Say I have a struct where all the memebers already are Equatable.
struct S {
let a : String
let b : Int
let c : Double
}
Would anyone know of any way to automatically make S equatable without manually and boringly defining ==?
In Swift 4.1, Types will now synthesize conformance to Equatable and Hashable if all of the Type's members conform to the protocol.
If all of the type's properties are Equatable, simply define your Struct as Equatable and the required equality methods will be synthesized.
struct S: Equatable {
let a : String
let b : Int
let c : Double
}
Proposal: SE-0185
Status: Implemented (Swift 4.1)
Developers have to write large amounts of boilerplate code to support equatability and hashability of complex types. This proposal offers a way for the compiler to automatically synthesize conformance to Equatable and Hashable to reduce this boilerplate, in a subset of scenarios where generating the correct implementation is known to be possible.
https://github.com/apple/swift-evolution/blob/master/proposals/0185-synthesize-equatable-hashable.md
No, you'll have to define it manually:
func == (l: S, r: S) -> Bool { return l.a == r.a && l.b == r.b && l.c == r.c }
That's about as short as you can make it. But it's pretty darn short (one line!), its meaning is clear, and it's ready for future complications you may add to your type. Be happy!
Related
Why doesn't this Swift code compile?
protocol P { }
struct S: P { }
let arr:[P] = [ S() ]
extension Array where Element : P {
func test<T>() -> [T] {
return []
}
}
let result : [S] = arr.test()
The compiler says: "Type P does not conform to protocol P" (or, in later versions of Swift, "Using 'P' as a concrete type conforming to protocol 'P' is not supported.").
Why not? This feels like a hole in the language, somehow. I realize that the problem stems from declaring the array arr as an array of a protocol type, but is that an unreasonable thing to do? I thought protocols were there exactly to help supply structs with something like a type hierarchy?
Why don't protocols conform to themselves?
Allowing protocols to conform to themselves in the general case is unsound. The problem lies with static protocol requirements.
These include:
static methods and properties
Initialisers
Associated types (although these currently prevent the use of a protocol as an actual type)
We can access these requirements on a generic placeholder T where T : P – however we cannot access them on the protocol type itself, as there's no concrete conforming type to forward onto. Therefore we cannot allow T to be P.
Consider what would happen in the following example if we allowed the Array extension to be applicable to [P]:
protocol P {
init()
}
struct S : P {}
struct S1 : P {}
extension Array where Element : P {
mutating func appendNew() {
// If Element is P, we cannot possibly construct a new instance of it, as you cannot
// construct an instance of a protocol.
append(Element())
}
}
var arr: [P] = [S(), S1()]
// error: Using 'P' as a concrete type conforming to protocol 'P' is not supported
arr.appendNew()
We cannot possibly call appendNew() on a [P], because P (the Element) is not a concrete type and therefore cannot be instantiated. It must be called on an array with concrete-typed elements, where that type conforms to P.
It's a similar story with static method and property requirements:
protocol P {
static func foo()
static var bar: Int { get }
}
struct SomeGeneric<T : P> {
func baz() {
// If T is P, what's the value of bar? There isn't one – because there's no
// implementation of bar's getter defined on P itself.
print(T.bar)
T.foo() // If T is P, what method are we calling here?
}
}
// error: Using 'P' as a concrete type conforming to protocol 'P' is not supported
SomeGeneric<P>().baz()
We cannot talk in terms of SomeGeneric<P>. We need concrete implementations of the static protocol requirements (notice how there are no implementations of foo() or bar defined in the above example). Although we can define implementations of these requirements in a P extension, these are defined only for the concrete types that conform to P – you still cannot call them on P itself.
Because of this, Swift just completely disallows us from using a protocol as a type that conforms to itself – because when that protocol has static requirements, it doesn't.
Instance protocol requirements aren't problematic, as you must call them on an actual instance that conforms to the protocol (and therefore must have implemented the requirements). So when calling a requirement on an instance typed as P, we can just forward that call onto the underlying concrete type's implementation of that requirement.
However making special exceptions for the rule in this case could lead to surprising inconsistencies in how protocols are treated by generic code. Although that being said, the situation isn't too dissimilar to associatedtype requirements – which (currently) prevent you from using a protocol as a type. Having a restriction that prevents you from using a protocol as a type that conforms to itself when it has static requirements could be an option for a future version of the language
Edit: And as explored below, this does look like what the Swift team are aiming for.
#objc protocols
And in fact, actually that's exactly how the language treats #objc protocols. When they don't have static requirements, they conform to themselves.
