I couldn't find any good explanation to my questions so I'd like to ask you directly. First of all I'd like to refine my code in this post.
My problem is the protocol AnyObject and the Self type. I didn't implement AnyObject into my code because it is marked with #objc and I don't want any Objective-C stuff involved in my code (don't judge me for that). I also couldn't find any explanation about the Self type. It just worked as expected, but Xcode does not replace Self with the type the static function is called at.
Here is some example:
extension Int : Instance {}
Int.singleton { (customInstanceName) -> Self in 0 } // Self shall be replaced with Int
As you can see Xcode produces a Self instead an Int. Is there any chance I could fix this? Am I right that Self does return the dynamicType and my implementation is fine as it is in my post above? I would really appreciate any good explanation about the Self type.
As you have seen in my code. I am using a custom protocol to check whether my instance is a class or not. Is there any other shiny implementation to check my instances if they are classes or structure types, or am I forced to use AnyObject if I want to get rid of my ClassInstance protocol?
Thank you for your time.
UPDATE:
protocol Test {}
class A : Test {}
struct B : Test {}
let aClass : Test = A()
let aStruct : Test = B()
if let someClass = aClass as? AnyObject {
print(someClass) // only this will print
}
if let someStruct = aStruct as? AnyObject {
print(someStruct)
}
This will work, but AnyObject is still marked as an #objc protocol.
The Self type can be only used in protocols where it is a implicit typealias of the type which conforms to it:
protocol Testable {
func test() -> Self
}
If you want to conform to this protocol you than have to replace Self with the name of the type. For instance:
struct Product: Testable {
func test() -> Product {
return Product()
}
}
Important Edit:
As DevAndArtist pointed out in the comments there is a working class check in Swift 1.2 (without automatic bridging to Objective C) but not Swift 2 (Xcode 7 beta 3; probably a bug):
if instance.dynamicType is AnyClass {
// instance is a class
} else {
// instance is not a class
}
You can see workaround (mainly) for Swift 2 below.
End Edit
With respect to classes you should use AnyObject if you want to keep it simple but you can also use reflection which would be much more effort.
Below you can see some reflection results of string interpolations (only the first few characters):
"\(reflect(classType))" // Swift._ClassMirror
"\(reflect(0))" // Swift._LeafMirror
"\(reflect(enumType))" // Swift._EnumMirror
"\(reflect(structure))" // Swift._StructMirror
"\(reflect([0, 4]))" // Swift._ArrayTypeMirror
"\(reflect(NSDate()))" // Foundation._NSDateMirror
"\(reflect(NSURLRelationship.Contains))" // Swift._EnumMirror
"\(reflect(Int?(2)))" // Swift._OptionalMirror
As you can see enums are consistent if they are not defined in the Swift standard library (unfortunately also Optional...). So you can distinguish also structs and enums:
public enum Type {
case Enum, Class, Struct
}
public func getType<T>(anything: T) -> Type {
if anything is AnyObject {
return .Class
}
if "\(reflect(anything))".hasPrefix("Swift._EnumMirror") {
return .Enum
}
return .Struct
}
So for a better result you have to put some effort into it to differentiate between all the different cases.
But the easiest way to distinguish only between reference types and value types (aka classes and structs/enums) is still (unfortunately only works for own declared structs and not built in types because they can be bridged to Objective C; I'm working on it...):
if instance is AnyObject {}
// or: if instance is of type Any
if let classInstance = instance as? AnyObject {}
Related
Since Swift 4, objects have gained subscript(keyPath:) which can be used to retrieve values using AnyKeyPath and its subclasses. According to the Swift book, the subscript is available on all types. For example, an instance of a class TestClass may be subscripted with an AnyKeyPath like so:
class TestClass {
let property = true
}
let anyKeyPath = \TestClass.property as AnyKeyPath
_ = TestClass()[keyPath: anyKeyPath]
This compiles correctly as expected. Use of any other valid subclass would also compile including PartialKeyPath<TestClass>, KeyPath<TestClass, Bool>, etc. This functionality is unavailable in a protocol extension. For example, the following is invalid:
class TestClass {
let property = true
}
protocol KeyPathSubscriptable {
}
extension KeyPathSubscriptable {
func test() {
let anyKeyPath = \TestClass.property as AnyKeyPath
_ = self[keyPath: anyKeyPath] // Value of type 'Self' has no subscripts
}
}
If we want to use that keyPath subscript in the protocol, we can include it in the protocol definition. However, the compiler will not resolve it automatically:
protocol KeyPathSubscriptable {
subscript(keyPath: AnyKeyPath) -> Any? { get }
}
extension KeyPathSubscriptable {
func test() {
let anyKeyPath = \TestClass.property as AnyKeyPath // This can be any valid KeyPath
_ = self[keyPath: anyKeyPath]
}
}
class TestClass: KeyPathSubscriptable { // Type 'TestObject' does not conform to protocol 'KeyPathSubscriptable'
let property = true
}
With this, we get a compile error: Type 'TestObject' does not conform to protocol 'KeyPathSubscriptable'. In order to resolve this, we must include a redundant implementation of that subscript in TestClass:
class TestClass: KeyPathSubscriptable {
let property = true
subscript(keyPath: AnyKeyPath) -> Any? {
fatalError() // This is never executed
}
}
This resolves the conformance issue and produces the goal result although it is seemingly unnecessary and illogical. I'm not sure how, but the subscript implementation is never even used. It's finding the expected implementation of subscript(keyPath:) and using that instead, but how? Where is that and is there any way to use it in a protocol? Why is this required by the compiler even though it's never used?
