Consider a base class A which defines a function f() which should be overridden in subclasses to customize behavior:
class A {
func f() -> String {
return "A"
}
}
class AA: A {}
class AAA: AA {}
class AAAA: A {}
Now I want to add some subclasses to the hierarchy tree where A is the root, and all of those subclasses return the same thing from f():
class B: AA {
override func f() -> String {
return "X"
}
}
class C: AAA {
override func f() -> String {
return "X"
}
}
Since I hate copy-pasting every time the same implementation of f() I would like to move it to a common place. Since base classes are different and out of my control, I thought maybe protocol extensions can help. I wanted to do something like this:
protocol P {
func f() -> String
}
extension P {
func f() -> String {
return "X"
}
}
and then just conform my classes to protocol P, hoping that I will get the default implementation in every class:
class B: AA, P {}
class C: AAA, P {}
However, this is not working:
let x: A = C() // the same with just "let x = C()"
print("result: \(x.f())") // prints "result: A"
Not sure why, maybe because of static dispatch of protocol extensions, or because the protocol brings the same function name as base class and the latter "wins", have no idea. Is there a way to achieve something similar to this in Swift?
Rationale: I am tired of copying the same code of making a controller portrait-only into each of them that is designed to have such a limitation:
override var shouldAutorotate: Bool { return false }
override var supportedInterfaceOrientations: UIInterfaceOrientationMask { return .portrait }
override var preferredInterfaceOrientationForPresentation: UIInterfaceOrientation { return .portrait }
Related
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.
How can we call class functions with a dynamic class name?
Assume the following example where I have two class with methods with same signature
class Foo{
class func doSomething()
}
class Foobar {
class func doSomething()
}
class ActualWork{
//call following method with a variable type so that it accepts dynamic class name
func callDynamicClassMethod(x: dynamicClass)
x.doSomething()
}
How can this be implemented so that x accepts values at run time
Edit: Sorry, I missed to mention that I was looking for any other ways other than protocol oriented approach. This is more of an exploratory question to explore if there is a more direct approach/pods/libraries to achieve this.
I liked this question, because it made me to think a lit'bit outside of the box.
I'll answer it, by dividing it into a few parts.
First
call class functions
Class function is basically a Type methods, which can be achieved using the static word inside the class context.
Taking that into account, you can get a simple solution, using protocol and passing the class reference (conforming to that protocol) like this:
protocol Aaa{
static func doSomething();
}
class Foo : Aaa{
static func doSomething() {
print("Foo doing something");
}
}
class FooBar : Aaa{
static func doSomething() {
print("FooBar doing something");
}
}
class ActualWork{
//Using class (static) method
func callDynamicClassMethod <T: Aaa> (x: T.Type) {
x.doSomething();
}
}
//This is how you can use it
func usage(){
let aw = ActualWork();
aw.callDynamicClassMethod(x: Foo.self);
aw.callDynamicClassMethod(x: Foo.self);
}
Second
In case you don't really need the method on the class context, you may consider using instance methods. In that case the solution would be even simpler, like this:
protocol Bbb{
func doSomething();
}
class Bar : Bbb{
func doSomething() {
print("Bar instance doing something");
}
}
class BarBar : Bbb{
func doSomething() {
print("BarBar instance doing something");
}
}
class ActualWork{
//Using instance (non-static) method
func callDynamicInstanceMethod <T: Bbb> (x: T){
x.doSomething();
}
}
//This is how you can use it
func usage(){
let aw = ActualWork();
aw.callDynamicInstanceMethod(x: Bar());
aw.callDynamicInstanceMethod(x: BarBar());
}
Third
If you need to use the class func syntax, as OP originally did:
class func doSomething()
You CANNOT simply use a protocol. Because protocol is not a class...
So compiler won't allow it.