The following compiles just fine:
import Foundation
#objc protocol P {
func foo()
}
class C : P {
func foo() {
print("C's foo called!")
}
}
func baz<T : P>(_ t: T) {
t.foo()
}
let c: P = C()
baz(c)
baz requires that T conforms to P; but we can substitute in P for T because P doesn't have static requirements. If we add a static requirement to P, the example no longer compiles:
import Foundation
#objc protocol P {
static func bar()
func foo()
}
class C : P {
static func bar() {
print("C's bar called")
}
func foo() {
print("C's foo called!")
}
}
func baz<T : P>(_ t: T) {
t.foo()
}
let c: P = C()
baz(c) // error: Cannot invoke 'baz' with an argument list of type '(P)'
So one workaround to to this problem is to make your protocol #objc. Granted, this isn't an ideal workaround in many cases, as it forces your conforming types to be classes, as well as requiring the Obj-C runtime, therefore not making it viable on non-Apple platforms such as Linux.
But I suspect that this limitation is (one of) the primary reasons why the language already implements 'protocol without static requirements conforms to itself' for #objc protocols. Generic code written around them can be significantly simplified by the compiler.
Why? Because #objc protocol-typed values are effectively just class references whose requirements are dispatched using objc_msgSend. On the flip side, non-#objc protocol-typed values are more complicated, as they carry around both value and witness tables in order to both manage the memory of their (potentially indirectly stored) wrapped value and to determine what implementations to call for the different requirements, respectively.
Because of this simplified representation for #objc protocols, a value of such a protocol type P can share the same memory representation as a 'generic value' of type some generic placeholder T : P, presumably making it easy for the Swift team to allow the self-conformance. The same isn't true for non-#objc protocols however as such generic values don't currently carry value or protocol witness tables.
However this feature is intentional and is hopefully to be rolled out to non-#objc protocols, as confirmed by Swift team member Slava Pestov in the comments of SR-55 in response to your query about it (prompted by this question):
Matt Neuburg added a comment - 7 Sep 2017 1:33 PM
This does compile:
#objc protocol P {}
class C: P {}
func process<T: P>(item: T) -> T { return item }
func f(image: P) { let processed: P = process(item:image) }
Adding #objc makes it compile; removing it makes it not compile again.
Some of us over on Stack Overflow find this surprising and would like
to know whether that's deliberate or a buggy edge-case.
Slava Pestov added a comment - 7 Sep 2017 1:53 PM
It's deliberate – lifting this restriction is what this bug is about.
Like I said it's tricky and we don't have any concrete plans yet.
So hopefully it's something that language will one day support for non-#objc protocols as well.
But what current solutions are there for non-#objc protocols?
Implementing extensions with protocol constraints
In Swift 3.1, if you want an extension with a constraint that a given generic placeholder or associated type must be a given protocol type (not just a concrete type that conforms to that protocol) – you can simply define this with an == constraint.
For example, we could write your array extension as:
extension Array where Element == P {
func test<T>() -> [T] {
return []
}
}
let arr: [P] = [S()]
let result: [S] = arr.test()
Of course, this now prevents us from calling it on an array with concrete type elements that conform to P. We could solve this by just defining an additional extension for when Element : P, and just forward onto the == P extension:
extension Array where Element : P {
func test<T>() -> [T] {
return (self as [P]).test()
}
}
let arr = [S()]
let result: [S] = arr.test()
However it's worth noting that this will perform an O(n) conversion of the array to a [P], as each element will have to be boxed in an existential container. If performance is an issue, you can simply solve this by re-implementing the extension method. This isn't an entirely satisfactory solution – hopefully a future version of the language will include a way to express a 'protocol type or conforms to protocol type' constraint.
Prior to Swift 3.1, the most general way of achieving this, as Rob shows in his answer, is to simply build a wrapper type for a [P], which you can then define your extension method(s) on.
Passing a protocol-typed instance to a constrained generic placeholder
Consider the following (contrived, but not uncommon) situation:
protocol P {
var bar: Int { get set }
func foo(str: String)
}
struct S : P {
var bar: Int
func foo(str: String) {/* ... */}
}
func takesConcreteP<T : P>(_ t: T) {/* ... */}
let p: P = S(bar: 5)
// error: Cannot invoke 'takesConcreteP' with an argument list of type '(P)'
takesConcreteP(p)
We cannot pass p to takesConcreteP(_:), as we cannot currently substitute P for a generic placeholder T : P. Let's take a look at a couple of ways in which we can solve this problem.
