(of: 42) // this should print 42

Let's say that each entity is represented by two things:

• Type: # entitiy name #

• Metatype: # entity name # .Type

A metatype type refers to the type of any type, including class types, structure types, enumeration types, and protocol types.

Source.

You can quickly notice that this is recursive and there can by types like (((T.Type).Type).Type) and so on.

.Type returns an instance of a metatype.

There are two ways we can get an instance of a metatype:

• Call .self on a concrete type like Int.self which will create a static metatype instance Int.Type.

• Get the dynamic metatype instance from any instance through type(of: someInstance).

Dangerous area:

struct S {}
protocol P {}

print("\(type(of: S.self))")      // S.Type
print("\(type(of: S.Type.self))") // S.Type.Type
print("\(type(of: P.self))")      // P.Protocol
print("\(type(of: P.Type.self))") // P.Type.Protocol

.Protocol is yet another metatype which only exisits in context of protocols. That said, there is no way how we can express that we want only P.Type. This prevents all generic algorithms to work with protocol metatypes and can lead to runtime crashes.

For more curious people:

The type(of:) function is actually handled by the compiler because of the inconsistency .Protocol creates.

// This implementation is never used, since calls to `Swift.type(of:)` are
// resolved as a special case by the type checker.
public func type<T, Metatype>(of value: T) -> Metatype { ... }

Answer 2

This was one of those topics that confused the hell out of me today.

I was writing a generic function:

func foo<T: Protocol>(ofType: T.Type) {
    T.bar()
}

And tried calling it as follows:

foo(ofType: ClassImplementingProtocol.Type) // Compiler error

Spent about 30 min researching why it wasn't working. Then I tried this:

foo(ofType: ClassImplementingProtocol.self) // Works

Turns out Xcode's code completion is very bad at showing the difference between meta types and types... From the code completion pop-up it looks like .self and .Type are the same thing:

But the "explain like im 5" of it is, when you have a method parameter of Class.Type, it is expecting an instance of Class.Type.

Class.self returns an instance of Class.Type, whereas Class.Type is referring to Class.Type...

Very unclear if you ask me.

Answer 3

Where is it used?

If you are writing/creating a function that accepts a type e.g. UIView.Type, not an instance e.g. UIView()then to you would write T.Type as the type of the parameter. What it expects as a parameter can be: String.self, CustomTableView.self, someOtherClass.self.

But why would a function ever need a type?

Normally a function which requires a type, is a function that instantiates objects for you. I can think of two good examples:

1. register function from tableview

tableView.register(CustomTableViewCell.self, forCellReuseIdentifier: "CustomTableViewCell")

Notice that you passed CustomTableViewCell.self. If later on you try to dequeue a tableView of type CustomTableViewCell but didn't register CustomTableViewCell type then it would crash because the tableView hasn't dequeued/instantiated any tableviewcells of CustomTableViewCell type.

2. decode function from JSONDecoder. Example is from the link

struct GroceryProduct: Codable {
    var name: String
    var points: Int
    var description: String?
}

let json = """
{
    "name": "Durian",
    "points": 600,
    "description": "A fruit with a distinctive scent."
}
""".data(using: .utf8)!

let decoder = JSONDecoder()
let product = try decoder.decode(GroceryProduct.self, from: json)

print(product.name)

Notice try decoder.decode(GroceryProduct.self, from: json). Because you passed GroceryProduct.self it knows that it needs to instantiate an object of type GroceryProduct. If it can't then it would throw an error. For more on JSONDecoder see this well written answer

3. As an alternate workaround for where types are needed see the following question: Swift can't infer generic type when generic type is being passed through a parameter. The accepted answer offers an intersting alternative.

---

More about the internals and how it works:

.Type

The metatype of a class, structure, or enumeration type is the name of that type followed by .Type. The metatype of a protocol type—not the concrete type that conforms to the protocol at runtime—is the name of that protocol followed by .Protocol. For example, the metatype of the class type SomeClass is SomeClass.Type and the metatype of the protocol SomeProtocol is SomeProtocol.Protocol.

_{From Apple : metaType Type}

Under the hood AnyClass is

typealias AnyClass = AnyObject.Type // which is why you see T.Type 

Basically where ever you see AnyClass, Any.Type, AnyObject.Type, its because it's in need of a type. A very very common place we see it is when we want to register a class for our tableView using register func.

func register(_ cellClass: Swift.AnyClass?, forCellReuseIdentifier identifier: String)

_{If you are confused as to what does 'Swift.' do then above, then see the comments from here}

The above could have also been written as:

func register(_ cellClass: AnyObject.Type, forCellReuseIdentifier identifier: String)

.self

You can use the postfix self expression to access a type as a value. For example, SomeClass.self returns SomeClass itself, not an instance of SomeClass. And SomeProtocol.self returns SomeProtocol itself, not an instance of a type that conforms to SomeProtocol at runtime. You can use a type(of:) expression with an instance of a type to access that instance’s dynamic, runtime type as a value, as the following example shows:

_{From Apple : metaType Type}

---

Playground code:

Easy example

struct Something {
    var x = 5
}

let a = Something()
type(of:a) == Something.self // true

Hard example

class BaseClass {
    class func printClassName() {
        print("BaseClass")
    }
}
class SubClass: BaseClass {
    override class func printClassName() {
        print("SubClass")
    }
}


let someInstance: BaseClass = SubClass()
/*                      |                |
                    compileTime       Runtime
                        |                | 
To extract, use:       .self          type(of)

  Check the runtime type of someInstance use `type(of:)`: */

print(type(of: someInstance) == SubClass.self) // True
print(type(of: someInstance) == BaseClass.self) // False

 /* Check the compile time type of someInstance use `is`: */

print(someInstance is SubClass) // True
print(someInstance is BaseClass) // True

I highly recommend to read Apple documentation on Types. Also see here

Answer 4

They appear in different places syntactically.

In a place syntactically where you have to specify a type, Something.Type is a valid type, corresponding to the type that is the metatype (which is metaclass for classes) of Something. Something.self is not a valid syntax for a type.

In a place syntactically where you have to write an expression, Something.self is a valid expression. It's an expression of type Something.Type, and the value is the thing ("class object" in the case of classes) that represents the type Something. Something.Type is not a valid expression syntax.