Swift - what's the difference between metatype .Type and .self?
What's the difference between metatype .Type
and .self
in Swift?
Do .self
and .Type
return a struct
?
I understand that .self
can be used to check with dynamicType
. How do you use .Type
?
Solution 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 likeInt.self
which will create a static metatype instanceInt.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 { ... }
Solution 2:
First and foremost see Apple docs on type(of:)
The functions signature is interesting:
func type<T, Metatype>(of value: T) -> Metatype
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:
- 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.
- 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
- 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
isSomeClass.Type
and the metatype of the protocolSomeProtocol
isSomeProtocol.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
Solution 3:
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.