In Swift 4 I've created the following protocol to determine if something has a + operator inplemented
protocol Addable { static func +(lhs: Self, rhs: Self) -> Self }
Now I've created a class named Vector<T> where T is of course a generic type.
class Vector<T: Addable>: Addable {
var values: [T]
init(values: [T]) {
self.values = values
}
static func +(lhs: Vector<T>, rhs: Vector<T>) -> Self {
return lhs
}
}
The return lhs part of the + operator implementation is just temporary. But for some reason this gives me the following error:
Cannot convert return expression of type 'Vector<T>' to return type 'Self'
Any idea what I'm doing wrong here? I haven't got a clue.
Moved from comments:
Problem is caused by class inheretability. It looks like Swift cannot infer returning Self type for non-final classes, because Self in current class and in it's subclasses mean different. But for some reason there's no such problem with Self in parameters.
And solutions to this problem are:
make class final, and set returning Self to the proper type and it will work
replace class with struct, and set proper type
add associatedtype that is Self by default:
protocol Addable {
associatedtype S = Self
static func + (lhs: Self, rhs: Self) -> S
}
Latter option will work with non-final classes, but associated type should be checked that it's still equal to Self.
Related
Here's an example:
protocol Feed {
func items<T>() -> [T]? where T: FeedItem
}
protocol FeedItem {}
class FeedModel: Feed, Decodable {
func items<T>() -> [T]? where T : FeedItem {
return [FeedItemModel]() // Error: Cannot convert return expression of type '[FeedItemModel]' to return type '[T]?'
}
}
class FeedItemModel: FeedItem, Decodable {}
Why does it:
A) try to convert to T when T is a generic, not a type?
B) does not recognize FeedItemModel as conforming to FeedItem?
func items<T>() -> [T]? where T : FeedItem
This says that the caller can define T to be whatever they want, as long as T conforms to FeedItemModel, and this function will return an optional array of those.
FeedItemModel is something that conforms to FeedItem, but it is not promised to be the type T that the caller requested.
As an example, consider:
class OtherModel: FeedItem {}
According to your function signature, I can do this:
let ms: [OtherModel]? = FeedModel().items()
But your function won't then return [OtherModel]? to me. I suspect you don't actually mean this to be generic. I expect you mean:
func items() -> [FeedItemModel]?
or possibly
func items() -> [FeedItem]?
(Though I would think very hard before doing the latter one and make sure that the protocol existential is really doing useful work here.)
A)
T is a type, a homogenous concrete type specified at runtime.
Imaging T is class Foo : FeedItem it's obvious that FeedItemModel cannot be converted to Foo
B)
FeedItemModel is recognized as conforming to FeedItem but this is irrelevant.
It's often a mix-up of generics and protocols. Generic types are not covariant. If you need covariant types use an associated type.
Either you can ignore generics because because it only applies to that one function and it isn't needed since directly saying that the return type is [FeedItem]? yields the same result
protocol Feed {
func items() -> [FeedItem]?
}
class FeedModel: Feed, Decodable {
func items() -> [FeedItem]? {
return [OtherModel]()
}
}
If you on the other hand want a generic protocol then you should use a associated type
protocol Feed2 {
associatedtype T: FeedItem
func items() -> [T]?
}
class FeedModel2: Feed2, Decodable {
typealias T = FeedItemModel
func items() -> [T]? {
return [FeedItemModel]()
}
}
I have a protocol P that returns a copy of the object:
protocol P {
func copy() -> Self
}
and a class C that implements P:
class C : P {
func copy() -> Self {
return C()
}
}
However, whether I put the return value as Self I get the following error:
Cannot convert return expression of type 'C' to return type 'Self'
I also tried returning C.
class C : P {
func copy() -> C {
return C()
}
}
That resulted in the following error:
Method 'copy()' in non-final class 'C' must return Self to conform
to protocol 'P'
Nothing works except for the case where I prefix class C with final ie do:
final class C : P {
func copy() -> C {
return C()
}
}
However if I want to subclass C then nothing would work. Is there any way around this?
