In trying to subclass my base class that conforms to a PAT, I get this compiler warning:
'MyType' is ambiguous for type lookup in this context
in my code as shown below:
protocol DataSourceProtocol {
associatedtype MyType
func get () -> MyType
}
class AnyDataSource: DataSourceProtocol {
typealias MyType = EntityProtocol
func get() -> MyType { fatalError("No implementation") }
}
class DataSource_A<T: EntityProtocol & DataSource_A_Compatible>: AnyDataSource {
typealias MyType = T
override func get() -> MyType { // <-- 'MyType' is ambiguous for type lookup in this context
return T()
}
}
class DataSource_B<T: EntityProtocol & DataSource_B_Compatible>: AnyDataSource {
override func get() -> MyType {
return T()
}
}
I am trying to override the method get() -> MyType so that the DataSource_A instance method returns a specialized type T, hence typealias MyType = T
If I change the DataSource_A method to override func get -> T, the compiler says that the function does not override any method in its superclass.
In DataSource_B, the overridden method is called but provides no type information for type T because the caller receives type EntityProtocol which is not sufficient.
If I remove the function declaration in DataSourceProtocol and in the base class AnyDataSource, and simply declare it in the subclass, it works fine, but I want to make the method's implementation mandatory for any subclass of AnyDataSource.
Why can't I specify that the overridden method returns an object of type T by changing the value of MyType in the subclass?
I want to make the method's implementation mandatory for any subclass
of AnyDataSource.
The way to express that is to make AnyDataSource generic.
class AnyDataSource<Entity: EntityProtocol>: DataSourceProtocol {
func get() -> Entity { .init() }
}
class DataSource_A<Entity: EntityProtocol & DataSource_A_Compatible>: AnyDataSource<Entity> {
// override if needed
}
Related
What I tried to do:
protocol HasElement {
associatedtype ItemType
func getElement() -> ItemType
func setElement(element: ItemType)
}
class Element {}
class BarElement: Element {}
class Foo: NSObject, HasElement {
typealias ItemType = Element
func getElement() -> Element { ... }
func setElement(element: Element) { ... }
}
class Bar: Foo {
typealias ItemType = BarElement
override func getElement() -> BarElement { ... } // This works.
override func setElement(element: BarElement) { ... } // This fails.
}
The error is:
Method does not override any method from its superclass
If I try to use ItemType instead:
override func setElement(element: ItemType) { ... } // Still fails.
The error is:
'ItemType' is ambiguous for type lookup in this context
Is there a way to make this work?
Here's a way to do what you want:
protocol HasElement {
associatedtype ItemType
func getElement() -> ItemType
func setElement(element: ItemType)
}
class Element {}
class BarElement: Element {}
class Foo: HasElement {
// no need for typealias, the associated type is inferred
func getElement() -> Element { return Element() }
func setElement(element: Element) { }
}
class Bar: Foo {
// no need for typealias, the associated type is inferred
override func getElement() -> BarElement { return BarElement() }
// hide the parent class method
#available(*, unavailable, message: "Use setElement(element: BarElement)")
override func setElement(element: Element) { }
// comply with protocol in this class
func setElement(element: BarElement) { }
}
// can't do this now:
let myElement = Element()
let myBar = Bar()
myBar.setElement(element: myElement) // Error: 'setElement(element: BarElement)' is unavailable: Use setElement(element: BarElement)
The problem here isn't associated types, it's that method inputs are contravariant. Therefore you cannot override a method that expects a given superclass instance input with a method that expects a subclass instance input.
In fact, you can simply boil your code down to:
class Element {}
class BarElement : Element {}
class Foo {
func setElement(element: Element) { }
}
class Bar : Foo {
// error: Method does not override any method from its superclass
override func setElement(element: BarElement) { }
}
You simply cannot override an (Element) -> Void method with a (BarElement) -> Void method. The reasons for this should be fairly obvious if you consider what would happen if you created a Bar instance and upcast it to Foo. You're now able to pass an Element instance to a method that expects a BarElement instance, which is illegal.
The reason it works for your getElement method is that method outputs are covariant. Therefore overriding a () -> Element method with a () -> BarElement method is perfectly legal, as even if you upcast a Bar instance to Foo, the BarElement instance returned from getElement can be freely upcast to Element.
As for solutions, it rather depends on your exact use case. It may well be that what you're trying to do doesn't require inheritance at all, and instead you can just conform Foo and Bar to HasElement separately.
This works:
override func getElement() -> BarElement { ... }
using override keyword is when function signature does not change (BarElement is still Element). You copy and paste function from super class with same function name, same parameter name and same return value type.
But
override func setElement(element: BarElement) { ... }
fails because when you change parameter type (with same parameter name).
This case is overloading not overriding.
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'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
Is it possible to have generic inside generic?