1. Opening existentials
Rather than attempting to substitute P for T : P, what if we could dig into the underlying concrete type that the P typed value was wrapping and substitute that instead? Unfortunately, this requires a language feature called opening existentials, which currently isn't directly available to users.
However, Swift does implicitly open existentials (protocol-typed values) when accessing members on them (i.e it digs out the runtime type and makes it accessible in the form of a generic placeholder). We can exploit this fact in a protocol extension on P:
extension P {
func callTakesConcreteP/*<Self : P>*/(/*self: Self*/) {
takesConcreteP(self)
}
}
Note the implicit generic Self placeholder that the extension method takes, which is used to type the implicit self parameter – this happens behind the scenes with all protocol extension members. When calling such a method on a protocol typed value P, Swift digs out the underlying concrete type, and uses this to satisfy the Self generic placeholder. This is why we're able to call takesConcreteP(_:) with self – we're satisfying T with Self.
This means that we can now say:
p.callTakesConcreteP()
And takesConcreteP(_:) gets called with its generic placeholder T being satisfied by the underlying concrete type (in this case S). Note that this isn't "protocols conforming to themselves", as we're substituting a concrete type rather than P – try adding a static requirement to the protocol and seeing what happens when you call it from within takesConcreteP(_:).
If Swift continues to disallow protocols from conforming to themselves, the next best alternative would be implicitly opening existentials when attempting to pass them as arguments to parameters of generic type – effectively doing exactly what our protocol extension trampoline did, just without the boilerplate.
However note that opening existentials isn't a general solution to the problem of protocols not conforming to themselves. It doesn't deal with heterogenous collections of protocol-typed values, which may all have different underlying concrete types. For example, consider:
struct Q : P {
var bar: Int
func foo(str: String) {}
}
// The placeholder `T` must be satisfied by a single type
func takesConcreteArrayOfP<T : P>(_ t: [T]) {}
// ...but an array of `P` could have elements of different underlying concrete types.
let array: [P] = [S(bar: 1), Q(bar: 2)]
// So there's no sensible concrete type we can substitute for `T`.
takesConcreteArrayOfP(array)
For the same reasons, a function with multiple T parameters would also be problematic, as the parameters must take arguments of the same type – however if we have two P values, there's no way we can guarantee at compile time that they both have the same underlying concrete type.
In order to solve this problem, we can use a type eraser.
2. Build a type eraser
As Rob says, a type eraser, is the most general solution to the problem of protocols not conforming to themselves. They allow us to wrap a protocol-typed instance in a concrete type that conforms to that protocol, by forwarding the instance requirements to the underlying instance.
So, let's build a type erasing box that forwards P's instance requirements onto an underlying arbitrary instance that conforms to P:
struct AnyP : P {
private var base: P
init(_ base: P) {
self.base = base
}
var bar: Int {
get { return base.bar }
set { base.bar = newValue }
}
func foo(str: String) { base.foo(str: str) }
}
Now we can just talk in terms of AnyP instead of P:
let p = AnyP(S(bar: 5))
takesConcreteP(p)
// example from #1...
let array = [AnyP(S(bar: 1)), AnyP(Q(bar: 2))]
takesConcreteArrayOfP(array)
Now, consider for a moment just why we had to build that box. As we discussed early, Swift needs a concrete type for cases where the protocol has static requirements. Consider if P had a static requirement – we would have needed to implement that in AnyP. But what should it have been implemented as? We're dealing with arbitrary instances that conform to P here – we don't know about how their underlying concrete types implement the static requirements, therefore we cannot meaningfully express this in AnyP.
Therefore, the solution in this case is only really useful in the case of instance protocol requirements. In the general case, we still cannot treat P as a concrete type that conforms to P.
EDIT: Eighteen more months of working w/ Swift, another major release (that provides a new diagnostic), and a comment from #AyBayBay makes me want to rewrite this answer. The new diagnostic is:
"Using 'P' as a concrete type conforming to protocol 'P' is not supported."
That actually makes this whole thing a lot clearer. This extension:
extension Array where Element : P {
doesn't apply when Element == P since P is not considered a concrete conformance of P. (The "put it in a box" solution below is still the most general solution.)