The problem is that you're making a promise that the compiler can't prove you'll keep.
So you created this promise: Calling copy() will return its own type, fully initialized.
But then you implemented copy() this way:
func copy() -> Self {
return C()
}
Now I'm a subclass that doesn't override copy(). And I return a C, not a fully-initialized Self (which I promised). So that's no good. How about:
func copy() -> Self {
return Self()
}
Well, that won't compile, but even if it did, it'd be no good. The subclass may have no trivial constructor, so D() might not even be legal. (Though see below.)
OK, well how about:
func copy() -> C {
return C()
}
Yes, but that doesn't return Self. It returns C. You're still not keeping your promise.
"But ObjC can do it!" Well, sort of. Mostly because it doesn't care if you keep your promise the way Swift does. If you fail to implement copyWithZone: in the subclass, you may fail to fully initialize your object. The compiler won't even warn you that you've done that.
"But most everything in ObjC can be translated to Swift, and ObjC has NSCopying." Yes it does, and here's how it's defined:
func copy() -> AnyObject!
So you can do the same (there's no reason for the ! here):
protocol Copyable {
func copy() -> AnyObject
}
That says "I'm not promising anything about what you get back." You could also say:
protocol Copyable {
func copy() -> Copyable
}
That's a promise you can make.
But we can think about C++ for a little while and remember that there's a promise we can make. We can promise that we and all our subclasses will implement specific kinds of initializers, and Swift will enforce that (and so can prove we're telling the truth):
protocol Copyable {
init(copy: Self)
}
class C : Copyable {
required init(copy: C) {
// Perform your copying here.
}
}
And that is how you should perform copies.
We can take this one step further, but it uses dynamicType, and I haven't tested it extensively to make sure that is always what we want, but it should be correct:
protocol Copyable {
func copy() -> Self
init(copy: Self)
}
class C : Copyable {
func copy() -> Self {
return self.dynamicType(copy: self)
}
required init(copy: C) {
// Perform your copying here.
}
}
Here we promise that there is an initializer that performs copies for us, and then we can at runtime determine which one to call, giving us the method syntax you were looking for.
With Swift 2, we can use protocol extensions for this.
protocol Copyable {
init(copy:Self)
}
extension Copyable {
func copy() -> Self {
return Self.init(copy: self)
}
}
There is another way to do what you want that involves taking advantage of Swift's associated type. Here's a simple example:
public protocol Creatable {
associatedtype ObjectType = Self
static func create() -> ObjectType
}
class MyClass {
// Your class stuff here
}
extension MyClass: Creatable {
// Define the protocol function to return class type
static func create() -> MyClass {
// Create an instance of your class however you want
return MyClass()
}
}
let obj = MyClass.create()
Actually, there is a trick that allows to easily return Self when required by a protocol (gist):
/// Cast the argument to the infered function return type.
func autocast<T>(some: Any) -> T? {
return some as? T
}
protocol Foo {
static func foo() -> Self
}
class Vehicle: Foo {
class func foo() -> Self {
return autocast(Vehicle())!
}
}
class Tractor: Vehicle {
override class func foo() -> Self {
return autocast(Tractor())!
}
}
func typeName(some: Any) -> String {
return (some is Any.Type) ? "\(some)" : "\(some.dynamicType)"
}
let vehicle = Vehicle.foo()
let tractor = Tractor.foo()
print(typeName(vehicle)) // Vehicle
print(typeName(tractor)) // Tractor
Swift 5.1 now allow a forced cast to Self, as! Self
1> protocol P {
2. func id() -> Self
3. }
9> class D : P {
10. func id() -> Self {
11. return D()
12. }
13. }
error: repl.swift:11:16: error: cannot convert return expression of type 'D' to return type 'Self'
return D()
^~~
as! Self
9> class D : P {
10. func id() -> Self {
11. return D() as! Self
12. }
13. } //works
Following on Rob's suggestion, this could be made more generic with associated types. I've changed the example a bit to demonstrate the benefits of the approach.
protocol Copyable: NSCopying {
associatedtype Prototype
init(copy: Prototype)
init(deepCopy: Prototype)
}
class C : Copyable {
typealias Prototype = C // <-- requires adding this line to classes
required init(copy: Prototype) {
// Perform your copying here.
