I have the following class:
class MyClass<T: BaseClass> {
let aThing = T()
func someMethod() {
configure()
}
}
whereas
class SubTypeAOfBaseClass: BaseClass { ... }
class SubTypeBOfBaseClass: BaseClass { ... }
In configure I want to configure aThing depending on its type. Therefore I have created a protocol Configurable and an extension:
protocol Configurable {
func configure()
}
extension MyClass: Configurable where T == SubTypeAOfBaseClass {
func configure() {
print("Configuring SubTypeAOfBaseClass")
aThing.doSomethingA()
}
}
The error that I get is in MyClass.someMethod where I call configure(): "Referencing instance method configure() requires types T and SubTypeAOfBaseClass be equivalent.
Another error when I add the following extension:
extension MyClass: Configurable where T == SubTypeBOfBaseClass {
func configure() {
print("Configuring SubTypeBOfBaseClass")
aThing.doSomethingB()
}
}
The error changes to "No exact matches in call to instance method configure" and in the line
extension MyClass: Configurable where T == SubTypeBOfBaseClass {
I get the error "Conflicting conformance of MyClass to protocol configurable, there cannot more than one conformance, even with different conditional bounds."
It should work, but apparently I am missing something or my understanding of how to achieve what I want is wrong.
Consider the case when T is BaseClass, or when T is AnotherSubclass that I defined as
class AnotherSubclass : BaseClass {
}
What would happen? You haven't declared a conformance to Configure when T is AnotherSubclass!
There's really only two (not bad) choices here.
You want configure to do nothing when T is neither SubTypeAOfBaseClass nor SubTypeBOfBaseClass
you only want MyClass<SubTypeAOfBaseClass> and MyClass<SubTypeBOfBaseClass> to be valid types - MyClass<BaseClass> and MyClass<AnotherSubclass> would give compiler errors.
Choice 2 is not possible in Swift. That would require something similar to the sealed types in Java or Kotlin.
Choice 1 can be done like this:
class BaseClass {
...
func configure() {
}
}
class SubTypeAOfBaseClass: BaseClass {
...
override func configure() {
print("Configuring SubTypeAOfBaseClass")
doSomethingA()
}
}
class SubTypeBOfBaseClass: BaseClass {
...
override func configure() {
print("Configuring SubTypeAOfBaseClass")
doSomethingB()
}
}
class MyClass<T: BaseClass> {
let aThing = T()
func someMethod() {
aThing.configure()
}
}
You might notice that the each implementation of configure has been moved to the base classes. If you want to implement them all in MyClass, you must check the type by hand:
class MyClass<T: BaseClass> {
let aThing = T()
func someMethod() {
if let selfA = self as? MyClass<SubTypeAOfBaseClass> {
selfA.configure()
} else if let selfB = self as? MyClass<SubTypeBOfBaseClass> {
selfB.configure()
}
}
}
extension MyClass where T == SubTypeAOfBaseClass {
func configure() {
print("Configuring SubTypeAOfBaseClass")
aThing.doSomethingA()
}
}
extension MyClass where T == SubTypeBOfBaseClass {
func configure() {
print("Configuring SubTypeBOfBaseClass")
aThing.doSomethingB()
}
}
This is because of the second problem in your code - different parameterisations of a generic type, MyClass<SubTypeAOfBaseClass> and MyClass<SubTypeBOfBaseClass>, can't conform to a protocol differently. This is a limitation of Swift, unfortunately. See here for more info.
I'm using Quick to test my Swift code.
However, I think it doesn't release objects defined in describe scope:
class MyClass {
deinit {
print(self, #function)
}
}
final class MyClassSpec: QuickSpec {
override func spec() {
describe("") {
let foo = MyClass()
it("") {
print(foo)
expect(true).to(beTrue())
}
}
}
}
I don't see any output from print inside deinit, and a debug breakpoint inside the deinit does not get catched.
If I move foo inside it, the deinit is called.
Is this a bug in Quick, or is it normal for deinit not to be called in a test suite?
Apparently the code I wrote was not only retaining the object but was also an anti-pattern.
Even a plain old XCTestCase retains an object:
class MyClass {
deinit {
print(self, #function)
}
}
final class MyClassTest: XCTestCase {
let foo = MyClass()
func testMyClass() {
print(foo)
XCTAssert(true)
}
}
deinit is not called for foo.
This is due to a nature of XCTestCase—it never really gets deinited.
So one should always use setUp & tearDown to manage everything (or more accurately, objects with reference semantics).
I believe this directly translates to QuickSpec as well, so I should always use beforeEach & afterEach in order to manage the objects.
