I've got a toy example here that I can't find any solutions for online.
protocol Tree {
associatedtype Element
// some nice tree functions
}
class BinaryTree<T> : Tree {
typealias Element = T
}
class RedBlackTree<T> : Tree {
typealias Element = T
}
enum TreeType {
case binaryTree
case redBlackTree
}
// Use a generic `F` because Tree has an associated type and it can no
// longer be referenced directly as a type. This is probably the source
// of the confusion.
class TreeManager<E, F:Tree> where F.Element == E {
let tree: F
init(use type: TreeType){
// Error, cannot assign value of type 'BinaryTree<E>' to type 'F'
switch(type){
case .binaryTree:
tree = BinaryTree<E>()
case .redBlackTree:
tree = RedBlackTree<E>()
}
}
}
I'm not sure what the problem here is or what I should be searching for in order to figure it out. I'm still thinking of protocols as I would interfaces in another language, and I view a BinaryTree as a valid implementation of a Tree, and the only constraint on F is that it must be a Tree. To confuse things even more, I'm not sure why the following snippet compiles given that the one above does not.
func weird<F:Tree>(_ f: F){ }
func test(){
// No error, BinaryTree can be assigned to an F:Tree
weird(BinaryTree<String>())
}
Any pointers or explanations would be greatly appreciated.
I don't understand the context of the situation this would be in. However, I have provided two solutions though:
1:
class Bar<F:Foo> {
let foo: FooClass.F
init(){
foo = FooClass.F()
}
}
2:
class Bar<F:Foo> {
let foo: FooClass
init(){
foo = FooClass()
}
}
What you are currently doing doesn't make logical sense, to whatever you are trying to achieve
I don't know what are you trying for, but of course it's not possible to do that. from your example, you attempt to create Bar class with generic. and that not the appropriate way to create a generic object, because the creation of generic object is to make the object to accept with any type.
Here's some brief explanation of the generic taken from Wikipedia.
On the first paragraph that says, "Generic programming is a style of computer programming in which algorithms are written in terms of types to-be-specified-later that are then instantiated when needed for specific types provided as parameters."
it's very clear about what are the meaning of to-be-specified-later, right :)
Back to your example :
class Bar<F: Foo> {
let foo: F
init() {
// Error, cannot assign value of type 'FooClass' to type 'F'
foo = FooClass()
}
}
From the above code, it is type parameter F which has a constraint to a type Foo. and, you try to create an instance for foo variable with a concrete implementation, that is FooClass. and that's not possible since the foo variable is a F type(which is abstract). of course we can downcast it, like this foo = FooClass() as! F, but then the foo is only limited to FooClass, so why even bother with generic then ?
Hope it help :)
Your approach to this is wrong. In your original example, you would have to specify the type of both the element (E) and the container (BinaryTree or RedBlackTree), in addition to the enum value. That make no sense.
Instead, you should construct the manager to take a tree as the constructor argument, allowing Swift to infer the generic arguments, i.e.
class TreeManager<E, F: Tree> where F.Element == E {
var tree: F
init(with tree: F) {
self.tree = tree
}
}
let manager = TreeManager(with: BinaryTree<String>())
Alternatively, you should look into using opaque return types in Swift 5.1 depending on what the final goal is (the example here is obviously not a real world scenario)
Something like this seems reasonable. The point was to try and have some piece of logic that would determine when the TreeManager uses one type of Tree vs another.
protocol TreePicker {
associatedtype TreeType : Tree
func createTree() -> TreeType
}
struct SomeUseCaseA<T>: TreePicker {
typealias TreeType = RedBlackTree<T>
func createTree() -> TreeType {
return RedBlackTree<T>()
}
}
struct SomeUseCaseB<T>: TreePicker {
typealias TreeType = BinaryTree<T>
func createTree() -> TreeType {
return BinaryTree<T>()
}
}
class TreeManager<T, Picker: TreePicker> where Picker.TreeType == T {
let tree: T
init(use picker: Picker){
tree = picker.createTree()
}
This introduces another protocol that cares about picking the tree implementation and the where clause specifies that the Picker will return a tree of type T.