}
required init(deepCopy: Prototype) {
// Perform your deep copying here.
}
#objc func copyWithZone(zone: NSZone) -> AnyObject {
return Prototype(copy: self)
}
}
I had a similar problem and came up with something that may be useful so I though i'd share it for future reference because this is one of the first places I found when searching for a solution.
As stated above, the problem is the ambiguity of the return type for the copy() function. This can be illustrated very clearly by separating the copy() -> C and copy() -> P functions:
So, assuming you define the protocol and class as follows:
protocol P
{
func copy() -> P
}
class C:P
{
func doCopy() -> C { return C() }
func copy() -> C { return doCopy() }
func copy() -> P { return doCopy() }
}
This compiles and produces the expected results when the type of the return value is explicit. Any time the compiler has to decide what the return type should be (on its own), it will find the situation ambiguous and fail for all concrete classes that implement the P protocol.
For example:
var aC:C = C() // aC is of type C
var aP:P = aC // aP is of type P (contains an instance of C)
var bC:C // this to test assignment to a C type variable
var bP:P // " " " P " "
bC = aC.copy() // OK copy()->C is used
bP = aC.copy() // Ambiguous.
// compiler could use either functions
bP = (aC as P).copy() // but this resolves the ambiguity.
bC = aP.copy() // Fails, obvious type incompatibility
bP = aP.copy() // OK copy()->P is used
In conclusion, this would work in situations where you're either, not using the base class's copy() function or you always have explicit type context.
I found that using the same function name as the concrete class made for unwieldy code everywhere, so I ended up using a different name for the protocol's copy() function.
The end result is more like:
protocol P
{
func copyAsP() -> P
}
class C:P
{
func copy() -> C
{
// there usually is a lot more code around here...
return C()
}
func copyAsP() -> P { return copy() }
}
Of course my context and functions are completely different but in spirit of the question, I tried to stay as close to the example given as possible.
Just throwing my hat into the ring here. We needed a protocol that returned an optional of the type the protocol was applied on. We also wanted the override to explicitly return the type, not just Self.
The trick is rather than using 'Self' as the return type, you instead define an associated type which you set equal to Self, then use that associated type.
Here's the old way, using Self...
protocol Mappable{
static func map() -> Self?
}
// Generated from Fix-it
extension SomeSpecificClass : Mappable{
static func map() -> Self? {
...
}
}
Here's the new way using the associated type. Note the return type is explicit now, not 'Self'.
protocol Mappable{
associatedtype ExplicitSelf = Self
static func map() -> ExplicitSelf?
}
// Generated from Fix-it
extension SomeSpecificClass : Mappable{
static func map() -> SomeSpecificClass? {
...
}
}
To add to the answers with the associatedtype way, I suggest to move the creating of the instance to a default implementation of the protocol extension. In that way the conforming classes won't have to implement it, thus sparing us from code duplication:
protocol Initializable {
init()
}
protocol Creatable: Initializable {
associatedtype Object: Initializable = Self
static func newInstance() -> Object
}
extension Creatable {
static func newInstance() -> Object {
return Object()
}
}
class MyClass: Creatable {
required init() {}
}
class MyOtherClass: Creatable {
required init() {}
}
// Any class (struct, etc.) conforming to Creatable
// can create new instances without having to implement newInstance()
let instance1 = MyClass.newInstance()
let instance2 = MyOtherClass.newInstance()
I wrote an extension onto Int as below.
