I'm trying to mark a variable of a generic type as weak:
class X<T> {
weak var t: T?
}
If I don't put in any constraints for T I get the error weak cannot be applied to non-class type 'T'.
If I would only use use this with NSObject derived classes this would work:
class X<T: NSObject> {
weak var t: T?
}
But I also want to be able to use pure Swift classes.
For protocols it is possible to require that the implementor is of class type by using the class keyword:
protocol ClassType: class {
}
With the ClassType protocol I can now mark the variable as weak:
class X<T: ClassType> {
weak var t: T?
}
But I cannot add the class keyword directly to the generic parameter:
class X<T: class> { // Compile error
weak var t: T?
}
I can make the protocol solution work for all NSObject derived classes with an extension:
extension NSObject: ClassType {
}
But for pure Swift classes there is no common superclass that I could add this extension to. Is there a way to make this work without adding the ClassType protocol to each class I want to use the X class with? E.g. some special qualifier for the generic parameter like class X<T:ThisIsAClass>?
You want AnyObject, which the Swift docs describe as:
The protocol to which all classes implicitly conform.
class X<T: AnyObject> {
weak var t: T?
}
class C { }
let x = X<C>() // works fine
x.t = C()
// error: type 'Int' does not conform to protocol ‘AnyObject’
// (because Int is a struct)
let y = X<Int>()
Related
This code fails to compile with Swift 5.1
import Foundation
protocol SomeClassProtocol: AnyObject {}
class SomeClass: SomeClassProtocol {}
class GenericClass<T:AnyObject> {
weak var t: T?
init(t: T) {
self.t = t
}
}
let test = GenericClass<SomeClassProtocol>(t: SomeClass())
The error is
'GenericClass' requires that 'SomeClassProtocol' be a class type
Does the compiler really need a class type here instead of a class-only protocol?
Yes. Though the syntax is the same (a colon), protocol inheritance is not the same thing as protocol conformance. Protocols do not conform to protocols; only types can conform to protocols. (AnyObject is not special in this regard; your question is good but the title isn't getting at the issue.)
In your example:
Your T needs to conform to AnyObject. SomeClassProtocol does not conform to AnyObject. But any types that conform to SomeClassProtocol will conform to AnyObject.
So you need to pick which of these you really want:
1.
let test = GenericClass( t: SomeClass() )
(test is a GenericClass<SomeClass>.)
2.
class Class {
weak var object: AnyObject?
init(object: AnyObject) {
self.object = object
}
}
Class( object: SomeClass() )
You do have the option of subclassing, if that would be useful.
class GenericClass<T: AnyObject>: Class {
var t: T? {
get { object as? T }
set { object = newValue }
}
}
Does the compiler really need a class type here instead of a class-only protocol?
Yes, it does. I think the problem is just understanding what this means:
class GenericClass<T:AnyObject>
That means: "To resolve GenericClass, the parameterized type T must be some type that is a class." Examples would be UIView, NSString, etc.
Okay, so:
let test = GenericClass<SomeClassProtocol>(t: SomeClass())
So, SomeClassProtocol is none of those; it isn't the name of a class. It's the name of a protocol.
A further difficulty may be understanding protocols as types. They are not really full-fledged types.
You don't need to specify T explicitly.
Change your code from this:
let test = GenericClass<SomeClassProtocol>(t: SomeClass())
To this:
let test = GenericClass(t: SomeClass())
I found a strange Swift compiler message while developing (I'm using Swift 4.1):
protocol Foo: class where Self: NSObject { // (1)
// Redundant constraint 'Self' : 'AnyObject'
}
What is happening here?
First, this is not redundant. When I write
protocol Foo: class { } // (2)
I have a protocol that any object could potentially conform to, even objects that don't derive from NSObject. But I can create weak references: weak var f: Foo? is okay.
On the other hand, when I write
protocol Foo where Self: NSObject { } // (3)
I have a protocol where I cannot produce weak references: weak var f: Foo? is a compile time error.
Second, where does the AnyObject come from? I'm asking for NSObject. The NSObject is respected though: I cannot declare a class MyFoo: Foo { } because it rightly complains that it must inherit from NSObject?