I think all of this was just the result of not being able to declare the tree of type Tree<T>. Its a bit more verbose but its basically what it would have had to look like with a generic interface instead. I think I also asked the question poorly. I should have just posted the version that didn't compile and asked for a solution instead of going one step further and confusing everyone.
protocol Foo {
associatedtype F
}
class FooClass : Foo {
typealias F = String
}
class Bar<M:Foo> {
let foo: M
init(){
foo = FooClass() as! M
}
}
Related
I am trying to create a method that will return a specialized instance of some generic type. Let's assume the following example:
class Base { }
class Foo: Base {}
class Bar: Base {}
class MyGenericView<T> where T: Base {
func show(with item: T) { }
}
class MySpecializedViewFoo: MyGenericView<Foo> { }
class MySpecializedViewBar: MyGenericView<Bar> { }
With the above given, I would like to have a function like:
func createView<T: Base>(for item: T) -> MyGenericView<T>
but when I try to implement it like
func createView<T: Base>(for item: T) -> MyGenericView<T> {
if item is Foo {
return MySpecializedViewFoo()
} else if item is Bar {
return MySpecializedViewBar()
}
fatalError("Unsupported item")
}
Then I receive
"Cannot convert return expression of type 'MyGenericView' to return type 'MyGenericView'" error.
Is that something that can be achieved? Could somebody please take a look and point an error in my understanding of that?
You can forcibly make the compiler trust you by doing:
func createView<T: Base>(for item: T) -> MyGenericView<T> {
if item is Foo {
return MySpecializedViewFoo() as! MyGenericView<T>
} else if item is Bar {
return MySpecializedViewBar() as! MyGenericView<T>
}
fatalError("Unsupported item")
}
The compiler doesn't let you do this because it is smart enough to do control flow analysis to determine the bounds of generic type parameters at any given point of a method. As far as it is concerned, return MySpecializedViewFoo() means the same thing inside and outside the if statement.
The compiler isn't designed like that because you are not supposed to checking the type of a generic type parameter anyway.
You should use two overloads instead:
func createView(for item: Foo) -> MyGenericView<Foo>
func createView(for item: Bar) -> MyGenericView<Bar>
In fact, your checks aren't adequate at all in the first place, so even if the compiler is smart enough, it will tell you that what you are doing isn't safe. Here's an example of how this could go wrong:
class FooSubclass: Foo {}
let foo = FooSubclass1()
let view: MyGenericView<FooSubclass1> = createView(foo)
Using the fix at the beginning, this will crash at runtime. createView will try to create and return a MySpecializedViewFoo, which is not a MyGenericView<FooSubclass1>. Swift generics are invariant. Note that in this case, it will go into the is Foo branch, not the fatalError branch.
Another case is using Base as T:
let foo: Base = Foo()
let view: MyGenericView<Base> = createView(for: foo)
This again will not go into the fatalError branch. If you make Base a protocol, you can make it produce an error when T is inferred to be Base. If you also make Foo and Bar final, then I think that should be safe, from the point of view of the hypothetical "smart" compiler.
New to Swift and trying to find nice ways of deciding which implementation of a protocol to use. Protocols with associated types don't seem to map onto generic interfaces in other languages at all. I thought something like the following would have worked.
protocol Foo {
associatedtype Item
}
class FooOffline<T>: Foo {
typealias Item = T
}
class FooOnline<T> : Foo {
typealias Item = T
}
func makeFoo<T, F: Foo>() -> F where F.Item == T{
return FooOnline<T>()
}
Am I going down the wrong path? Even if, could someone offer an explanation for why this wouldn't work?
I'm a developer on Java and I'm trying to write in Swift the same solution that I have in Java code.
Is it possible to do this on Swift?
Example Java:
public interface Converter<S,T> {
T convert(S in)
}
public class CarConverterToDTO implements Converter<Car, CarDTO> {
#Override
public CarDTO convert(Car in) {
.....
}
}
Example Swift:
protocol Converter {
func convert<IN, OUT>(in: IN) -> OUT
}
How it would be the implementation?
Thanks!!!
What appears to be a simple question is actually the tip of a rather large and unpleasant iceberg…
I'm going to start by giving you what is probably the real solution to your problem:
class Converter<Input, Output> {
func convert(_ input: Input) -> Output {
fatalError("subclass responsibility")
}
}
struct Car { }
struct CarDTO { }
class DTOCarConverter: Converter<Car, CarDTO> {
override func convert(_ input: Car) -> CarDTO {
return CarDTO()
}
}
Above, I've translated your Java Converter interface into a Swift class instead of a Swift protocol. That's probably what you want.