To "fix" the problem, I should test like:
final class MyClassSpec: QuickSpec {
override func spec() {
describe("") {
let foo: MyClass!
beforeEach { foo = MyClass() }
afterEach { foo = nil }
it("") {
print(foo)
expect(true).to(beTrue())
}
}
}
}
ArtistVC is a subclass of CategoryVC which is a subclass of BrowserVC.
Most of ArtistVC's UITableView Data Source methods wind up calling CategoryVC's implementation of those methods with return super.tableView(tableView, cellForRowAt: adjustedIndexPath)(for example), after some other things happen of course, like modifying the indexPath passed to the method.
But sometimes I actually want to bypass CategoryVC's implementation and use BrowserVC's implementation.
super.super.someMethod() apparently isn't the answer. Is this possible?
There's a (convoluted) way to do this but it requires that the intermediate class "contributes" to the bypassing of its override.
For example:
class A
{
func doIt(_ parameter:Int) { print("A.doIt(\(parameter))") }
}
class B:A
{
override func doIt(_ parameter:Int) { print("B.doIt(\(parameter))") }
var super_doIt:(Int)->() { return super.doIt }
}
class C:B
{
override func doIt(_ parameter:Int)
{ super.super_doIt(parameter) }
}
let c = C()
c.doIt(3) // A.doIt(3)
Is there a way to create an abstract class in the Swift Language, or is this a limitation just like Objective-C? I'd like to create a abstract class comparable to what Java defines as an abstract class.
There are no abstract classes in Swift (just like Objective-C). Your best bet is going to be to use a Protocol, which is like a Java Interface.
With Swift 2.0, you can then add method implementations and calculated property implementations using protocol extensions. Your only restrictions are that you can't provide member variables or constants and there is no dynamic dispatch.
An example of this technique would be:
protocol Employee {
var annualSalary: Int {get}
}
extension Employee {
var biweeklySalary: Int {
return self.annualSalary / 26
}
func logSalary() {
print("$\(self.annualSalary) per year or $\(self.biweeklySalary) biweekly")
}
}
struct SoftwareEngineer: Employee {
var annualSalary: Int
func logSalary() {
print("overridden")
}
}
let sarah = SoftwareEngineer(annualSalary: 100000)
sarah.logSalary() // prints: overridden
(sarah as Employee).logSalary() // prints: $100000 per year or $3846 biweekly
Notice that this is providing "abstract class" like features even for structs, but classes can also implement the same protocol.
Also notice that every class or struct that implements the Employee protocol will have to declare the annualSalary property again.
Most importantly, notice that there is no dynamic dispatch. When logSalary is called on the instance that is stored as a SoftwareEngineer it calls the overridden version of the method. When logSalary is called on the instance after it has been cast to an Employee, it calls the original implementation (it doesn't not dynamically dispatch to the overridden version even though the instance is actually a Software Engineer.
For more information, check great WWDC video about that feature: Building Better Apps with Value Types in Swift
Note that this answer is targeted at Swift 2.0 and above
You can achieve the same behaviour with protocols and protocol extensions.
First, you write a protocol that acts as an interface for all the methods that have to be implemented in all types that conform to it.
protocol Drivable {
var speed: Float { get set }
}
Then you can add default behaviour to all types that conform to it
extension Drivable {
func accelerate(by: Float) {
speed += by
}
}
You can now create new types by implementing Drivable.
struct Car: Drivable {
var speed: Float = 0.0
init() {}
}
let c = Car()
c.accelerate(10)
So basically you get:
Compile time checks that guarantee that all Drivables implement speed
You can implement default-behaviour for all types that conform to Drivable (accelerate)
Drivable is guaranteed not to be instantiated since it's just a protocol
This model actually behaves much more like traits, meaning you can conform to multiple protocols and take on default implementations of any of them, whereas with an abstract superclass you're limited to a simple class hierarchy.
I think this is the closest to Java's abstract or C#'s abstract:
class AbstractClass {
private init() {
}
}
Note that, in order for the private modifiers to work, you must define this class in a separate Swift file.
EDIT: Still, this code doesn't allow to declare an abstract method and thus force its implementation.