I have this protocol
public protocol ListViewModelProtocol {
typealias ViewModel
typealias Cell
func titleForHeaderInSection(section: Int) -> String?
func numberOfSections() -> Int
func numberOfRowsInSection(section: Int) -> Int
func viewModelAtIndexPath(indexPath: NSIndexPath) -> ViewModel
}
I also have base ListViewModel that implements this protocol
public class BaseListViewModel<T, U> : ListViewModelProtocol {
}
But already here it says that my ListViewModelProtocol is not implemented. How can I set T and U to be of specific class inside protocol? Because if I write this in protocol
typealias ViewModel: CustomClass
typealias Cell: CustomCell
Its still not working.
My goal is to subclass BaseListViewModel like
public class TestListViewModel : BaseListViewModel<TestCellViewModel, TestTableViewCell> {
}
Then I could just do this in my BaseViewController
public class BaseViewController<T: ListViewModelProtocol>: UITableViewController {
}
And in some subclass ViewController do this:
public class CustomViewController: BaseViewController<TestListViewModel> {
}
and that way CustomViewController would "get" TestCellViewModel and TestTableViewCell (actually its BaseViewController).
But of course this is not working as I expected. What am I missing? Or I have to define typealias for ListViewModelProtocol in every class that implements it or uses it as generic type? Which means I would have to define ViewModel and Cell of ListViewModelProtocol in both BaseListViewModel class and BaseViewController class, but thats not so generic since I just want to put base types of those in protocol and thats it.
Or maybe there is something wrong with my approach and I should implement this differently?
Any suggestions are useful. Thanks
EDIT
I have managed to fix this but I have another problem.
public class BaseViewController<T: ListViewModelProtocol>: UITableViewController {
var dataSource: T?
}
This datasource is used inside UITableViewDataSource methods by calling its own methods (see ListViewModelProtocol methods). Everything is working fine but when some custom controller:
Controller: BaseViewController<TestListViewModel>
is being deinitialized I get EXC_BAD_ACCESS error. If I put
deinit {
self.dataSource = nil
}
it works but I would like to know why I need to set it to nil.
Thanks.
typealias keyword has more than one meaning ...
// protocol can't be generic
protocol P {
// here typealias is just placeholder, alias
// for some unknown type
typealias A
func foo(a:A)->String
}
// C is generic
class C<T>:P {
// here typealias define the associated type
// in this example it is some generic type
typealias A = T
func foo(a: A) -> String {
return String(a)
}
}
let c1 = C<Int>()
print(c1.foo(1)) // 1
let c2 = C<Double>()
print(c2.foo(1)) // 1.0
// D is not generic!!!
class D: C<Double> {}
let d = D()
print(d.foo(1)) // 1.0
Update, to answer the question from discussion
class Dummy {}
protocol P {
// here typealias is just placeholder, alias
// for some inknown type
typealias A : Dummy
func foo(a:A)->String
}
// C is generic
class C<T where T:Dummy>:P {
// here typealias define the associated type
// in this example it is some generic type
typealias SomeType = T
func foo(a: SomeType) -> String {
return String(a)
}
}
class D:Dummy {}
let c = C<D>()
print(c.foo(D())) // D
and
// now next line doesn't compile
let c1 = C<Int>() // error: 'C' requires that 'Int' inherit from 'Dummy'
If you want to implement a protocol with associated types you have to set these associated types in the your generic implementation:
public class BaseListViewModel<T, U> : ListViewModelProtocol {
typealias ViewModel = T
typealias Cell = U
// implement the methods as well
}
Consider the following:
protocol Foo {
typealias A
func hello() -> A
}
protocol FooBar: Foo {
func hi() -> A
}
extension FooBar {
func hello() -> A {
return hi()
}
}
class FooBarClass: FooBar {
typealias A = String
func hi() -> String {
return "hello world"
}
}
This code compiles. But if I comment out explicit definition of associated type typealias A = String, then for some reason, swiftc fails to infer the type.
I'm sensing this has to do with two protocols sharing the same associated type but without a direct assertion through, for example, type parameterization (maybe associated type is not powerful/mature enough?), which makes it ambiguous for type inference.
I'm not sure if this is a bug / immaturity of the language, or maybe, I'm missing some nuances in protocol extension which rightfully lead to this behaviour.
Can someone shed some light on this?
look at this example
protocol Foo {
typealias A
func hello() -> A
}
protocol FooBar: Foo {
typealias B
func hi() -> B
}
extension FooBar {
func hello() -> B {
return hi()
}
}
class FooBarClass: FooBar {
//typealias A = String
func hi() -> String {
return "hello world"
}
}
with generics
class FooBarClass<T>: FooBar {
var t: T?
func hi() -> T? {
return t
}
}
let fbc: FooBarClass<Int> = FooBarClass()
fbc.t = 10
fbc.hello() // 10
fbc.hi() // 10
Providing explicit values for associated types in a protocol is required for conformance to said protocol. This can be accomplished by hard coding a type, as you've done with typealias A = String, or using a parameterized type as you mentioned, such as below:
class FooBarClass<T>: FooBar {
typealias A = T
...
}
Swift will not infer your associated type from an implemented method of the protocol, as there could be ambiguity with multiple methods with mismatching types. This is why the typealias must be explicitly resolved in your implementing class.