Old Answer:
It's yet another case of metatypes. Swift really wants you to get to a concrete type for most non-trivial things. [P] isn't a concrete type (you can't allocate a block of memory of known size for P). (I don't think that's actually true; you can absolutely create something of size P because it's done via indirection.) I don't think there's any evidence that this is a case of "shouldn't" work. This looks very much like one of their "doesn't work yet" cases. (Unfortunately it's almost impossible to get Apple to confirm the difference between those cases.) The fact that Array<P> can be a variable type (where Array cannot) indicates that they've already done some work in this direction, but Swift metatypes have lots of sharp edges and unimplemented cases. I don't think you're going to get a better "why" answer than that. "Because the compiler doesn't allow it." (Unsatisfying, I know. My whole Swift life…)
The solution is almost always to put things in a box. We build a type-eraser.
protocol P { }
struct S: P { }
struct AnyPArray {
var array: [P]
init(_ array:[P]) { self.array = array }
}
extension AnyPArray {
func test<T>() -> [T] {
return []
}
}
let arr = AnyPArray([S()])
let result: [S] = arr.test()
When Swift allows you to do this directly (which I do expect eventually), it will likely just be by creating this box for you automatically. Recursive enums had exactly this history. You had to box them and it was incredibly annoying and restricting, and then finally the compiler added indirect to do the same thing more automatically.
If you extend CollectionType protocol instead of Array and constraint by protocol as a concrete type, you can rewrite the previous code as follows.
protocol P { }
struct S: P { }
let arr:[P] = [ S() ]
extension CollectionType where Generator.Element == P {
func test<T>() -> [T] {
return []
}
}
let result : [S] = arr.test()
Why doesn't this Swift code compile?
protocol P { }
struct S: P { }
let arr:[P] = [ S() ]
extension Array where Element : P {
func test<T>() -> [T] {
return []
}
}
let result : [S] = arr.test()
The compiler says: "Type P does not conform to protocol P" (or, in later versions of Swift, "Using 'P' as a concrete type conforming to protocol 'P' is not supported.").
Why not? This feels like a hole in the language, somehow. I realize that the problem stems from declaring the array arr as an array of a protocol type, but is that an unreasonable thing to do? I thought protocols were there exactly to help supply structs with something like a type hierarchy?
Why don't protocols conform to themselves?
Allowing protocols to conform to themselves in the general case is unsound. The problem lies with static protocol requirements.
These include:
static methods and properties
Initialisers
Associated types (although these currently prevent the use of a protocol as an actual type)
We can access these requirements on a generic placeholder T where T : P – however we cannot access them on the protocol type itself, as there's no concrete conforming type to forward onto. Therefore we cannot allow T to be P.
Consider what would happen in the following example if we allowed the Array extension to be applicable to [P]:
protocol P {
init()
}
struct S : P {}
struct S1 : P {}
extension Array where Element : P {
mutating func appendNew() {
// If Element is P, we cannot possibly construct a new instance of it, as you cannot
// construct an instance of a protocol.
append(Element())
}
}
var arr: [P] = [S(), S1()]
// error: Using 'P' as a concrete type conforming to protocol 'P' is not supported
arr.appendNew()
We cannot possibly call appendNew() on a [P], because P (the Element) is not a concrete type and therefore cannot be instantiated. It must be called on an array with concrete-typed elements, where that type conforms to P.
It's a similar story with static method and property requirements:
protocol P {
static func foo()
static var bar: Int { get }
}
struct SomeGeneric<T : P> {
func baz() {
// If T is P, what's the value of bar? There isn't one – because there's no
// implementation of bar's getter defined on P itself.
print(T.bar)
T.foo() // If T is P, what method are we calling here?
}
}
// error: Using 'P' as a concrete type conforming to protocol 'P' is not supported
SomeGeneric<P>().baz()
We cannot talk in terms of SomeGeneric<P>. We need concrete implementations of the static protocol requirements (notice how there are no implementations of foo() or bar defined in the above example). Although we can define implementations of these requirements in a P extension, these are defined only for the concrete types that conform to P – you still cannot call them on P itself.
Because of this, Swift just completely disallows us from using a protocol as a type that conforms to itself – because when that protocol has static requirements, it doesn't.
Instance protocol requirements aren't problematic, as you must call them on an actual instance that conforms to the protocol (and therefore must have implemented the requirements). So when calling a requirement on an instance typed as P, we can just forward that call onto the underlying concrete type's implementation of that requirement.
However making special exceptions for the rule in this case could lead to surprising inconsistencies in how protocols are treated by generic code. Although that being said, the situation isn't too dissimilar to associatedtype requirements – which (currently) prevent you from using a protocol as a type. Having a restriction that prevents you from using a protocol as a type that conforms to itself when it has static requirements could be an option for a future version of the language
Edit: And as explored below, this does look like what the Swift team are aiming for.