The context of this use case is in a logging module. The goal is that an object should be able to adopt a particular protocol which, with no additional setup on the object, would provide a human readable description of the object, instead of the default for many objects which is a memory address. The protocol would use Mirror to fetch KeyPaths of an object, read the values, and print them to the console. It is intended for debugging purposes and would not run in any production environment.
Please let me know if I can make any clarifications. I may post this to the Swift team if others think that this could potentially be a bug of sorts. All help is appreciated. Thanks in advance.
Full gist located here.
It's easy to check protocol conformance of a type at runtime:
guard type is Encodable.Type else { ... }
but this technique fails if type is not a struct or class, but instead a protocol that inherits from Encodable. Is there a way to make a similar check for protocols?
This is a tad hacky, but it will probably solve your problem:
Given these example types:
protocol Animal: Encodable {}
struct Person: Animal {}
protocol SomeOtherProtocol {}
struct Demon: SomeOtherProtocol {}
I'm simplifing the init logic for the example by assuming:
typealias Group = Array
let animals: Group<Animal> = [Person(), Person(), Person()]
let notAnimals: Group<SomeOtherProtocol> = [Demon(), Demon(), Demon()]
This should work for a custom collection class however... Continuing, define the following extension for your custom Collection
extension Group {
func asEncodables() -> Group<Encodable>?{
return self as? Group<Encodable>
}
var isElementEncodable: Bool {
return self is Group<Encodable>
}
}
You now have access to the following:
animals.asEncodables() //returns some
notAnimals.asEncodables() // returns none
animals.isElementEncodable //true
notAnimals.isElementEncodable //false
And so for your original question, you could make your check as follows:
guard notAnimals.isElementEncodable else { return }
Hope this helps you. Currently I know of no way to make the comparison with something similar to an if X is Y :/
class SuperDelegate <T: AnyObject> {
func addDelegate(delegate: T)
{
}
}
My question is about T key, does it mean the same as id in Objective-c? I mean about case of uses.
how to understand the first line class SuperDelegate <T: AnyObject> Sorry I am new in Swift.
As Objective C program for me this line means that we make class to conform a protocol that has to implement all required method. But I don't understand func addDelegate(delegate: T) is this the same like
- (void)addDelegate:(id)delegate which is a property id <T> delegate.
Yes you are correct in your assumptions that AnyObject behaves like id:
You can call any Objective-C method and access any property on an
AnyObject value without casting to a more specific class type. This
includes Objective-C compatible methods and properties marked with the
#objc attribute.
but you have used it here as a generic type rather than as a concrete type that should be cast to. The class is requiring a type that adheres to the AnyObject protocol but it isn't forcing it to be AnyObject (see header files: cmd + click on AnyObject inside Xcode).
So your instance could be instantiated SuperDelegate<AnyObject> but it could also be instantiated SuperDelegate<NSDate>. This means that the whole subset of ObjC methods and properties cannot be guaranteed as they can with a cast to AnyObject as a concrete type because at runtime T might represent NSDate or NSNumber or any other class.
To achieve what you want you would need to write:
class SuperDelegate {
func addDelegate(delegate: AnyObject)
{
}
}
But Swift is a strongly-typed language and it would normally be the case that you had a delegate protocol and that the delegate for your type adhered to the delegate protocol:
protocol DelegateProtocol {
func somethingHappened()
}
struct MyTypeDelegate:DelegateProtocol {
func somethingHappened() {
print("Thanks for telling me!")