Now I'll explain why.
A programmer coming from Java to Swift might think that a Swift protocol is the equivalent of a Java interface. So you might write this:
protocol Converter {
associatedtype Input
associatedtype Output
func convert(_ input: Input) -> Output
}
struct Car { }
struct CarDTO { }
class /* or struct */ DTOCarConverter: Converter {
func convert(_ input: Car) -> CarDTO {
return CarDTO()
}
}
Okay, now you can create a converter and convert something:
let converter = DTOCarConverter()
let car = Car()
let dto = converter.convert(car)
But you're going to run into a problem as soon as you want to write a function that takes a Converter as an argument:
func useConverter(_ converter: Converter) { }
// ^
// error: protocol 'Converter' can only be used as a generic constraint because it has Self or associated type requirements
“Well, duh,” you say, “you forgot the type arguments!” But no, I didn't. Swift doesn't allow explicit type arguments after a protocol name:
func useConverter(_ converter: Converter<Car, CarDTO>) { }
// ^ ~~~~~~~~~~~~~
// error: cannot specialize non-generic type 'Converter'
I don't want to get into why you can't do this. Just accept that a Swift protocol is not generally equivalent to a Java interface.
A Swift protocol with no associated types and no mention of Self is, generally, equivalent to a non-generic Java interface. But a Swift protocol with associated types (or that mentions Self) is not really equivalent to any Java construct.
When discussing this problem, we often use the acronym “PAT”, which stands for “Protocol with Associated Types” (and includes protocols that mention Self). A PAT doesn't define a type that you can use as a function argument, return value, or property value. There's not much you can do with a PAT:
You can define a subprotocol. For example, Equatable is a PAT because it defines the == operator to take two arguments of type Self. Hashable is a subprotocol of Equatable.
You can use a PAT as a type constraint. For example, Set is a generic type. Set's type parameter is named Element. Set constrains its Element to be Hashable.
So you can't write a function that takes a plain Converter as an argument. But you can write a function that takes any implementation of Converter as an argument, by making the function generic:
func useConverter<MyConverter: Converter>(_ converter: MyConverter)
where MyConverter.Input == Car, MyConverter.Output == CarDTO
{ }
That compiles just fine. But sometimes it's inconvenient to make your function generic.
And there's another problem that this doesn't solve. You might want a container that holds various Converters from Car to CarDTO. That is, you might want something like this:
var converters: [Converter<Car, CarDTO>] = []
// ^ ~~~~~~~~~~~~~
// error: cannot specialize non-generic type 'Converter'
We can't fix this by making converters generic, like we did with the useConverter function.
What you end up needing is a “type-erased wrapper”. Note that “type-erased” here has a different meaning that Java's “type erasure”. In fact it's almost the opposite of Java's type erasure. Let me explain.
If you look in the Swift standard library, you'll find types whose names start with Any, like AnyCollection. These are mostly “type-erased wrappers” for PATs. An AnyCollection conforms to Collection (which is a PAT), and wraps any type that conforms to Collection. For example:
var carArray = Array<Car>()
let carDictionary = Dictionary<String, Car>()
let carValues = carDictionary.values
// carValues has type Dictionary<String, Car>.Values, which is not an array but conforms to Collection
// This doesn't compile:
carArray = carValues
// ^~~~~~~~~
// error: cannot assign value of type 'Dictionary<String, Car>.Values' to type '[Car]'
// But we can wrap both carArray and carValues in AnyCollection:
var anyCars: AnyCollection<Car> = AnyCollection(carArray)
anyCars = AnyCollection(carValues)
Note that we have to explicitly wrap our other collections in AnyCollection. The wrapping is not automatic.
Here's why I say this is almost the opposite of Java's type erasure:
Java preserves the generic type but erases the type parameter. A java.util.ArrayList<Car> in your source code turns into a java.util.ArrayList<_> at runtime, and a java.util.ArrayList<Truck> also becomes a java.util.ArrayList<_> at runtime. In both cases, we preserve the container type (ArrayList) but erase the element type (Car or Truck).