#objc protocols
And in fact, actually that's exactly how the language treats #objc protocols. When they don't have static requirements, they conform to themselves.
The following compiles just fine:
import Foundation
#objc protocol P {
func foo()
}
class C : P {
func foo() {
print("C's foo called!")
}
}
func baz<T : P>(_ t: T) {
t.foo()
}
let c: P = C()
baz(c)
baz requires that T conforms to P; but we can substitute in P for T because P doesn't have static requirements. If we add a static requirement to P, the example no longer compiles:
import Foundation
#objc protocol P {
static func bar()
func foo()
}
class C : P {
static func bar() {
print("C's bar called")
}
func foo() {
print("C's foo called!")
}
}
func baz<T : P>(_ t: T) {
t.foo()
}
let c: P = C()
baz(c) // error: Cannot invoke 'baz' with an argument list of type '(P)'
So one workaround to to this problem is to make your protocol #objc. Granted, this isn't an ideal workaround in many cases, as it forces your conforming types to be classes, as well as requiring the Obj-C runtime, therefore not making it viable on non-Apple platforms such as Linux.
But I suspect that this limitation is (one of) the primary reasons why the language already implements 'protocol without static requirements conforms to itself' for #objc protocols. Generic code written around them can be significantly simplified by the compiler.
Why? Because #objc protocol-typed values are effectively just class references whose requirements are dispatched using objc_msgSend. On the flip side, non-#objc protocol-typed values are more complicated, as they carry around both value and witness tables in order to both manage the memory of their (potentially indirectly stored) wrapped value and to determine what implementations to call for the different requirements, respectively.
Because of this simplified representation for #objc protocols, a value of such a protocol type P can share the same memory representation as a 'generic value' of type some generic placeholder T : P, presumably making it easy for the Swift team to allow the self-conformance. The same isn't true for non-#objc protocols however as such generic values don't currently carry value or protocol witness tables.
However this feature is intentional and is hopefully to be rolled out to non-#objc protocols, as confirmed by Swift team member Slava Pestov in the comments of SR-55 in response to your query about it (prompted by this question):
Matt Neuburg added a comment - 7 Sep 2017 1:33 PM
This does compile:
#objc protocol P {}
class C: P {}
func process<T: P>(item: T) -> T { return item }
func f(image: P) { let processed: P = process(item:image) }
Adding #objc makes it compile; removing it makes it not compile again.
Some of us over on Stack Overflow find this surprising and would like
to know whether that's deliberate or a buggy edge-case.
Slava Pestov added a comment - 7 Sep 2017 1:53 PM
It's deliberate – lifting this restriction is what this bug is about.
Like I said it's tricky and we don't have any concrete plans yet.
So hopefully it's something that language will one day support for non-#objc protocols as well.
But what current solutions are there for non-#objc protocols?
Implementing extensions with protocol constraints
In Swift 3.1, if you want an extension with a constraint that a given generic placeholder or associated type must be a given protocol type (not just a concrete type that conforms to that protocol) – you can simply define this with an == constraint.
For example, we could write your array extension as:
extension Array where Element == P {
func test<T>() -> [T] {
return []
}
}
let arr: [P] = [S()]
let result: [S] = arr.test()
Of course, this now prevents us from calling it on an array with concrete type elements that conform to P. We could solve this by just defining an additional extension for when Element : P, and just forward onto the == P extension:
extension Array where Element : P {
func test<T>() -> [T] {
return (self as [P]).test()
}
}
let arr = [S()]
let result: [S] = arr.test()
However it's worth noting that this will perform an O(n) conversion of the array to a [P], as each element will have to be boxed in an existential container. If performance is an issue, you can simply solve this by re-implementing the extension method. This isn't an entirely satisfactory solution – hopefully a future version of the language will include a way to express a 'protocol type or conforms to protocol type' constraint.
Prior to Swift 3.1, the most general way of achieving this, as Rob shows in his answer, is to simply build a wrapper type for a [P], which you can then define your extension method(s) on.
Passing a protocol-typed instance to a constrained generic placeholder
Consider the following (contrived, but not uncommon) situation:
protocol P {
var bar: Int { get set }
func foo(str: String)
}
struct S : P {
var bar: Int
func foo(str: String) {/* ... */}
}
func takesConcreteP<T : P>(_ t: T) {/* ... */}
let p: P = S(bar: 5)
// error: Cannot invoke 'takesConcreteP' with an argument list of type '(P)'
takesConcreteP(p)
We cannot pass p to takesConcreteP(_:), as we cannot currently substitute P for a generic placeholder T : P. Let's take a look at a couple of ways in which we can solve this problem.