Is this a bug in Swift? Or am I missing something? If it is a bug: is it a bug because snippet (3) does not let me take weak references? Or because of the compiler warning? Or both? And if I am missing something: what is it?
It is not possible to constrain a protocol to be subclasses of a specific class in Swift 4.1. You can inherit Foo from NSObjectProtocol, which probably matches your intent.
protocol Foo: NSObjectProtocol {
// ....
}
In Swift 4.2, what you've written is legal Swift and does what you expect.
From the Swift public API:
public typealias AnyObject
/// The protocol to which all class types implicitly conform.
So by declaring your protocol to conform to class it automatically conforms to AnyObject
I have this kind of code:
protocol MyProtocol {
}
class P1: MyProtocol {
}
class P2: MyProtocol {
}
class C <T: MyProtocol> {
}
Then i need to define a variable to delegate all kinds of C<MyProtocol>:
var obj: C <MyProtocol>
But compile error comes:
Using 'MyProtocol' as a concrete type conforming to protocol 'MyProtocol' is not supported
How can I do?
This code:
class C <T: MyProtocol> { }
Means that C is a generic class that can specialize on any type T that conforms to MyProtocol.
When you declare:
var obj: C <MyProtocol>
You are (I think) trying to say that the var obj will be an instance of C specialized to some type that conforms to MyProtocol,but you can't specialize a generic class on a protocol type, because there is no such thing as a direct concrete instance of a protocol. There can only be instances of a type conforming to the protocol. And there can theoretically be many different types that conform to the protocol. So that notation doesn't really tell the compiler which specific specialization of C to use.
This shouldn't be a problem though, because you can write things like:
var obj: C<P1> = C<P1>()
or
var obj = C<P2>() // type is inferred
And within your class C you can still treat any uses of T as conforming to MyProtocol. So I think this should give you everything you need, as long as you remember that an instance of a generic class must be specialized to a single specific concrete type, not a protocol which could represent many possible concrete types.
You usually don't need to declare type for a Swift's variable. The compiler can infer type in most cases. Also, for generic class, you should let the compiler figure out what the generic classes resolve to:
class C<T: MyProtocol> {
var value: T
init (value: T) {
self.value = value
}
}
var obj = C(value: P1()) // type: C<P1>
// or:
var obj = C(value: P2()) // type: C<P2>
In my swift project I have a case where I use protocol inheritance as follow
protocol A : class{
}
protocol B : A{
}
What Im trying to achieve next is declaring another protocol with associated type, a type which must inherit from protocol A. If I try to declare it as :
protocol AnotherProtocol{
associatedtype Type : A
weak var type : Type?{get set}
}
it compiles without errors but when trying to adopt AnotherProtocol in the following scenario:
class SomeClass : AnotherProtocol{
typealias Type = B
weak var type : Type?
}
compilation fails with error claiming that SomeClass does not conform to AnotherProtocol. If I understood this correctly it means that B
does not adopt A while Im trying to declare and asking you on how to declare an associated type which inherits from protocol A?
I made the above assumption based on fact that the following scenario compiles just fine
class SomeDummyClass : B{
}
class SomeClass : AnotherProtocol{
typealias Type = SomeDummyClass
weak var type : Type?
}
This is pretty interesting. It appears that once you constrain the type of an associatedtype in a given protocol, you need to provide a concrete type in the implementation of that protocol (instead of another protocol type) – which is why your second example worked.
If you remove the A constraint on the associated type, your first example will work (minus the error about not being able to use weak on a non-class type, but that doesn’t seem to be related).
That all being said, I can't seem to find any documentation in order to corroborate this. If anyone can find something to back this up with (or completely dispute it), I’d love to know!
To get your current code working, you can use generics. This will actually kill two birds with one stone, as both your code will now compile and you'll benefit from the increased type safety that generics bring (by inferring the type you pass to them).
For example:
protocol A : class {}
protocol B : A {}
protocol AnotherProtocol{
associatedtype Type : A
weak var type : Type? {get set}
}
class SomeClass<T:B> : AnotherProtocol {
typealias Type = T
weak var type : Type?
}
Edit: It appears the above solution won't work in your particular case, as you want to avoid using concrete types. I'll leave it here in case it's useful for anyone else.