extension Int {
func squared () -> Int {
return self * self
}
}
print(10.squared()) // works
The above code works. Now I want to extend the IntegerType protocol so that Int, UInt, Int64, etc would all conform. My code is as below.
extension IntegerType {
func squared () -> IntegerType { // this line creates error
return self * self
}
}
I get error:
Protocol 'IntegerType' can only be used as a generic constraint because it has Self or associated type requirements
I already saw this question and its video & this question, still couldn't understand. I only understood that there is some associatedType which in this case is Self but couldn't connect the dots. I feel like also my lack of knowledge on the Generics subject is also a reason...
Can someone elaborate a bit on the subject and why does the extension create an error?
You just have to return Self
edit/update:
Note: You can extend all numeric types (Integer & FloatingPoint) in Swift 4 extending the Numeric Protocol
Swift 4
extension Numeric {
func squared() -> Self {
return self * self
}
}
Swift 3
extension Integer {
func squared() -> Self {
return self * self
}
}
A function return type can only be a concrete Type.
The point is Type. Anything struct, class or Protocols that are completely defined in themselves are pure Type. However when a protocol or struct depend on another Generic Type Placeholder such as T, then this is a partial type.
Type are a data construct that compiler has to allocate certain memory.
So something like this:
let a = Array<T>() or let b = T is not sufficient information for the compiler to deduce at compile time.
Hence, this wont work.
extension IntegerType {
func squared () -> IntegerType { // this line creates error
return self * self
}
}
Here, IntegerType is a partial type. It is a generic protocol that only when conformed can then we know the exact type. Similar to Array. Array itself is not a type. Its a generic container. Only when someone creates it with Array() or Array()... then it has a type.
The same happened with you.
public protocol IntegerType : _IntegerType, RandomAccessIndexType {
then again,
public protocol RandomAccessIndexType : BidirectionalIndexType, Strideable, _RandomAccessAmbiguity {
#warn_unused_result
public func advancedBy(n: Self.Distance) -> Self
then again,
public protocol _RandomAccessAmbiguity {
associatedtype Distance : _SignedIntegerType = Int
}
Hence, as RandomAccessIndexType has Self requirement meaning until and unless someone conforms to it, Self is unknown placeholder. It is partial Type.
Since IntegerType conforms to the RandomAccessIndexType and _RandomAccessAmbuiguity which requires Distance associated type too.
Hence you cant do this too
let a: IntegerType = 12
Again IntegerType needs to know Self and Distance (associatedType).
Int however provides the details like so
public struct Int : SignedIntegerType, Comparable, Equatable {
/// A type that can represent the number of steps between pairs of
/// values.
public typealias Distance = Int
Hence you can do such
let a:Int = 10
because it provides Self for SignedIntegerType and Distance for its other counterpart.
Simply put it:
A partial type cannot be used where a concrete type can be. A partial type are good for other generics and constraining them.
I'm trying to add functionality to an NSManagedObject via a protocol. I added a default implementation which works fine, but as soon as I try to extend my subclass with the protocol it tells me that parts of it are not implemented, even though I added the default implementation.
Anyone having Ideas of what I'm doing wrong?
class Case: NSManagedObject {
}
protocol ObjectByIdFetchable {
typealias T
typealias I
static var idName: String { get }
static func entityName() -> String
static func objectWithId(ids:[I], context: NSManagedObjectContext) -> [T]
}
extension ObjectByIdFetchable where T: NSManagedObject, I: AnyObject {
static func objectWithId(ids:[I], context: NSManagedObjectContext) -> [T] {
let r = NSFetchRequest(entityName: self.entityName())
r.predicate = NSPredicate(format: "%K IN %#", idName, ids)
return context.typedFetchRequest(r)
}
}
extension Case: ObjectByIdFetchable {
typealias T = Case
typealias I = Int
class var idName: String {
return "id"
}
override class func entityName() -> String {
return "Case"
}
}
The error I get is Type Case doesn't conform to protocol ObjectByIdFetchable
Help very much appreciated.