1. Opening existentials
Rather than attempting to substitute P for T : P, what if we could dig into the underlying concrete type that the P typed value was wrapping and substitute that instead? Unfortunately, this requires a language feature called opening existentials, which currently isn't directly available to users.
However, Swift does implicitly open existentials (protocol-typed values) when accessing members on them (i.e it digs out the runtime type and makes it accessible in the form of a generic placeholder). We can exploit this fact in a protocol extension on P:
extension P {
func callTakesConcreteP/*<Self : P>*/(/*self: Self*/) {
takesConcreteP(self)
}
}
Note the implicit generic Self placeholder that the extension method takes, which is used to type the implicit self parameter – this happens behind the scenes with all protocol extension members. When calling such a method on a protocol typed value P, Swift digs out the underlying concrete type, and uses this to satisfy the Self generic placeholder. This is why we're able to call takesConcreteP(_:) with self – we're satisfying T with Self.
This means that we can now say:
p.callTakesConcreteP()
And takesConcreteP(_:) gets called with its generic placeholder T being satisfied by the underlying concrete type (in this case S). Note that this isn't "protocols conforming to themselves", as we're substituting a concrete type rather than P – try adding a static requirement to the protocol and seeing what happens when you call it from within takesConcreteP(_:).
If Swift continues to disallow protocols from conforming to themselves, the next best alternative would be implicitly opening existentials when attempting to pass them as arguments to parameters of generic type – effectively doing exactly what our protocol extension trampoline did, just without the boilerplate.
However note that opening existentials isn't a general solution to the problem of protocols not conforming to themselves. It doesn't deal with heterogenous collections of protocol-typed values, which may all have different underlying concrete types. For example, consider:
struct Q : P {
var bar: Int
func foo(str: String) {}
}
// The placeholder `T` must be satisfied by a single type
func takesConcreteArrayOfP<T : P>(_ t: [T]) {}
// ...but an array of `P` could have elements of different underlying concrete types.
let array: [P] = [S(bar: 1), Q(bar: 2)]
// So there's no sensible concrete type we can substitute for `T`.
takesConcreteArrayOfP(array)
For the same reasons, a function with multiple T parameters would also be problematic, as the parameters must take arguments of the same type – however if we have two P values, there's no way we can guarantee at compile time that they both have the same underlying concrete type.
In order to solve this problem, we can use a type eraser.
2. Build a type eraser
As Rob says, a type eraser, is the most general solution to the problem of protocols not conforming to themselves. They allow us to wrap a protocol-typed instance in a concrete type that conforms to that protocol, by forwarding the instance requirements to the underlying instance.
So, let's build a type erasing box that forwards P's instance requirements onto an underlying arbitrary instance that conforms to P:
struct AnyP : P {
private var base: P
init(_ base: P) {
self.base = base
}
var bar: Int {
get { return base.bar }
set { base.bar = newValue }
}
func foo(str: String) { base.foo(str: str) }
}
Now we can just talk in terms of AnyP instead of P:
let p = AnyP(S(bar: 5))
takesConcreteP(p)
// example from #1...
let array = [AnyP(S(bar: 1)), AnyP(Q(bar: 2))]
takesConcreteArrayOfP(array)
Now, consider for a moment just why we had to build that box. As we discussed early, Swift needs a concrete type for cases where the protocol has static requirements. Consider if P had a static requirement – we would have needed to implement that in AnyP. But what should it have been implemented as? We're dealing with arbitrary instances that conform to P here – we don't know about how their underlying concrete types implement the static requirements, therefore we cannot meaningfully express this in AnyP.
Therefore, the solution in this case is only really useful in the case of instance protocol requirements. In the general case, we still cannot treat P as a concrete type that conforms to P.
EDIT: Eighteen more months of working w/ Swift, another major release (that provides a new diagnostic), and a comment from #AyBayBay makes me want to rewrite this answer. The new diagnostic is:
"Using 'P' as a concrete type conforming to protocol 'P' is not supported."
That actually makes this whole thing a lot clearer. This extension:
extension Array where Element : P {
doesn't apply when Element == P since P is not considered a concrete conformance of P. (The "put it in a box" solution below is still the most general solution.)