The Swift type-erasing wrapper erases the generic type but preserves the type parameter. We turn an Array<Car> into an AnyCollection<Car>. We also turn a Dictionary<String, Car>.Values into an AnyCollection<Car>. In both cases, we lose the original container type (Array or Dictionary.Values) but preserve the element type (Car).
So anyway, for your Converter type, one solution to storing Converters in a container is to write an AnyConverter type-erased wrapper. For example:
struct AnyConverter<Input, Output>: Converter {
init<Wrapped: Converter>(_ wrapped: Wrapped) where Wrapped.Input == Input, Wrapped.Output == Output {
self.convertFunction = { wrapped.convert($0) }
}
func convert(_ input: Input) -> Output { return convertFunction(input) }
private let convertFunction: (Input) -> Output
}
(There are multiple ways to implement type-erased wrappers. That is just one way.)
You can then use AnyConverter in property types and function arguments, like this:
var converters: [AnyConverter<Car, CarDTO>] = [AnyConverter(converter)]
func useConverters(_ converters: [AnyConverter<Car, CarDTO>]) {
let car = Car()
for c in converters {
print("dto = \(c.convert(car))")
}
}
But now you should ask: what's the point? Why bother making Converter a protocol at all, if I'm going to have to use a type-erased wrapper? Why not just use a base class to define the interface, with subclasses implementing it? Or a struct with some closures provided at initialization (like the AnyConverter example above)?
Sometimes, there's not a good reason to use a protocol, and it's more sensible to just use a class hierarchy or a struct. So you should take a good look at how you're implementing and using your Converter type and see if a non-protocol approach is simpler. If it is, try a design like I showed at the top of this answer: a base class defining the interface, and subclasses implementing it.
The Swift equivalent to your Java code looks like this.
protocol Converter {
associatedtype Input
associatedtype Output
func convert(input: Input) -> Output
}
class CarConverterToDTO: Converter {
typealias Input = Car
typealias Output = CarDTO
func convert(input: Car) -> CarDTO {
return CarDTO()
}
}
Explanation
The equivalent to a generic Java interface in Swift, would be a protocol with associatedtypes.
protocol Converter {
associatedtype Input
associatedtype Output
}
To create an implementation of that protocol, the implementation must specify which types the associated types maps to, using typealias.
class CarConverterToDTO: Converter {
typealias Input = Car
typealias Output = CarDTO
}
Type Erasure
If you try to use to this approach, you may run into the issue of trying to store an instance of your generic protocol in a variable or property, in which case you will get the compiler error:
protocol 'Converter' can only be used as a generic constraint because it has Self or associated type requirements
The way to solve this issue in Swift, is by using type erasure, where you create a new implementation of your generic protocol, that itself is a generic type (struct or class), and uses a constructor accepting a generic argument, matching your protocol, like so:
struct AnyConverter<Input, Output>: Converter {
// We don't need to specify type aliases for associated types, when the type
// itself has generic parameters, whose name matches the associated types.
/// A reference to the `convert(input:)` method of a converter.
private let _convert: (Input) -> Output
init<C>(_ converter: C) where C: Converter, C.Input == Input, C.Output == Output {
self._convert = converter.convert(input:)
}
func convert(input: Input) -> Output {
return self._convert(input)
}
}
This is usually accompanied by an extension function on the generic protocol, that performs the type erasure by creating an instance of AnyConverter<Input, Output> using self, like so:
extension Converter {
func asConverter() -> AnyConverter<Input, Output> {
return AnyConverter(self)
}
}
Using type erasure, you can now create code that accepts a generic Converter (by using AnyConverter<Input, Output>), that maps Car to CarDTO:
let car: Car = ...
let converter: AnyConverter<Car, CarDTO> = ...
let dto: CarDTO = converter.convert(input: car)
Let's say that I create a protocol like this:
protocol A {
associatedtype T
func computeSomething(with:T) -> Double
}
In my generic typed class, I would like to do something like this:
class B<U> {
var doSomething:A<U>
}
This thing is that this generates an error, but I would like to accept any type that would support computeSomething on my type U but I don't know at all how to do that?
Edit for clarity
Basically if A was a generic struct or class, that would be possible, but what if no default implementation (provided by class or struct) makes sense here and the only thing I want is to ensure that the type does what I want?
