Why Does Type(Of:) Return Metatype, Rather Than T.Type

Why does type(of:) return Metatype, rather than T.Type?

tl;dr: The behavior of type(of:) depends on whether T is existential or concrete, and the type system can't effectively reflect the actual return type syntactically, so it's handled directly in the type checking system. Metatype is specifically not bound in code to be the same as T so that the effective behavior can be specialized. Metatype and T are not necessarily related.

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type(of:) is special in that its behavior differs depending on the type passed into it. Specifically, it has special behavior for existential types by being able to reach through the existential box to get the underlying type of the value passed in. For example:

func myType<T>(of value: T) -> T.Type {
    return T.self
}

protocol Foo {}
struct X: Foo {}

let x = X()
print(type(of: x), "vs.", myType(of: x)) // => X vs. X

let f: Foo = X()
print(type(of: f), "vs.", myType(of: f)) // => X vs. Foo

When given an existential type like Foo, a return type of T.Type could only return the metatype of the existential itself (i.e. Foo.self), as opposed to the metatype of the value inside of the existential container (X.self). So instead of returning T.Type, type(of:) returns an unrelated type Metadata which is bound to the correct type in the type checker itself. This is the edge case you were looking for:

My guess is that there is some edge case where type(of:) will return a completely unrelated type to T, but I have no idea what that is.

If you look in lib/Sema/TypeChecker.h, you can see some special semantics declarations for several stdlib function types:

/// Special-case type checking semantics for certain declarations.
enum class DeclTypeCheckingSemantics {
  /// A normal declaration.
  Normal,

  /// The type(of:) declaration, which performs a "dynamic type" operation,
  /// with different behavior for existential and non-existential arguments.
  TypeOf,

  /// The withoutActuallyEscaping(_:do:) declaration, which makes a nonescaping
  /// closure temporarily escapable.
  WithoutActuallyEscaping,

  /// The _openExistential(_:do:) declaration, which extracts the value inside
  /// an existential and passes it as a value of its own dynamic type.
  OpenExistential,
};

The key one here is TypeOf, which is indeed returned for functions with the @_semantics("typechecker.type(of:)") attribute you noted. (You can see how that attribute is checked in TypeChecker::getDeclTypeCheckingSemantics)

If you go looking for usages of TypeOf, there are two key locations in type-checking:

1. getTypeOfReferenceWithSpecialTypeCheckingSemantics which injects the type constraint in the type checker constraint system. type(of:) is handled here as an overload, because Metadata isn't actually bound; the constraint solver here applies an effective type checking constraint which constrains Metadata to be the actual type of value. The key here is that type(of:) is written in this way so that it would be an overload, and handled here.
2. ExprRewriter::finishApply which performs the actual expression rewriting in the AST to replace the return type with the effective actual type of the value

From (1):

// Proceed with a "DynamicType" operation. This produces an existential
// metatype from existentials, or a concrete metatype from non-
// existentials (as seen from the current abstraction level), which can't
// be expressed in the type system currently.

Pulling back some history — this was implemented back in commit 1889fde2284916e2c368c9c7cc87906adae9155b. The commit message from Joe is illuminating:

Resolve type(of:) by overload resolution rather than parse hackery.

type(of:) has behavior whose type isn't directly representable in Swift's type system, since it produces both concrete and existential metatypes. In Swift 3 we put in a parser hack to turn type(of: <expr>) into a DynamicTypeExpr, but this effectively made type(of:) a reserved name. It's a bit more principled to put Swift.type(of:) on the same level as other declarations, even with its special-case type system behavior, and we can do this by special-casing the type system we produce during overload resolution if Swift.type(of:) shows up in an overload set. This also lays groundwork for handling other declarations we want to ostensibly behave like normal declarations but with otherwise inexpressible types, viz. withoutActuallyEscaping from SE-0110.

Since then, as we can see from WithoutActuallyEscaping and OpenExistential, other special functions have been rewritten to take advantage of this.

Why does Swift infer the metatype 'Any.Type' instead of Type 'Any'?

Why T is not Any

This is simple. There is no == operator defined for Any? and Any?, as Any is not Equatable. T cannot possibly be Any.

Why T can be Any.Type

This is also simple. There is such a built in operator that matches this signature exactly.

func == (t0: Any.Type?, t1: Any.Type?) -> Bool

Why it is not ambiguous

This is more subtle. You'd think that T could be String or Int or Float, right? After all, they all are Equatable, and for all Equatable types, Optional defines a == and != for them. However, it seems like that operators declared in the global scope are considered "better" than operators declared in a conditional extension of Optional, so operator overload resolution always chooses the non-extension operators first. This is reasonable - after all, adding extensions shouldn't cause old code to break.

As an example, adding this code would make your code produce an "ambiguous" error:

struct Foo {}

func ==(lhs: Foo?, rhs: Foo?) -> Bool {
    true
}

func !=(lhs: Foo?, rhs: Foo?) -> Bool {
    false
}

The newly added != operator for Foo? and the built in operator for Any.Type? are considered equally good candidates for the call giveMeZero() != nil.

However, this would not:

struct Foo { }
extension Optional where Wrapped == Foo {
    static func ==(lhs: Foo?, rhs: Foo?) -> Bool {
        true
    }

    static func !=(lhs: Foo?, rhs: Foo?) -> Bool {
        false
    }
}

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

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 { ... }

Why does Swift's typechecking system allow a function that returns a type to not return anything?