Old Answer:
It's yet another case of metatypes. Swift really wants you to get to a concrete type for most non-trivial things. [P] isn't a concrete type (you can't allocate a block of memory of known size for P). (I don't think that's actually true; you can absolutely create something of size P because it's done via indirection.) I don't think there's any evidence that this is a case of "shouldn't" work. This looks very much like one of their "doesn't work yet" cases. (Unfortunately it's almost impossible to get Apple to confirm the difference between those cases.) The fact that Array<P> can be a variable type (where Array cannot) indicates that they've already done some work in this direction, but Swift metatypes have lots of sharp edges and unimplemented cases. I don't think you're going to get a better "why" answer than that. "Because the compiler doesn't allow it." (Unsatisfying, I know. My whole Swift life…)
The solution is almost always to put things in a box. We build a type-eraser.
protocol P { }
struct S: P { }
struct AnyPArray {
var array: [P]
init(_ array:[P]) { self.array = array }
}
extension AnyPArray {
func test<T>() -> [T] {
return []
}
}
let arr = AnyPArray([S()])
let result: [S] = arr.test()
When Swift allows you to do this directly (which I do expect eventually), it will likely just be by creating this box for you automatically. Recursive enums had exactly this history. You had to box them and it was incredibly annoying and restricting, and then finally the compiler added indirect to do the same thing more automatically.
If you extend CollectionType protocol instead of Array and constraint by protocol as a concrete type, you can rewrite the previous code as follows.
protocol P { }
struct S: P { }
let arr:[P] = [ S() ]
extension CollectionType where Generator.Element == P {
func test<T>() -> [T] {
return []
}
}
let result : [S] = arr.test()
Say I have two variables, and I don't know what type they are (and it's not possible to know what type they are until runtime):
var a: Any
var b: Any
How can I test if they are equal, using the Equatable protocol? I can't just do a == b because that requires that both of the items are the same, Equatable type, and the compiler can't prove that because they could be different types (and one or both might not even be Equatable).
So, is it possible to tell the compiler to check if they both have the same type, and if that type conforms to Equatable, then to use the == operator on them and return the result, otherwise returning false?
If there is no way to do this, is there a good reason Swift prevents this, or is it a current limitation of Swift that could be fixed in the future?
Under the assumption that if a and b differ in type then they are never equal, you can use a generic function with constraint to achieve the goal.
func isEqual<T : Equatable>(a: T, b: T) -> Bool {
return a == b;
}
You cannot have a and b differ in type, as the Equatable protocol assumes that the LHS and RHS of the comparison are of the same type. This seems a reasonable constraint, but one can certainly write a notion of equality that doesn't require this. In these cases, you'll need your own equality protocol.
You use generics and function overload:
func isEqual<T: Equatable>(a: T, b: T) -> Bool {
return a == b
}
func isEqual<T, U>(a: T, b: U) -> Bool {
return false
}
If both variables have the same type, as that type conforms to Equatable, then the compiler will choose the first function, otherwise will go the second one.
This will work for Objective-C objects too, providing you cast to NSObject before calling the function: isEqual(var1 as? NSObject, var2 as? NSObject)
So I want to equate two dictionaries of the same type [AnyLanguage: SortedList<String>].
AnyLanguage conforms to Hashable, and therefore also Equatable, so it's not a problem. SortedList on the other hand does not conform to Equatable, as it is generic over Element, which has no restrictions:
public struct SortedList<Element> { ... }
Now, I have implemented the equality operator for SortedList, when its Elements conform to Equatable:
public func ==<T: Equatable>(left: SortedList<T>, right: SortedList<T>) -> Bool
Now, when comparing two of the aforementioned dictionaries:
let dictA: [AnyLanguage: SortedList<String>] = [:]
let dictB: [AnyLanguage: SortedList<String>] = [:]
dictA == dictB
I get the following error:
I assume that the compiler is complaining because SortedList does not conform to Equatable, even though I have implemented the comparison operator for equatable Elements (which SortedList<String> would be able to use, as String is equatable).
How can I compare the two dictionaries?
The == operator for dictionaries requires that both Key and Value
type conform to Equatable.
You have implemented a (restricted) == operator for SortedList<Element>.
But implementing a == operator for a type does not make that type
Equatable automatically. The conformance must be declared explicitly.
Unfortunately, it is (currently) not possible to make SortedList<Element> conform to Equatable only if Element
is Equatable. The same problem exists for arrays, compare
Why can't I make Array conform to Equatable? for a discussion
in the Apple developer forum.
The only solutions – as far as I know – are to make SortedList<Element>
conform to Equatable unconditionally (as in Cristik's answer), or to define
a custom comparison operator for dictionaries
func ==<Key : Equatable, T : Equatable>(lhs: [Key : SortedList<T>], rhs: [Key : SortedList<T>]) -> Bool {
// ...