Edit #2 (with concrete example)
I wanted to simplify my question which makes it pretty hard to understand so here is a still simplified and fictional problem that probably matches better the issue I am encountering:
I am writing a generic class that processes its generic type T:
class Process<T> { ... }
The class Process itself includes code that processes T, but in order for this code to work, it needs T to conform to some protocols, for instance:
protocol A {
func mixWith(other:A) -> A
}
protocol B {
var isFoo:Bool { get set }
}
So my first approach was to simply require T to conform to those protocols:
class Process<T:<A,B>> { ... }
This looks like the simplest approach and probably is in many cases, but in my case I think that this actually is problematic, for this reason:
First, I may need to process the same type in many different ways, and changing a way a type is being processed often requires changing the actual implementation of protocols A and B for instance in this case, fooProcess and barProcess are both of type Process with generic type MyType:
let fooProcess = Process<MyType>()
let barProcess = Process<MyType>()
But I want fooProcess and barProcess to do different operations which in many cases would require to change the implementation of the A and B protocols of my MyType type and that's simply not possible.
So my first idea was to simply require some closures and variables to be defined so that I wouldn't need protocols anymore and would define the way data is being processed only in my Process class, a little bit like this:
class Process<T> {
//
var mix:(_ lhs:T, _ rhs:T)->(T)
var isFoo:(_ a:T)->(Bool)
...
}
There all of the processing would be directly implemented in my Processing class, again this would have looked like the right solution but now comes another issue, which led me to my associated type approach: it turns out that in many cases, the user of my Process class would want to get some default behaviour implemented by my framework, for instance, I could automatically implement protocol A and B for them as long as their class conformed to protocol X, here is how it did it:
protocol X:A,B {
...
}
extension protocol X {
// Here was my default implementation of A and B, which enabled my user to directly get A and B implemented as long as their type conformed to X
}
By using this method, I would let my user directly choose what they wanted to implement themselves, by conforming to protocol X they would only need to write a little bit of code and let my framework to all of the rest by itself, and if they wanted to implement themselves A or B they still could.
So if I am right, there is no way to do such a thing with my closures implementation.
So for this reason, I thought that an associated type protocol would be a good solution because here I could let my users easily get some default behaviour or write their own, so now we are getting back to my original question:
protocol AProcessing {
associatedtype U
func mix(_ lhs:U, _ rhs:U) -> U
}
protocol BProcessing {
associatedtype U
func isFoo(_ a:U) -> Bool
}
And then do something like that:
class Process<T> {
var aProcessor:AProcessing<T>
var bProcessor:BProcessing<T>
}
Here the advantage compared to closures is that I could write a special class conforming to AProcessing that could provide default implementation, this way:
class AutomaticAProcessing<T:X>:AProcessing { ... }
That would have enabled my users to so something like that:
var processData = Process<SomeType>()
processData.aProcessor = AutomaticAProcessing<SomeType>()
processData.bProcessor = TheirOwnImplemtation
Not only is this not possible in Swift, but it also feels like I am using too many "hacks" to get things done and there should be an easier language feature to do that, unfortunately I don't know what I should use.
I don't think it is possible, because the generic type of the protocol is specified in the class that implements it.
You could write something like this:
class B<U, P: A> where P.T == U {
var someVar: P?
}
But then you would need to specify a second parameter with the specific class. For example:
class C: A {
typealias T = String
func computeSomething(with: String) -> Double {
return 0.0
}
}
let b = B<String, C>()
let c = b.someVar
But it can't return a protocol with specific type in its associatedtype
One way would be to start with an empty generic struct, and then extend it on types where it makes sense:
struct A<T> {}
extension A where T: Base {
func computeSomething(with: T) -> Double {
return 1
}
}
Usage:
protocol Base {}
class B<U: Base> {
let doSomething = A<U>()
func foo(x: U) -> Double {
return doSomething.computeSomething(with: x)
}
}
class Z : Base {}
let x = B<Z>()
let y = x.foo(x: Z())
print(y)
To EDIT #2:
Remove associated types, and it should be workable:
protocol A {}
protocol B {}
protocol AProcessing {
func mix(_ lhs: A, _ rhs: A) -> A
}
protocol BProcessing {
func isFoo(_ a: B) -> Bool
}
Then, your processor:
class Process<T: A & B> {
var aProcessor: AProcessing!