}
}
struct MyType {
var delegate:DelegateProtocol?
func tellDelegateSomethingHappened() {
delegate?.somethingHappened()
}
}
let del = MyTypeDelegate()
var type = MyType()
type.delegate = del
type.tellDelegateSomethingHappened()
I've run into a problem where I have some protocol:
protocol Baz {
func bar<T>(input:T)
}
The function bar is made generic because I don't want the protocol itself to have a Self(it needs to be useable in a collection). I have an implementation of the protocol defined as:
class Foo<S>: Baz {
var value:S
init(value:S) {
self.value = value
}
func bar<T>(input:T) {
value = input
}
}
This gives an error because the compiler doesn't know that S and T are the same type. Ideally I should be able to write something like:
func bar<T where T==S>(input:T) {
value = input
}
or
func bar<T:S>(input:T) {
value = input
}
The first form gives a "Same-type requirement makes generic parameter 'S' and 'T' equivalent" error (which is exactly what I'm trying to do, so not sure why it's an error). The second form gives me a "Inheritance from non-protocol, non-class type 'S'".
Any ideas of on either how to get this to work, or a better design pattern in Swift?
Update: As #luk2302 pointed out, I forgot to make Foo adhere to the Baz protocol
#luk2302 has hinted at much of this in the comments, but just to make it explicit for future searchers.
protocol Baz {
func bar<T>(input:T)
}
This protocol is almost certainly useless as written. It is effectively identical to the following protocol (which is also almost completely useless):
protocol Baz {
func bar(input:Any)
}
You very likely mean (and hint that you mean):
protocol Baz {
typealias T
func bar(input: T)
}
As you note, this makes the protocol a PAT (protocol with associated type), which means you cannot put it directly into a collection. As you note, the usual solution to that, if you really need a collection of them, is a type eraser. It would be nice if Swift would automatically write the eraser for you, which it likely will be able to do in the future, and would eliminate the problem. That said, though slightly tedious, writing type erasers is very straightforward.
Now while you cannot put a PAT directly into a collection, you can put a generically-constrained PAT into a collection. So as long as you wrap the collection into a type that constrains T, it's still no problem.
If these become complex, the constraint code can become tedious and very repetitive. This can be improved through a number of techniques, however.
Generic structs with static methods can be used to avoid repeatedly providing constraints on free-functions.
The protocol can be converted into a generic struct (this formalizes the type eraser as the primary type rather than "as needed").
Protocols can be replaced with functions in many cases. For example, given this:
protocol Bar {
typealias T
func bar(input: T)
}
struct Foo : Bar {
func bar(input: Int) {}
}
You can't do this:
let bars: [Bar] = [Foo()] // error: protocol 'Bar' can only be used as a generic constraint because it has Self or associated type requirements
But you can easily do this, which is just as good:
let bars = [(Int) -> Void] = [Foo().bar]
This is particularly powerful for single-method protocols.
A mix of protocols, generics, and functions is much more powerful than trying to force everything into the protocol box, at least until protocols add a few more missing features to fulfill their promise.
(It would be easier to give specific advice to a specific problem. There is no one answer that solves all issues.)
EDITED (Workaround for "... an error because the compiler doesn't know that S and T are the same type.")
First of all: This is just an separate note (and perhaps an attempt at redemption for my previous answer that ended up being myself chasing my own tail to compute lots and lots of redundant code) in addition to Robs excellent answer.
The following workaround will possibly let your implementation protocol Foo ... / class Bas : Foo ... mimic the behaviour you initially asked for, in the sense that class method bar(...) will know if generics S and T are actually the same type, while Foo still conforms to the protocol also in the case where S is not of the same type as T.
protocol Baz {
func bar<T>(input:T)
}
class Foo<S>: Baz {
var value:S
init(value:S) {
self.value = value
}
func bar<T>(input:T) {
if input is S {
value = input as! S
}
else {
print("Incompatible types. Throw...")
}
}
}
// ok
var a = Foo(value: 1.0) // Foo<Double>
print(a.value) // 1.0
a.bar(2.0)
print(a.value) // 2.0
let myInt = 1
a.bar(myInt) // Incompatible types. Throw...
print(a.value) // 2.0
// perhaps not a loophole we indended
let myAny : Any = 3.0
a.bar(myAny)
print(a.value) // 3.0
The Any and AnyObject loophole here could be redeemed by creating a dummy type constraint that you extend all types (that you wish for to use the generics) to, however not extending Any and AnyObject.
protocol NotAnyType {}
extension Int : NotAnyType {}
extension Double : NotAnyType {}
extension Optional : NotAnyType {}
// ...
protocol Baz {
func bar<T: NotAnyType>(input:T)
}
class Foo<S: NotAnyType>: Baz {
var value:S
init(value:S) {
self.value = value
}
func bar<T: NotAnyType>(input:T) {
if input is S {
value = input as! S
}
else {
print("Incompatible types. Throw...")
}
}
}
// ok
var a = Foo(value: 1.0) // Foo<Double>
// ...