Because fatalError returns Never. From the documentation, Never is,

The return type of functions that do not return normally, that is, a type with no values.

Through compiler magic, Swift understands that fatalError "does not return normally", namely that it crashes the app. This is why it does not complain about square not returning an Int.

If a path of your function call a function that "does not return normally", that path does not return normally either. "Does not return normally" means that the caller won't be able to use the return value, so there is no point in enforcing the "you must return a value" rule in that path of your function.

Check whether Swift object is an instance of a given metatype

Unfortunately, you can currently only use a named type with the is operator, you cannot yet use an arbitrary metatype value with it (although really IMO you ought to be able to).

Assuming you have control over the creation of the metatypes that you want to compare against, one solution that achieves the same result would be to create a wrapper type with an initialiser that stores a closure that performs the is check against a generic placeholder:

struct AnyType {

  let base: Any.Type
  private let _canCast: (Any) -> Bool

  /// Creates a new AnyType wrapper from a given metatype.
  /// The passed metatype's value **must** match its static value,
  /// i.e `T.self == base`.
  init<T>(_ base: T.Type) {
    precondition(T.self == base, """
      The static value \(T.self) and dynamic value \(base) of the passed \
      metatype do not match
      """)

    self.base = T.self
    self._canCast = { $0 is T }
  }

  func canCast<T>(_ x: T) -> Bool {
    return _canCast(x)
  }
}

protocol P {}
class C : P {}
class D : C {}

let types = [
  AnyType(P.self), AnyType(C.self), AnyType(D.self), AnyType(String.self)
]

for type in types {
  print("C instance can be typed as \(type.base): \(type.canCast(C()))")
  print("D instance can be typed as \(type.base): \(type.canCast(D()))")
}

// C instance can be typed as P: true
// D instance can be typed as P: true
// C instance can be typed as C: true
// D instance can be typed as C: true
// C instance can be typed as D: false
// D instance can be typed as D: true
// C instance can be typed as String: false
// D instance can be typed as String: false

The only limitation of this approach is that given we're performing the is check with T.self, we have to enforce that T.self == base. For example, we cannot accept AnyType(D.self as C.Type), as then T.self would be C.self while base would be D.self.

However this shouldn't be a problem in your case, as we're just constructing AnyType from metatypes that are known at compile time.

---

If however you don't have control over the creation of the metatypes (i.e you get handed them from an API), then you're quite a bit more limited with what you can do with them.

As @adev says, you can use type(of:) in order to get the dynamic metatype of a given instance, and the == operator to determine if two metatypes are equivalent. However, one problem with this approach is that it disregards both class hierarchies and protocols, as a subtype metatypes will not compare equal with a supertype metatypes.

One solution in the case of classes is to use Mirror, as also shown in this Q&A:

/// Returns `true` iff the given value can be typed as the given
/// **concrete** metatype value, `false` otherwise.
func canCast(_ x: Any, toConcreteType destType: Any.Type) -> Bool {
  return sequence(
    first: Mirror(reflecting: x), next: { $0.superclassMirror }
  )
  .contains { $0.subjectType == destType }
}

class C {}
class D : C {}

print(canCast(D(), toConcreteType: C.self)) // true
print(canCast(C(), toConcreteType: C.self)) // true
print(canCast(C(), toConcreteType: D.self)) // false
print(canCast(7, toConcreteType: Int.self)) // true
print(canCast(7, toConcreteType: String.self)) // false

We're using sequence(first:next:) to create a sequence of metatypes from the dynamic type of x through any superclass metatypes it might have.

However this method still won't work with protocols. Hopefully a future version of the language will provide much richer reflection APIs that allow you to compare the relationship between two metatype values.

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However, given the above knowledge of being able to use Mirror, we can use it to lift the aforementioned restriction of T.self == base from our AnyType wrapper on by handling class metatypes separately:

struct AnyType {

  let base: Any.Type
  private let _canCast: (Any) -> Bool

  /// Creates a new AnyType wrapper from a given metatype.
  init<T>(_ base: T.Type) {

    self.base = base

    // handle class metatypes separately in order to allow T.self != base.
    if base is AnyClass {
      self._canCast = { x in
        sequence(
          first: Mirror(reflecting: x), next: { $0.superclassMirror }
        )
        .contains { $0.subjectType == base }
      }
    } else {
      // sanity check – this should never be triggered,
      // as we handle the case where base is a class metatype.
      precondition(T.self == base, """
        The static value \(T.self) and dynamic value \(base) of the passed \
        metatype do not match
        """)

      self._canCast = { $0 is T }
    }
  }

  func canCast<T>(_ x: T) -> Bool {
    return _canCast(x)
  }
}

print(AnyType(D.self as C.Type).canCast(D())) // true

The case where T.self is a class metatype should be the only case where T.self != base, as with protocols, when T is some protocol P, T.Type is P.Protocol, which is the type of the protocol itself. And currently, this type can only hold the value P.self.

Getting Type identity of Swift type by name

The Metatype is actually a generic argument of the type(of:) function (i.e., not a nominal type). The dynamic type returned is a concrete metatype (T.Type).

For instance, try using Any.Type instead ;)

Storing and then casting to Metatypes in Swift

Ahmm. Very nice question. I had same 3-hours-pain few days ago. Finally, i found this answer, that main idea is:

Swift's static typing means the type of a variable must be known at
compile time.

So, the way you coding unfortunately (or fortunately :) ) not allowed.

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