Overview
I would like to use switch statement by matching array type. I have the following classes.
Class:
class A {}
class B : A {}
Switch on single value works:
let a : A = B()
switch a {
case let b as B:
print("b = \(b)")
default:
print("unknown type")
}
Switch on array (Compilation Error):
let aArray : [A] = [B(), B()]
switch aArray {
case let bArray as [B] :
print("bArray = \(bArray)")
default:
print("unknown type")
}
Error:
Downcast pattern value of type '[B]' cannot be used
Note: Tested on Swift 4
Question:
How can I achieve this ?
In Swift 4.1, you get a better error message (thanks to #11441):
Collection downcast in cast pattern is not implemented; use an explicit downcast to '[B]' instead
In short, you're hitting a bit of the compiler that isn't fully implemented yet, the progress of which is tracked by the bug SR-5671.
You can however workaround this limitation by coercing to Any before performing the cast:
class A {}
class B : A {}
let aArray : [A] = [B(), B()]
switch aArray /* or you could say 'as Any' here depending on the other cases */ {
case let (bArray as [B]) as Any:
print("bArray = \(bArray)")
default:
print("unknown type")
}
// bArray = [B, B]
Why does this work? Well first, a bit of background. Arrays, dictionaries and sets are treated specially by Swift's casting mechanism – despite being generic types (which are invariant by default), Swift allows you to cast between collections of different element types (see this Q&A for more info).
The functions that implement these conversions reside in the standard library (for example, Array's implementation is here). At compile time, Swift will try to identify collection downcasts (e.g [A] to [B] in your example) so it can directly call the aforementioned conversion functions, and avoid having to do a full dynamic cast through the Swift runtime.
However the problem is that this specialised logic isn't implemented for collection downcasting patterns (such as in your example), so the compiler emits an error. By first coercing to Any, we force Swift to perform a fully dynamic cast, which dispatches through the runtime, which will eventually wind up calling the aforementioned conversion functions.
Although why the compiler can't temporarily treat such casts as fully dynamic casts until the necessary specialised logic is in place, I'm not too sure.
Related
I am implementing a generic collection and I would like to make its usage with types/protocol hierarchy more convenient. I stumbled on the following question.
How one does implement a collection
class SpecialSet<ObjectType> {
Such that given two types B : A one can perform implicite conversion of SpecialSet<B> into SpecialSet<A> ?
I am suprised but I wasn't able to find any documentation explaining this simple concept that indeed exist for the Array collection. This swift code is indeed valid:
let b = [B(), B(), B()]
let a: [A] = b
I probably do not used the right vocabulary in my searches.
I tried to define a constructor as follow but it seams not possible to enforce inheritance between two generic.
init<T>(_ specialSet: SpecialSet<T>) where T: ObjectType {
It has always been impossible.
Swift generics are normally invariant, but the Swift standard library
collection types — even though those types appear to be regular
generic types — use some sort of magic inaccessible to mere mortals
that lets them be covariant.
Potential workaround. 🤷♂️
protocol Subclass: AnyObject {
associatedtype Superclass
}
extension B: Subclass {
typealias Superclass = A
}
class SpecialSet<Object>: ExpressibleByArrayLiteral {
required init(arrayLiteral _: Object...) { }
init<T: Subclass>(_: SpecialSet<T>) where T.Superclass == Object { }
}
let specialSetB: SpecialSet = [B(), B()]
let specialSetA = SpecialSet<A>(specialSetB)
Background
I have an enum which, simplified, goes like this:
enum Container<T> {
case a(T)
case b(T)
case c
}
I want to be able to instantiate this with a few different types for the generic and use typealias for that:
typealias IntContainer = Container<Int>
typealias FloatContainer = Container<Float>
Problem
So far, this is all fine. However, I also want to create a recursive instance of this:
typealias CyclicContainer = Container<CyclicContainer>
Swift reports this with a compile error:
Type alias 'CyclicContainer' references itself
… This error is still reported when I change the Container declaration to:
indirect enum Container<T>
This is a bit annoying, because this would be wholesome and compiling Swift:
indirect enum CyclicContainer {
case a(CyclicContainer)
case b(CyclicContainer)
case c
}
Workaround 1
There is a work around for this. I can declare, for example:
indirect enum CyclicContainer {
case container(Container<CyclicContainer>)
}
… However, this becomes awkward for concise descriptions of a CyclicContainer instance:
let cyclicContainer: CyclicContainer = .container(.a(.container(.b(.container(.c)))))
Instead of just:
let cyclicContainer: CyclicContainer = .a(.b(.c))
Workaround 2
I could also simply create a separate enum for the recursive case, as shown earlier:
indirect enum CyclicContainer {
case a(CyclicContainer)
case b(CyclicContainer)
case c
}
However, I'll need to: create functions to convert between CyclicContainer and Container; re-implement various member functions that exist on Container<T>; and keep the two types in sync in future if I add a new case or change a case name.