In your specific case, you may be able to use a type erasure in order to create a pseudo concrete type for your B protocol. Rob Napier has a great article about type erasures.
It's a bit of a weird solution in this case (as type erasures are normally used to wrap protocols with associatedtypes), and it's also definitely less preferred than the above solution, as you have to re-implement a 'proxy' method for each method in your A & B protocols – but it should work for you.
For example:
protocol A:class {
func doSomethingInA() -> String
}
protocol B : A {
func doSomethingInB(foo:Int)
func doSomethingElseInB(foo:Int)->Int
}
// a pseudo concrete type to wrap a class that conforms to B,
// by storing the methods that it implements.
class AnyB:B {
// proxy method storage
private let _doSomethingInA:(Void)->String
private let _doSomethingInB:(Int)->Void
private let _doSomethingElseInB:(Int)->Int
// initialise proxy methods
init<Base:B>(_ base:Base) {
_doSomethingInA = base.doSomethingInA
_doSomethingInB = base.doSomethingInB
_doSomethingElseInB = base.doSomethingElseInB
}
// implement the proxy methods
func doSomethingInA() -> String {return _doSomethingInA()}
func doSomethingInB(foo: Int) {_doSomethingInB(foo)}
func doSomethingElseInB(foo: Int) -> Int {return _doSomethingElseInB(foo)}
}
protocol AnotherProtocol{
associatedtype Type:A
weak var type : Type? {get set}
}
class SomeClass : AnotherProtocol {
typealias Type = AnyB
weak var type : Type?
}
class AType:B {
// implement the methods here..
}
class AnotherType:B {
// implement the methods here..
}
// your SomeClass instance
let c = SomeClass()
// set it to an AType instance
c.type = AnyB(AType())
// set it to an AnotherType instance
c.type = AnyB(AnotherType())
// call your methods like normal
c.type?.doSomethingInA()
c.type?.doSomethingInB(5)
c.type?.doSomethingElseInB(4)
You can now use the AnyB type in place of using the B protocol type, without making it any more type restrictive.
What's the difference between Protocols and class-bound Protocols, and which one we should use in Swift?
protocol A : class { ... }
protocol A { ... }
We get an error when attempting to add a weak delegate when the Protocol is not defined as : class:
protocol A { ... }
weak var delegate: A
Gives the error:
'weak' cannot be applied to non-class type
or
'weak' must not be applied to non-class-bound 'A'; consider adding a protocol conformance that has a class bound
Swift >= 4:
protocol A : AnyObject { ... {
Swift < 4:
protocol A : class { ... }
defines a "class-only protocol": Only class types (and not structures or enumerations) can adopt this protocol.
Weak references are only defined for reference types. Classes
are reference types, structures and enumerations are value types.
(Closures are reference types as well, but closures cannot adopt
a protocol, so they are irrelevant in this context.)
Therefore, if the object conforming to the protocol needs to be stored in a weak property then the protocol must be a class-only protocol.
Here is another example which requires a class-only protocol:
protocol A {
var name : String { get set }
}
func foo(a : A) {
a.name = "bar" // error: cannot assign to property: 'a' is a 'let' constant
}
This does not compile because for instances of structures and enumerations, a.name = "bar" is a mutation of a. If you define
the protocol as
protocol A : class {
var name : String { get set }
}
then the compiler knows that a is an instance of a class type to that
a is a reference to the object storage,
and a.name = "bar" modifies the referenced object, but not a.
So generally, you would define a class-only protocol if you need
the types adopting the protocol to be reference types and not value types.
If you are using Swift 4 or later, use AnyObject:
protocol A : AnyObject { ... }
Using class as before gives the warning and fix-it:
Using 'class' keyword to define a class-constrained protocol is deprecated; use 'AnyObject' instead
Replace 'class' with 'AnyObject'
You can make the protocol derive from any class type like NSObject or AnyObject:
protocol TopNewsTableDelegate : AnyObject {
func topNewsTableDidLoadedStories()
}
Or you can type like this
#objc protocol A { ... }
then you can make a weak delegate reference
protocol CustomProtocolName : NSObjectProtocol {
// ...
}