This is a valid protocol declaration in Swift:
protocol Proto1: class {
func method1() -> Self
func method2() -> [Proto1]
}
But this isn't:
protocol Proto2: class {
func method1(param: Self)
func method2() -> [Proto2]
}
The error message is:
Protocol 'Proto2' can only be used as a generic constraint because it
has Self or associated type requirements
So it seems that, when using Self as function's return type, Swift doesn't consider that as a constraint of the protocol being defined and hence it's OK to use the protocol itself as functions' return type. But when using Self as function's parameter type, the behavior is completely different.
I wonder why such a difference?
Because it's not useful to have a Self as a parameter type of a method. Suppose that you can do:
protocol P {
func f(_ x: Self)
}
class A: P {
func f(_ x: Self) {
// not relevant
}
}
class B: A { }
Now suppose I have:
func g(x: A) {
// what can I do with x?
}
The fact is, there is no way to call x.f. Because I can pass an instance of B to x, in which case x.f would accept a B instead. x could be an instance of any subclass of A that I have no way of knowing at compile time, so I don't know what I can pass to x.f.
Compare that to Self used as the return type:
protocol P {
func f() -> Self
}
// Implementation:
class A: P {
func f() -> Self {
self
}
}
class B: A { }
func g(x: A) {
let foo: A = x.f()
}
Here, we know that I can at least assign the return value of x.f to a variable of type A. Even if x is an instance of B, which means that f returns a B, we can still assign that to a variable of type A.
how to return protocol with associated type?
protocol AProtocol {
}
class A: AProtocol {
}
class Main {
func sendA() -> AProtocol {
return A()
}
}
it works.
but
protocol BProtocol {
associatedtype B
}
class B: BProtocol {
typealias B = Int
}
class Main {
func sendA() -> AProtocol {
return A()
}
func sendB() -> BProtocol { // error
return B()
}
// function1
func sendB_<T: BProtocol>() -> T{
return B() as! T
}
}
i want to return 'return B()' in function 1
is it possible?
You are trying to return BProtocol in case 1. The thing is PAT (Protocol with Associated Types) are not exactly types. They act as sort of placeholder for types. So you cannot return BProtocol directly.
Swift 5.1
I am not 100% sure but I think in the next iteration of swift (5.1), they have introduced opaque types which allow the functionality you desire.
In that case you would be able to call it like this:
func sendB() -> some BProtocol {
return B()
}
In this function
func sendB_<T: BProtocol>() -> T{
return B() as! T
}
you cannot return a B as a T because the person who uses the function defines what T is, not you and T can be any type that conforms to Protocol For example, I could do this:
class C: BProtocol
{
typealias B = Float
}
let c: C = Main().sendB_()
By doing that, I am setting T to be a C and the forced typecast in sendB_() will fail.
Unfortunately, protocols with associated types cannot, themselves, be treated like a concrete type, so the approach you took with AProtocol will not work.
As I see it, you have two options. Change your function return type to B. After all, you always do return a B
func sendB_() -> B {
return B()
}
If you want to keep it generic, try
protocol BProtocol
{
associatedtype B
init() // Needed to be able to do T() in generic function
}
func sendB_<T: BProtocol>() -> T{
return T()
}
You need to add the initialiser to the protocol to make sure that an instance of type T always exists.
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()
Is it possible to return a type of generic that conforms to protocol for later use with class functions using Swift 1.2? Take a look:
protocol SomeProtocol
{
static func start(kind: Kind)
}
class A: SomeProtocol
{
class func start(kind: Kind)
{
print("A started")
}
}
class B: SomeProtocol
{
class func start(kind: Kind)
{
print("B started")
}
}
enum Kind {
case Akind
case Bkind
private func classKind<T: SomeProtocol>() -> T.Type
{
switch self {
case .Akind: return A.self
case .Bkind: return B.self
}
}
func doSomething() {
self.classKind().start(self)
}
}
I tried various methods but every of them ended with some errors. Currently I got 'A' is not a subtype of 'T' in classKind method (same for 'B') and cannot invoke 'start' with an argument list of type '(Kind)' in doSomething.
I'm sure I'm pretty close but can't solve it...
If you're using Swift 2, to achieve what you want you only need to change:
private func classKind<T: SomeProtocol>() -> T.Type { ... }
to
private func classKind() -> SomeProtocol.Type { ... }
Now back to the not-working code to see where the errors were coming from. You don't need to make the changes I'm now detailing, this is just to explain the errors.
First examine your doSomething method:
func doSomething() {
self.classKind().start(self)
// Error: Argument for generic parameter 'T' could not be inferred.
//
// (I'm using Xcode 7 b6, which may explain the differing error messages)
}
For the type returned by classKind to be inferred, you'd have to do:
let type: A.Type = self.classKind() // Or you could use `B.Type`.
type.start(self)
Which obviously defeats the point of your goal, since you have to specify the type you want.
Secondly, the errors in classKind:
private func classKind<T: SomeProtocol>() -> T.Type
{
switch self {
case .Akind: return A.self
// Cannot convert return expression of type 'A.Type' to return type 'T.Type'.
case .Bkind: return B.self
// Cannot convert return expression of type 'B.Type' to return type 'T.Type'.
}
}
To see why this doesn't work consider the following example, in which I have another type that conforms to SomeProtocol:
struct C: SomeProtocol { ... }
Then in doSomething:
func doSomething() {
let type: C.Type = self.classKind()
type.start(self)
}
The errors you're getting can now be read as: Cannot convert return expression of type 'A.Type'/'B.Type' to return type 'C.Type'.
I'm working with an Observer API (ObserverSet) which have the following function :
public func add<T: AnyObject>(object: T, _ f: T -> Parameters -> Void) -> ObserverSetEntry<Parameters>
It simply register an object then call the instance method f on the object when notification triggers
In one of my manager, I need to hide the previous function with one of mine so I can force an observer to call a predefine function implemented via a protocol.
Here's what I've done so far :
#objc protocol Observer : NSObjectProtocol {
func observe(param: String) -> Void
}
func addObserver<T: AnyObject where T: Observer>(observer: T) {
let f: T -> String -> Void = observer.dynamicType.observe
entries.addObserver(observer, f)
}
Unfortunately, I have the following error showing up Partial application of generic method is not allowed
I've found a possible workaround somewhere on SO which look like that :
let f: T -> String -> Void = { (obs: T) in obs.dynamicType.observe(obs) }
But this line of code drives my XCode crazy with some Segmentation Fault: 11 on compilation (and Communication interrupted with Playground ..)
Is there any workaround for what I'm trying to do ?
I haven't tested but you can try:
#objc protocol Observer : NSObjectProtocol {
func observe(param: String) -> Void
}
func addObserver<T: AnyObject where T: Observer>(observer: T) {
let f: T -> String -> Void = { ($0 as AnyObject).observe }
entries.addObserver(observer, f)
}
At least, this compiles because AnyObject has all methods from ObjC - including #objc - classes/protocols, as ImplicitlyUnwrappedOptional.
So, this compiles:
let str = NSString(string: "test")
(str as AnyObject).observe("foo")
Of course this causes runtime error because NSString has no observe(_:) method. But, in your case, T is guaranteed to be Observer, it should works.