I have a abstract base class (DataSource, in my simplified scenario below) and a handful of subclasses (IntSource and StringSource). The base class is using generics.
I want to have a variable (say, currentDataSource) typed as the base class so it can refer to an instance of any of the subclasses. I can't figure out how to declare it, though. The declaration requires a concrete type which immediately breaks the abstraction.
class DataSource<Element> where Element:Equatable {
var elements = [Element]()
func description() {
print("must override")
}
}
class IntSource: DataSource<Int> {
override func description() {
print("Ints!")
}
}
class StringSource: DataSource<String> {
override func description() {
print("Strings!")
}
}
let a = IntSource()
let b = StringSource()
var current: DataSource<???> // this is where I'm stuck
current = a
print(current.description())
current = b
print(current.description())
In swift, an instance func can't call a static/class func without prefixing the method call with the class name. OR you can use type(of: self), e.g
class Foo {
static func doIt() { }
func callIt() {
Foo.doIt() // This works
type(of: self).doIt() // Or this
doIt() // This doesn't compile (unresolved identifier)
}
}
My question is, what's the difference here?
Is it just a matter of coding style, or is there some difference e.g. static or dynamic dispatch going on?
If it is just coding style, what's the preferred style?
There are two main differences.
1. The value of self inside the static method
The metatype that you call the static method on is available to you in the method as self (it's simply passed as an implicit parameter). Therefore if you call doIt() on type(of: self), self will be the dynamic metatype of the instance. If you call it on Foo, self will be Foo.self.
class Foo {
static func doIt() {
print("hey I'm of type \(self)")
}
func callDoItOnDynamicType() {
type(of: self).doIt() // call on the dynamic metatype of the instance.
}
func classDoItOnFoo() {
Foo.doIt() // call on the metatype Foo.self.
}
}
class Bar : Foo {}
let f: Foo = Bar()
f.callDoItOnDynamicType() // hey I'm of type Bar
f.classDoItOnFoo() // hey I'm of type Foo
This difference can be really important for factory methods, as it determines the type of instance you create.
class Foo {
required init() {}
static func create() -> Self {
return self.init()
}
func createDynamic() -> Foo {
return type(of: self).create()
}
func createFoo() -> Foo {
return Foo.create()
}
}
class Bar : Foo {}
let f: Foo = Bar()
print(f.createDynamic()) // Bar
print(f.createFoo()) // Foo
2. The dispatching of the static method
(Martin has already covered this, but I thought I would add it for the sake of completion.)
For class methods that are overridden in subclasses, the value of the metatype that you call the method on determines which implementation to call.
If called on a metatype that is known at compile time (e.g Foo.doIt()), Swift is able to statically dispatch the call. However, if you call the method on a metatype that isn't known until runtime (e.g type(of: self)), the method call will be dynamically dispatched to the correct implementation for the metatype value.
class Foo {
class func doIt() {
print("Foo's doIt")
}
func callDoItOnDynamicType() {
type(of: self).doIt() // the call to doIt() will be dynamically dispatched.
}
func classDoItOnFoo() {
Foo.doIt() // will be statically dispatched.
}
}
class Bar : Foo {
override class func doIt() {
print("Bar's doIt")
}
}
let f: Foo = Bar()
f.callDoItOnDynamicType() // Bar's doIt
f.classDoItOnFoo() // Foo's doIt
For a class method it makes a difference if the method is
overridden in a subclass:
class Foo {
class func doIt() {
print("Foo doit")
}
func callViaClassname() {
Foo.doIt()
}
func callViaTypeOf() {
type(of: self).doIt()
}
}
class Bar: Foo {
override class func doIt() {
print("Bar doit")
}
}
Bar().callViaClassname() // Foo doit
Bar().callViaTypeOf() // Bar doit
This is also documented in "Types" in the Swift Language Reference:
You can use a type(of:) expression with an instance of a type to access that instance’s dynamic, runtime type as a value, ...
I don't know a difference for a static method (which is final
and cannot be overridden in a subclass).
Correction: See Hamish's answer for the difference
in both static and class methods.