We'll use a more scaled-down example (below) to shed light on what goes wrong here. The key "error", however, is that Case cannot make use of the default implementation of objectWithId() for ... where T: NSManagedObject, I: AnyObject; since type Int does not conform to the type constraint AnyObject. The latter is used to represent instances of class types, whereas Int is a value type.
AnyObject can represent an instance of any class type.
Any can represent an instance of any type at all, including function types.
From the Language Guide - Type casting.
Subsequently, Case does not have access to any implementation of the blueprinted objectWithId() method, and does hence not conform to protocol ObjectByIdFetchable.
Default extension of Foo to T:s conforming to Any works, since Int conforms to Any:
protocol Foo {
typealias T
static func bar()
static func baz()
}
extension Foo where T: Any {
static func bar() { print ("bar") }
}
class Case : Foo {
typealias T = Int
class func baz() {
print("baz")
}
}
The same is, however, not true for extending Foo to T:s conforming to AnyObject, as Int does not conform to the class-type general AnyObject:
protocol Foo {
typealias T
static func bar()
static func baz()
}
/* This will not be usable by Case below */
extension Foo where T: AnyObject {
static func bar() { print ("bar") }
}
/* Hence, Case does not conform to Foo, as it contains no
implementation for the blueprinted method bar() */
class Case : Foo {
typealias T = Int
class func baz() {
print("baz")
}
}
Edit addition: note that if you change (as you've posted in you own answer)
typealias T = Int
into
typealias T = NSNumber
then naturally Case has access to the default implementation of objectWithId() for ... where T: NSManagedObject, I: AnyObject, as NSNumber is class type, which conforms to AnyObject.
Finally, note from the examples above that the keyword override is not needed for implementing methods blueprinted in a protocol (e.g., entityName() method in your example above). The extension of Case is an protocol extension (conforming to ObjectByIdFetchable by implementing blueprinted types and methods), and not really comparable to subclassing Case by a superclass (in which case you might want to override superclass methods).
I found the solution to the problem. I thought it's the typealias T which is the reason for not compiling. That's actually not true, it's I which I said to AnyObject, the interesting thing is that Int is not AnyObject. I had to change Int to NSNumber
Update: Swift 3 permits Self to be used from other types, thanks to SE-0068 – Expanding Swift Self to class members and value types.
You can return "Self" from a class function:
extension NSObject {
class func makeOne() -> Self {
return self()
}
}
So you can do:
let set : NSCountedSet = NSCountedSet.makeOne()
However, the following two don't compile:
extension NSObject {
class func makeTwo() -> (Self, Self) {
return (self(), self())
}
class func makeMany() -> [Self] {
return [self(), self(), self(), self(), self()]
}
}
The error is:
<REPL>:11:34: error: 'Self' is only available in a protocol or as the result of a class method; did you mean 'NSObject'?
class func makeTwo() -> (Self, Self) {
^~~~
NSObject
<REPL>:11:40: error: 'Self' is only available in a protocol or as the result of a class method; did you mean 'NSObject'?
class func makeTwo() -> (Self, Self) {
^~~~
NSObject
<REPL>:15:35: error: 'Self' is only available in a protocol or as the result of a class method; did you mean 'NSObject'?
class func makeMany() -> [Self] {
^~~~
NSObject
Does anyone know of any way to declare that a class function returns multiple instances of the class itself?
The problem, I suspect, is that Self is ambiguous; it means "this class or a subclass, whatever the thing happens to be at the time we are called". In other words, Self is polymorphic. But you can't make an array consisting of two different classes, for example. And although the class may permit a certain initializer, we cannot know in advance that its subclass will.
The solution is to use the class name itself. Here's an example for a struct Thing:
extension Thing {
static func makeTwo() -> (Thing, Thing) {
return (Thing(), Thing())
}
}