// no longer a loophole
let myAny : Any = 3.0
a.bar(myAny) // compile time error
let myAnyObject : AnyObject = 3.0
a.bar(myAnyObject) // compile time error
This, however, excludes Any and AnyObject from the generic in full (not only for "loophole casting"), which is perhaps not a sought after behaviour.
Both this declaration
protocol SomeProtocol : AnyObject {
}
and this declaration
protocol SomeProtocol : class {
}
seem to make it so that only classes can conform to this protocol (i.e. that the instances of the protocol are references to objects), and have no other effects.
Is there any difference between them? Should one be preferred over the other? If not, why is there two ways to do the same thing?
I am using the latest released Xcode 6.3.1.
This was answered by an official Swift developer (Slava_Pestov) on the Swift forums. Here is the summary:
You should use AnyObject (protocol SomeProtocol: AnyObject).
AnyObject and class are equivalent. There is no difference.
class will eventually be deprecated.
Regarding the answer https://forums.swift.org/t/class-only-protocols-class-vs-anyobject/11507/4, this answer is deprecated. These words are the same now.
DEPRECATED
Update: After consulting with the powers that be, the two definitions are supposed to be equivalent, with AnyObject being used as a stand-in while class was being finished. In the future the latter will obviate the former but, for now, they do present a few minor differences.
The difference lies in the semantics of #objc declarations. With AnyObject, the expectation is that conforming classes may or may not be proper Objective-C objects, but the language treats them as such anyway (in that you lose static dispatch sometimes). The takeaway from this is that you can treat an AnyObject et al. protocol constraint as a way to ask for #objc member functions as shown in the example in documentation for AnyObject in the STL:
import Foundation
class C {
#objc func getCValue() -> Int { return 42 }
}
// If x has a method #objc getValue()->Int, call it and
// return the result. Otherwise, return nil.
func getCValue1(x: AnyObject) -> Int? {
if let f: ()->Int = x.getCValue { // <===
return f()
}
return nil
}
// A more idiomatic implementation using "optional chaining"
func getCValue2(x: AnyObject) -> Int? {
return x.getCValue?() // <===
}
// An implementation that assumes the required method is present
func getCValue3(x: AnyObject) -> Int { // <===
return x.getCValue() // x.getCValue is implicitly unwrapped. // <===
}
The same example falls over immediately if you change that to a class-deriving protocol:
import Foundation
protocol SomeClass : class {}
class C : SomeClass {
#objc func getCValue() -> Int { return 42 }
}
// If x has a method #objc getValue()->Int, call it and
// return the result. Otherwise, return nil.
func getCValue1(x: SomeClass) -> Int? {
if let f: ()->Int = x.getCValue { // <=== SomeClass has no member 'getCValue'
return f()
}
return nil
}
// A more idiomatic implementation using "optional chaining"
func getCValue2(x: SomeClass) -> Int? {
return x.getCValue?() // <=== SomeClass has no member 'getCValue'
}
// An implementation that assumes the required method is present
func getCValue3(x: SomeClass) -> Int { // <===
return x.getCValue() // <=== SomeClass has no member 'getCValue'
}
So it seems class is a more conservative version of AnyObject that should be used when you only care about reference semantics and not about dynamic member lookups or Objective-C bridging.
In the Swift programming language guide for protocols, under the Class-Only Protocols section. It only mentioned AnyObject, but not class.
You can limit protocol adoption to class types (and not structures or enumerations) by adding the AnyObject protocol to a protocol’s inheritance list.
protocol SomeClassOnlyProtocol: AnyObject, SomeInheritedProtocol {
// class-only protocol definition goes here
}
For that reason, I will suggest using AnyObject over class for new code or new project. Other than that, I don't see any obvious difference between them.
From 2021, Xcode 12.5, Big Sur OS:
Usage of class is deprecated by apple.
Use AnyObject instead.
Happy Coding.
AnyObject is a protocol to which all classes implicitly conform (source). So I would say there is no difference: you can use either to require class constraint.
If you open the help (alt-click) in Xcode 9 for class in a line such as protocol P: class {}, you will get typealias AnyObject.
Thus, the code compiled (in Swift 4) will be the same whether you constrain the protocol to class or AnyObject.
That said, there is also the question of style and future options — a future Swift version might want to treat class and AnyObject differently in some subtle way, even if that is not the case right now.
(Edit: This has finally happened in Swift 5.4/Xcode 12.5.)
I misspoke before. #MartinR should really answer this, since he's the one who corrected me and provided the correct information.
The real difference is that a protocol with the class qualifier can only be applied to a class, not a struct or enum.
Martin, why don't you answer and the OP can accept your answer?