}
SortedList is not Equatable, so Dictionary can't compare the elements. You need to declare as Equatable, and implement == without the generic constraint. This will work:
public struct SortedList<Element>: Equatable { ... }
public func ==<T>(left: SortedList<T>, right: SortedList<T>) -> Bool
If you want to restrict your SortedList to Equatable elements, you can declare it to accept only stuff that conforms to Equatable
public struct SortedList<Element: Equatable>: Equatable
however for your SortedList I think Comparable would suit better.
I’m trying to mix generics with protocols and I’m getting a really hard time xD
I have certain architecture implemented in an Android/Java project, and I’m trying to rewrite it to fit it in a swift/iOS project. But I’ve found this limitation.
ProtocolA
protocol ProtocolA {
}
ProtocolB
protocol ProtocolB : ProtocolA {
}
ImplementProtocolA
class ImplementProtocolA <P : ProtocolA> {
let currentProtocol : P
init(currentProtocol : P) {
self.currentProtocol = currentProtocol
}
}
ImplementProtocolB
class ImplementProtocolB : ImplementProtocolA<ProtocolB> {
}
So, when I try to set ProtocolB as the concrete type that implements ProtocolA, I get this error:
Using 'ProtocolB' as a concrete type conforming to protocol 'ProtocolA' is not supported
1 Is there any reason for this “limitation”?
2 Is there any workaround to get this implemented?
3 Will it be supported at some point?
--UPDATED--
Another variant of the same problem, I think:
View protocols
protocol View {
}
protocol GetUserView : View {
func showProgress()
func hideProgress()
func showError(message:String)
func showUser(userDemo:UserDemo)
}
Presenter protocols
protocol Presenter {
typealias V : View
}
class UserDemoPresenter : Presenter {
typealias V = GetUserView
}
Error:
UserDemoPresenter.swift Possibly intended match 'V' (aka
'GetUserView') does not conform to 'View’
What is that?? It conforms!
Even if I use View instead of GetUserView, it does not compile.
class UserDemoPresenter : Presenter {
typealias V = View
}
UserDemoPresenter.swift Possibly intended match 'V' (aka 'View') does
not conform to 'View'
xxDD I don’t get it, really.
--UPDATED--
With the solution proposed by Rob Napier the problem is not fixed, instead, it is just delayed.
When a try to define a reference to UserDemoPresenter, I need to specify the generic type, so I get the same error:
private var presenter : UserDemoPresenter<GetUserView>
Using 'GetUserView' as a concrete type conforming to protocol
'GetUserView' is not supported
The underlying reason for the limitation is that Swift doesn't have first-class metatypes. The simplest example is that this doesn't work:
func isEmpty(xs: Array) -> Bool {
return xs.count == 0
}
In theory, this code could work, and if it did there would be a lot of other types I could make (like Functor and Monad, which really can't be expressed in Swift today). But you can't. You need to help Swift nail this down to a concrete type. Often we do that with generics:
func isEmpty<T>(xs: [T]) -> Bool {
return xs.count == 0
}
Notice that T is totally redundant here. There is no reason I should have to express it; it's never used. But Swift requires it so it can turn the abstract Array into the concrete [T]. The same is true in your case.
This is a concrete type (well, it's an abstract type that will be turned into a concrete type any time it's instantiated and P is filled in):
class ImplementProtocolA<P : ProtocolA>
This is a fully abstract type that Swift doesn't have any rule to turn into a concrete type:
class ImplementProtocolB : ImplementProtocolA<ProtocolB>
You need to make it concrete. This will compile:
class ImplementProtocolB<T: ProtocolB> : ImplementProtocolA<T> {}
And also:
class UserDemoPresenter<T: GetUserView> : Presenter {
typealias V = T
}
Just because you're likely to run into the issue later: your life will go much easier if you'll make these structs or final classes. Mixing protocols, generics, and class polymorphism is full of very sharp edges. Sometimes you're lucky and it just won't compile. Sometimes it will call things you don't expect.
You may be interested in A Little Respect for AnySequence which details some related issues.
private var presenter : UserDemoPresenter<GetUserView>
This is still an abstract type. You mean:
final class Something<T: GetUserView> {
private var presenter: UserDemoPresenter<T>
}
If that creates a problem, you'll need to create a box. See Protocol doesn't conform to itself? for discussion of how you type-erase so that you can hold abstract types. But you need to work in concrete types. You can't ultimately specialize on a protocol. You must eventually specialize on something concrete in the majority of cases.