But it's still possible, you can achieve that by using
Selector with NSObject.perform method
like this:
class ActualWork : NSObject{
func callDynamicClassMethod<T: NSObject>(x: T.Type, methodName: String){
x.perform(Selector(methodName));
}
}
class Ccc : NSObject{
#objc class func doSomething(){
print("Ccc class Doing something ");
}
}
class Ddd : NSObject{
#objc class func doSomething(){
print("Ccc class Doing something ");
}
#objc class func doOther(){
print("Ccc class Doing something ");
}
}
//This is how you can use it
func usage() {
let aw = ActualWork();
aw.callDynamicClassMethod(x: Ccc.self, methodName: "doSomething");
aw.callDynamicClassMethod(x: Ddd.self, methodName: "doSomething");
aw.callDynamicClassMethod(x: Ddd.self, methodName: "doOther");
}
Generics and Protocol oriented programming will do the job:
protocol Doable {
static func doSomething()
}
class Foo: Doable {
static func doSomething() {
debugPrint("Foo")
}
}
class Foobar: Doable {
static func doSomething() {
debugPrint("Foobar")
}
}
class ActualWork {
func callDynamicClassMethod<T: Doable>(x: T.Type) {
x.doSomething()
}
}
let work = ActualWork()
work.callDynamicClassMethod(x: Foo.self)
work.callDynamicClassMethod(x: Foobar.self)
you can achieve this with help of Protocol
protocol common {
static func doSomething()
}
class Foo : common{
static func doSomething() {
print("Foo")
}
}
class Foobar : common {
static func doSomething() {
print("Foobar")
}
}
class ActualWork{
//call following method with a variable type so that it accepts dynamic class name
func callDynamicClassMethod(x: common.Type) {
x.doSomething()
}
}
let fooObj : common = Foo()
let Foobarobj : common = Foobar()
let workObk = ActualWork()
workObk.callDynamicClassMethod(x:Foo.self)
workObk.callDynamicClassMethod(x:Foobar.self)
I think, there are three solutions. I shared an sample below.
Use "protocol" that has "doSomething()" function requirements.
Create a function which gets function definition as a parameter.
Use reflection. you can use EVReflection that is good Api for reflection.
sample code:
protocol FooProtocol {
static func doSomething()
}
class Foo: FooProtocol {
class func doSomething() {
print("Foo:doSomething")
}
}
class Foobar: FooProtocol {
class func doSomething() {
print("Foobar:doSomething")
}
}
class ActualWork {
func callDynamicClassMethod<T: FooProtocol>(x: T.Type) {
x.doSomething()
}
func callDynamicClassMethod(x: #autoclosure () -> Void) {
x()
}
func callDynamicClassMethod(x: () -> Void) {
x()
}
}
ActualWork().callDynamicClassMethod(x: Foo.self)
ActualWork().callDynamicClassMethod(x: Foobar.self)
print("\n")
ActualWork().callDynamicClassMethod(x: Foo.doSomething())
ActualWork().callDynamicClassMethod(x: Foobar.doSomething())
print("\n")
ActualWork().callDynamicClassMethod(x: Foo.doSomething)
ActualWork().callDynamicClassMethod(x: Foobar.doSomething)
Looks like you are searching for duck typing, and this is harder to achieve in a statically typed language (with some exceptions, listed in the linked Wikipedia page).
This is because dynamically calling a method requires knowledge about the layout of the target object, thus either inheritance of the class declaring the method, or conformance to a protocol that requires that method.
Starting with Swift 4.2, and the introduction of dynamic member lookup, there is another approach to solve your problem, however it also involves some ceremony:
// This needs to be used as base of all classes that you want to pass
// as arguments
#dynamicMemberLookup
class BaseDynamicClass {
subscript(dynamicMember member: String) -> () -> Void {
return { /* empty closure do nothing */ }
}
}
// subclasses can choose to respond to member queries any way they like
class Foo: BaseDynamicClass {
override subscript(dynamicMember member: String) -> () -> Void {
if member == "doSomething" { return doSomething }
return super[dynamicMember: member]
}
func doSomething() {
print("Dynamic from Foo")
}
}
class Bar: BaseDynamicClass {
override subscript(dynamicMember member: String) -> () -> Void {
if member == "doSomething" { return doSomething }
return super[dynamicMember: member]
}
func doSomething() {
print("Dynamic from Bar")
}
}
func test(receiver: BaseDynamicClass) {
receiver.doSomething()
}
test(receiver: Bar()) // Dynamic from Bar
To conclude, in the current Swift version there is no way to have both the argument and the method dynamic, some common ground needs to be set.
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()
I'm facing a problem regarding protocols methods dispatch.
I have a class hierarchy that looks like that:
protocol E {
func test()
}
extension E {
func test() {
print("jello")
}
}
class A: E {
}
class B: A {
func test() {
print("hello")
}
}
But when I call test on an instance of class B statically forced to be typed A, "jello" gets printed, not "hello".
let b: A = B() // prints "jello" not "hello"
b.test()
My understanding is that test method printing "jello" gets "integrated" into instances of A (since A conforms to E protocol). I'm then providing another implementation of test inside B (that inherits form A). I thought polymorphism would work here and calling test on B instance that are stored inside A references would print hello. What's happening here?
It's perfectly working when not using any protocol:
class A {
func test() {
print("jello")
}
}
class B: A {
override func test() {
print("hello")
}
}
let b: A = B() // prints "hello"
b.test()
What's different from adopting a protocol that adds new methods to my parent class and providing a new implementation in a subclass, than having directly written this method in the parent class and then overriding it in a subclass?