var bProcessor: BProcessing!
func foo(_ a: T) -> Bool {
let x = aProcessor.mix(a, a)
guard let b = x as? B else { return false }
return bProcessor.isFoo(b)
}
}
And usage:
struct AutomaticAProcessing : AProcessing {
func mix(_ lhs: A, _ rhs: A) -> A { return lhs }
}
struct TheirOwnImplemtation : BProcessing {
func isFoo(_ a: B) -> Bool { return false }
}
struct SomeType : A, B {}
var processData = Process<SomeType>()
processData.aProcessor = AutomaticAProcessing()
processData.bProcessor = TheirOwnImplemtation()
let x = SomeType()
let y = processData.foo(x)
print(y)
I have the following structure:
class A
class A<T: SomeInterface> : C {
}
class B
class B : SomeInterface {
func someFunc(param: C){
// check if param is of class A with `self.dynamicType` generic type
if let typedC = param as? A<self.dynamicType> { // does not work
}
}
}
I would like to check if param has a class of type B as a generic constraint. Is there a way to specify a wildcard generic type ? It is not of importance whether the generic constraint in fact is of type B or not, but this wouldn't be possible either:
class B : SomeInterface {
func someFunc(param: C){
// check if param is of type A
if let typedC = param as? A { // don't check generic constraint type
}
}
}
or (in case there is a wildcard):
class B : SomeInterface {
func someFunc(param: C){
// check if param is of type A with wildcard (Is there such a thing as a wildcard?)
if let typedC = param as? A<?> { // don't require specific generic type, just enter if param is of type A
}
}
}
EDIT - Playground sample
import UIKit
protocol SomeInterface{
func someFunc(param: C)
}
class C {
}
class A<T: SomeInterface> : C{
}
class B : SomeInterface {
func someFunc(param: C) {
// A is a subclass of C, but I want the typed generic subclass
if let typed = param as? A<self.dynamicType> {
}
}
}
let b = B()
let a = A<B>()
b.someFunc(a)
It is hard to express how complicated the world gets once you start allowing non-final subclasses. If you can possibly make these final classes or protocols, many little sharp edges will go away.
What you say you want and the code you're writing don't really line up (which is very common when dealing with subclasses). So it matters what you really want.
I would like to check if param has a class of type B as a generic constraint.
Sure. That's easy.
class B : SomeInterface {
func someFunc(param: C) {
if let typed = param as? A<B> {
print(typed)
} else {
print("NO")
}
}
}
This will work anytime we get an A<B>. It will not work if we receive A<SubClassOfB>, even if that's passed to SubclassOfB.someFunc(). (See what I meant about it being subtle?)
From your code, you seem to actually want "if I'm passed A<Self>" where "Self" means "exactly my class, not a superclass or subclass." The only way I've ever seen that done is with a protocol extension:
extension SomeInterface {
func someFunc(param: C) {
if let typed = param as? A<Self> {
print(typed)
} else {
print("NO")
}
}
}
This allows Self to be replaced with "my class." But note that this means someFunc(A<MySubclass>) won't work. A<Self> is not the same thing as A<Subclass>. Note that you can add a where clause to the extension to limit its scope:
extension SomeInterface where Self: B { ... }
If you mean "T is either Self or some subclass of Self", maybe it's possible, but in my experience it gets more and more tricky and fragile. You can't make generic types covariant on their parameters in Swift, so A<Subclass> is not itself a subtype of A<Superclass> (that's a much trickier problem than it sounds like, and you wouldn't always want it to be true). This is the kind of crazy corner case that goes away if you'll just mark B as final. Then most of the rat's nest goes away.
When you define:
class A<T: SomeInterface>{
}
you are not defining a class, and so a type, but a generic type A, that must be instantiated to become some type.
So, you cannot define inside another class a function in this way:
func someFunc(paramA: A)
This will give always a type error. Instead you should define:
func someFunc(paramA: A<SomeTypeSubtypeOfSomeInterface>)
For instance, since you have defined B as subtype of SomeInterface, you could define:
func someFunc(paramA: A<B>)