This isn't terribly arduous and it's the approach I'm leaning towards. But it seems a shame to have to do this when Swift can handle an indirect enum completely happily, but not when its induced by instantiation of a generic argument on the enum.
Question
Is there a better way?
I'm struggling to understand the different between Types and Metatypes in swift 4. In particular I am looking to create an array something like this:
class A { ... }
class B {
func doStuff() {
let otherType = A.self
let itemArr : [otherType] = // Objects of type A
}
}
This throws a compile time error of Use of undeclared type 'otherType' which I think is occurring because otherType is actually A.Type. I think this may have something to do with Generics, but the catch is the type might not be known at compile time...
Is there a solution to this problem?
Swift is powerful because you can pass types as parameters and create variable of types. This is attractive, but yet, there is better solution for your issue.
Define protocol CustomArrayPopulatable {} And add this to all subclasses/classes that could be added to the array. E.G. A1: CustomArrayPopulatable ; A2: CusstomArrayPopulatable.
With generic types, you could achieve great abstraction. However, have in mind, if you need to cast to specific subtype later on, you would need to cast it yourself such as: if type as? A1 {...}, so use generics carefully and think before you start, do you really need them.
As you may know, Swift can infer types from usage. For example, you can have overloaded methods that differ only in return type and freely use them as long as compiler is able to infer type. For example, with help of additional explicitly typed variable that will hold return value of such method.
I've found some funny moments. Imagine this class:
class MyClass {
enum MyError: Error {
case notImplemented
case someException
}
func fun1() throws -> Any {
throw MyError.notImplemented
}
func fun1() -> Int {
return 1
}
func fun2() throws -> Any {
throw MyError.notImplemented
}
func fun2() throws -> Int {
if false {
throw MyError.someException
} else {
return 2
}
}
}
Of course, it will work like:
let myClass = MyClass()
// let resul1 = myClass.fun1() // error: ambiguous use of 'fun1()'
let result1: Int = myClass.fun1() // OK
But next you can write something like:
// print(myClass.fun1()) // error: call can throw but is not marked with 'try'
// BUT
print(try? myClass.fun1()) // warning: no calls to throwing functions occur within 'try' expression
so it looks like mutual exclusive statements. Compiler tries to choose right function; with first call it tries to coerce cast from Int to Any, but what it's trying to do with second one?
Moreover, code like
if let result2 = try? myClass.fun2() { // No warnings
print(result2)
}
will have no warning, so one may assume that compiler is able to choose right overload here (maybe based on fact, that one of the overloads actually returns nothing and only throws).
Am I right with my last assumption? Are warnings for fun1() logical? Do we have some tricks to fool compiler or to help it with type inference?
Obviously you should never, ever write code like this. It's has way too many ways it can bite you, and as you see, it is. But let's see why.
First, try is just a decoration in Swift. It's not for the compiler. It's for you. The compiler works out all the types, and then determines whether a try was necessary. It doesn't use try to figure out the types. You can see this in practice here:
class X {
func x() throws -> X {
return self
}
}
let y = try X().x().x()
You only need try one time, even though there are multiple throwing calls in the chain. Imagine how this would work if you'd created overloads on x() based on throws vs non-throws. The answer is "it doesn't matter" because the compiler doesn't care about the try.
Next there's the issue of type inference vs type coercion. This is type inference:
let resul1 = myClass.fun1() // error: ambiguous use of 'fun1()'
Swift will never infer an ambiguous type. This could be Any or it could beInt`, so it gives up.