What I tried to do:
protocol HasElement {
associatedtype ItemType
func getElement() -> ItemType
func setElement(element: ItemType)
}
class Element {}
class BarElement: Element {}
class Foo: NSObject, HasElement {
typealias ItemType = Element
func getElement() -> Element { ... }
func setElement(element: Element) { ... }
}
class Bar: Foo {
typealias ItemType = BarElement
override func getElement() -> BarElement { ... } // This works.
override func setElement(element: BarElement) { ... } // This fails.
}
The error is:
Method does not override any method from its superclass
If I try to use ItemType instead:
override func setElement(element: ItemType) { ... } // Still fails.
The error is:
'ItemType' is ambiguous for type lookup in this context
Is there a way to make this work?
Here's a way to do what you want:
protocol HasElement {
associatedtype ItemType
func getElement() -> ItemType
func setElement(element: ItemType)
}
class Element {}
class BarElement: Element {}
class Foo: HasElement {
// no need for typealias, the associated type is inferred
func getElement() -> Element { return Element() }
func setElement(element: Element) { }
}
class Bar: Foo {
// no need for typealias, the associated type is inferred
override func getElement() -> BarElement { return BarElement() }
// hide the parent class method
#available(*, unavailable, message: "Use setElement(element: BarElement)")
override func setElement(element: Element) { }
// comply with protocol in this class
func setElement(element: BarElement) { }
}
// can't do this now:
let myElement = Element()
let myBar = Bar()
myBar.setElement(element: myElement) // Error: 'setElement(element: BarElement)' is unavailable: Use setElement(element: BarElement)
The problem here isn't associated types, it's that method inputs are contravariant. Therefore you cannot override a method that expects a given superclass instance input with a method that expects a subclass instance input.
In fact, you can simply boil your code down to:
class Element {}
class BarElement : Element {}
class Foo {
func setElement(element: Element) { }
}
class Bar : Foo {
// error: Method does not override any method from its superclass
override func setElement(element: BarElement) { }
}
You simply cannot override an (Element) -> Void method with a (BarElement) -> Void method. The reasons for this should be fairly obvious if you consider what would happen if you created a Bar instance and upcast it to Foo. You're now able to pass an Element instance to a method that expects a BarElement instance, which is illegal.
The reason it works for your getElement method is that method outputs are covariant. Therefore overriding a () -> Element method with a () -> BarElement method is perfectly legal, as even if you upcast a Bar instance to Foo, the BarElement instance returned from getElement can be freely upcast to Element.
As for solutions, it rather depends on your exact use case. It may well be that what you're trying to do doesn't require inheritance at all, and instead you can just conform Foo and Bar to HasElement separately.
This works:
override func getElement() -> BarElement { ... }
using override keyword is when function signature does not change (BarElement is still Element). You copy and paste function from super class with same function name, same parameter name and same return value type.
But
override func setElement(element: BarElement) { ... }
fails because when you change parameter type (with same parameter name).
This case is overloading not overriding.
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()
The question is better explained in code:
class A {
class func thefunc() -> String {
/* Can I here know if thefunc was called using
A.thefunc() or
B.thefunc()?
*/
return "A" /* or "B"= */
}
}
class B: A {
}
You can use self in a static method to refer to the type (as compared to the instance for using self in an instance method)
class A {
class func thefunc() -> A.Type {
return self
}
}
class B: A { }
let metaTypeA = A.thefunc() // A.Type
let metaTypeB = B.thefunc() // B.Type
Similarly, you can use runtime introspection, specifically the subjectType property of the Mirror representation of self.
Instance Variables
...
var subjectType: Any.Type
The static type of the subject being reflected.
From the swiftdoc.org reference of Mirror structure.
E.g.:
class A {
class func thefunc() {
print(Mirror(reflecting: self).subjectType)
}
}
class B: A { }
A.thefunc() // A.Type
B.thefunc() // B.Type
Alternatively, if you needn't actually make use of the meta-type (just differ between the "static caller"), you could use the String representation of self.
class A {
class func thefunc() -> String {
return String(self)
}
}
class B: A { }
print(A.thefunc()) // A
print(B.thefunc()) // B