Do you guys have any workaround?
Smells like a bug.
The only workaround I came up with was very ugly...
protocol E {
func test()
}
func E_test(_s: E) {
print("jello")
}
extension E {
func test() { E_test(self) }
}
class A: E {
func test() { E_test(self) }
}
class B: A {
override func test() {
print("hello")
}
}
let b: A = B()
b.test()
It is a bug indeed. Here is the link to it: https://bugs.swift.org/browse/SR-103
I have an extension on UIView implementing a protocol
protocol SomeProtocol {
var property : Int
}
extension UIView : SomeProtocol {
var property : Int {
get {
return 0
}
set {
// do nothing
}
}
}
in a concrete subclass I want to override this extension method:
class Subclass : UIView, SomeProtocol {
var _property : Int = 1
var property : Int {
get { return _property}
set(val) {_property = val}
}
}
I set breakpoints and see that the extension method is called and not the concrete subclass method:
var subclassObject = Subclass()
someObject.doSomethingWithConcreteSubclassObject(subclassObject)
// other code;
fun doSomethingWithConcreteSuclassObject(object : UIView) {
var value = object.property // always goes to extension class get/set
}
As others have noted, Swift does not (yet) allow you to override a method declared in a class extension. However, I'm not sure whether you'll ever get the behavior you want even if/when Swift someday allows you to override these methods.
Consider how Swift deals with protocols and protocol extensions. Given a protocol to print some metasyntactic variable names:
protocol Metasyntactic {
func foo() -> String
func bar() -> String
}
An extension to provide default implementations:
extension Metasyntactic {
func foo() -> String {
return "foo"
}
func bar() -> String {
return "bar"
}
}
And a class that conforms to the protocol:
class FooBar : Metasyntactic {
func foo() -> String {
return "FOO"
}
func bar() -> String {
return "BAR"
}
}
Swift will use dynamic dispatch to call the appropriate implementations of foo() and bar() based on each variable's runtime type rather than on the type inferred by the compiler:
let a = FooBar()
a.foo() // Prints "FOO"
a.bar() // Prints "BAR"
let b: Metasyntactic = FooBar()
b.foo() // Prints "FOO"
b.bar() // Prints "BAR"
If, however, we extend the protocol further to add a new method:
extension Metasyntactic {
func baz() -> String {
return "baz"
}
}
And if we override our new method in a class that conforms to the protocol:
class FooBarBaz : Metasyntactic {
func foo() -> String {
return "FOO"
}
func bar() -> String {
return "BAR"
}
func baz() -> String {
return "BAZ"
}
}
Swift will now use static dispatch to call the appropriate implementation of baz() based on the type inferred by the compiler:
let a = FooBarBaz()
a.baz() // Prints "BAZ"
let b: Metasyntactic = FooBarBaz()
b.baz() // Prints "baz"
Alexandros Salazar has a fantastic blog post explaining this behavior in depth, but suffice it to say that Swift only uses dynamic dispatch for methods declared in the original protocol, not for methods declared in protocol extensions. I imagine the same would be true of class extensions, as well.
I know this question has been asked a while ago. But this will be handy for someone who looking for an easier way. There is a way of overriding an extension methods. I know its bit hacky but it does the job beautifully.
If you declare your protocol with #objc
#objc protocol MethodOverridable {
func overrideMe()
}
In Extension
extension MainClass: MethodOverridable {
func overrideMe() {
print("Something useful")
}
}
Subclass - You can able to override it in your subclass. It works like a magic. Well, not really when adding #objc it exposes your protocol to Objective-C and its Runtime. That allows your subclass to override.
class SubClass: MainClass {
override func overrideMe() {
print("Something more useful")
}
}
Swift 5
class Class
{
#objc dynamic func make() { print("make from class") }
}
class SubClass: Class {}
extension SubClass {
override func make() {
print("override")
}
}
It looks like you can override property for 2nd super class property. For example, you can access UIView property by making extension to the UILabel wanting to override frame property of UIView. This sample works for me in Xcode 6.3.2
extension UILabel {
override public var frame: CGRect {
didSet {
println("\(frame)")
}
}
}
You can't do this through normal means.
It's in Apple's docs that you can't override a method in an extension in a subclass.
Also, extensions can add new functionality to a type, but they cannot override existing functionality.
https://docs.swift.org/swift-book/LanguageGuide/Extensions.html
I think you forgot to override the superclass property in your subclass:
class Subclass : UIView {
var _property : Int = 1
override var property : Int {
get { return _property}
set(val) {_property = val}
}
}