This is not type inference (the type is known):
let result1: Int = myClass.fun1() // OK
This also has a known, unambiguous type (note no ?):
let x : Any = try myClass.fun1()
But this requires type coercion (much like your print example)
let x : Any = try? myClass.fun1() // Expression implicitly coerced from `Int?` to `Any`
// No calls to throwing function occur within 'try' expression
Why does this call the Int version? try? return an Optional (which is an Any). So Swift has the option here of an expression that returns Int? and coercing that to Any or Any? and coercing that to Any. Swift pretty much always prefers real types to Any (and it properly hates Any?). This is one of the many reasons to avoid Any in your code. It interacts with Optional in bizarre ways. It's arguable that this should be an error instead, but Any is such a squirrelly type that it's very hard to nail down all its corner cases.
So how does this apply to print? The parameter of print is Any, so this is like the let x: Any =... example rather than like the let x =... example.
A few automatic coercions to keep in mind when thinking about these things:
Every T can be trivially coerced to T?
Every T can be explicitly coerced to Any
Every T? can also be explicitly coerce to Any
Any can be trivially coerced to Any? (also Any??, Any???, and Any????, etc)
Any? (Any??, Any???, etc) can be explicitly coerced to Any
Every non-throwing function can be trivially coerced to a throwing version
So overloading purely on "throws" is dangerous
So mixing throws/non-throws conversions with Any/Any? conversions, and throwing try? into the mix (which promotes everything into an optional), you've created a perfect storm of confusion.
Obviously you should never, ever write code like this.
The Swift compiler always tries to call the most specific overloaded function is there are several overloaded implementations.
The behaviour shown in your question is expected, since any type in Swift can be represented as Any, so even if you type annotate the result value as Any, like let result2: Any = try? myClass.fun1(), the compiler will actually call the implementation of fun1 returning an Int and then cast the return value to Any, since that is the more specific overloaded implementation of fun1.
You can get the compiler to call the version returning Any by casting the return value to Any rather than type annotating it.
let result2 = try? myClass.fun1() as Any //nil, since the function throws an error
This behaviour can be even better observed if you add another overloaded version of fun1 to your class, such as
func fun1() throws -> String {
return ""
}
With fun1 having 3 overloaded versions, the outputs will be the following:
let result1: Int = myClass.fun1() // 1
print(try? myClass.fun1()) //error: ambiguous use of 'fun1()'
let result2: Any = try? myClass.fun1() //error: ambiguous use of 'fun1()'
let stringResult2: String? = try? myClass.fun1() // ""
As you can see, in this example, the compiler simply cannot decide which overloaded version of fun1 to use even if you add the Any type annotation, since the versions returning Int and String are both more specialized versions than the version returning Any, so the version returning Any won't be called, but since both specialized versions would be correct, the compiler cannot decide which one to call.
import Foundation
class A: NSObject {
}
class B: A {
}
let array = [B(),B(),B(),B()]
array as? [A] //this does not works
B() as? A // this works
as? is a conditional downcasting operation (which can fail, and therefore returns an optional) – and you're trying to upcast (which can never fail, and therefore you can do freely).
You therefore can just use as in order to freely upcast the array – or rely on type inference, as Casey notes.
let arrayOfB = [B(),B(),B(),B()]
let arrayOfA = arrayOfB as [A] // cast of [B] to [A] via 'as'
let explicitArrayOfA : [A] = arrayOfB // cast of [B] to [A] via inference of explicit type
func somethingThatExpectsA(a:[A]) {...}
somethingThatExpectsA(arrayOfB) // implicit conversion of [B] to [A]
Although the compiler's error of A is not a subtype of B is certainly unhelpful in the context. A more suitable error (or maybe just a warning) could be 'as?' will always succeed in casting '[B]' to '[A]' – did you mean to use 'as'?.
You should also note that for your example:
B() as? A // this works
You'll get a compiler warning saying Conditional cast from 'B' to 'A' always succeeds. Although why you get a compiler warning, rather than error for this – I can't say. I suspect it's due to the special way in which array casting happens.
Because B is an A, you can drop the conditional cast and
let thing = array as [A]
Or declare the destination as the desired type and forget about casting
let thing : [A] = array