The simplest way is to use a call to fatalError("Not Implemented") into the abstract method (not variable) on the protocol extension.
protocol MyInterface {
func myMethod() -> String
}
extension MyInterface {
func myMethod() -> String {
fatalError("Not Implemented")
}
}
class MyConcreteClass: MyInterface {
func myMethod() -> String {
return "The output"
}
}
MyConcreteClass().myMethod()
After I struggled for several weeks, I finally realized how to translate a Java/PHP abstract class to Swift:
public class AbstractClass: NSObject {
internal override init(){}
public func getFoodToEat()->String
{
if(self._iAmHungry())
{
return self._myFavoriteFood();
}else{
return "";
}
}
private func _myFavoriteFood()->String
{
return "Sandwich";
}
internal func _iAmHungry()->Bool
{
fatalError(__FUNCTION__ + "Must be overridden");
return false;
}
}
public class ConcreteClass: AbstractClass, IConcreteClass {
private var _hungry: Bool = false;
public override init() {
super.init();
}
public func starve()->Void
{
self._hungry = true;
}
public override func _iAmHungry()->Bool
{
return self._hungry;
}
}
public protocol IConcreteClass
{
func _iAmHungry()->Bool;
}
class ConcreteClassTest: XCTestCase {
func testExample() {
var concreteClass: ConcreteClass = ConcreteClass();
XCTAssertEqual("", concreteClass.getFoodToEat());
concreteClass.starve();
XCTAssertEqual("Sandwich", concreteClass.getFoodToEat());
}
}
However I think Apple did not implement abstract classes because it generally uses the delegate+protocol pattern instead. For example the same pattern above would be better done like this:
import UIKit
public class GoldenSpoonChild
{
private var delegate: IStomach!;
internal init(){}
internal func setup(delegate: IStomach)
{
self.delegate = delegate;
}
public func getFoodToEat()->String
{
if(self.delegate.iAmHungry())
{
return self._myFavoriteFood();
}else{
return "";
}
}
private func _myFavoriteFood()->String
{
return "Sandwich";
}
}
public class Mother: GoldenSpoonChild, IStomach
{
private var _hungry: Bool = false;
public override init()
{
super.init();
super.setup(self);
}
public func makeFamilyHungry()->Void
{
self._hungry = true;
}
public func iAmHungry()->Bool
{
return self._hungry;
}
}
protocol IStomach
{
func iAmHungry()->Bool;
}
class DelegateTest: XCTestCase {
func testGetFood() {
var concreteClass: Mother = Mother();
XCTAssertEqual("", concreteClass.getFoodToEat());
concreteClass.makeFamilyHungry();
XCTAssertEqual("Sandwich", concreteClass.getFoodToEat());
}
}
I needed this kind of pattern because I wanted to commonize some methods in UITableViewController such as viewWillAppear etc. Was this helpful?
There is a way for simulating abstract classes using Protocols.
This is an example:
protocol MyProtocol {
func doIt()
}
class BaseClass {
weak var myDelegate: MyProtocol?
init() {
...
}
func myFunc() {
...
self.myDelegate?.doIt()
...
}
}
class ChildClass: BaseClass, MyProtocol {
override init(){
super.init()
self.myDelegate = self
}
func doIt() {
// Custom implementation
}
}
One more way how you can implement abstract class is to block initializer.
I've done it this way:
class Element:CALayer { // IT'S ABSTRACT CLASS
override init(){
super.init()
if self.dynamicType === Element.self {
fatalError("Element is abstract class, do not try to create instance of this class")
}
}
}
It's a really old question but still… Here's a snippet of actual code that compiles on Swift 5.2 and works as intended:
protocol Context {
init() throws
func out(_ aStr: String) throws
// Other stuff
}
class AbstractContext: Context {
required init() throws {
if Self.self === AbstractContext.self {
preconditionFailure("Call to abstract method \(Self.self).\(#function)")
}
}
func out(_ aStr: String) throws {
preconditionFailure("Call to abstract method \(Self.self).\(#function)")
}
// Other stuff
}
class CompileContext: AbstractContext {
required init() throws {}
override func out(_ aStr: String) throws {
print(aStr)
}
// Other stuff
}
And here's what I get once I remove CompileContext.out:
Fatal error: Call to abstract method CompileContext.out(_:): file swiftpg/contexts.swift, line 28
With the limitation of no dynamic dispatch, you could do something like this:
import Foundation
protocol foo {
static var instance: foo? { get }
func prt()
}
extension foo {
func prt() {
if Thread.callStackSymbols.count > 30 {
print("super")
} else {
Self.instance?.prt()
}
}
}
class foo1 : foo {
static var instance : foo? = nil
init() {
foo1.instance = self
}
func prt() {
print("foo1")
}
}
class foo2 : foo {
static var instance : foo? = nil
init() {
foo2.instance = self
}
func prt() {
print("foo2")
}
}
class foo3 : foo {
static var instance : foo? = nil
init() {
foo3.instance = self
}
}
var f1 : foo = foo1()
f1.prt()
var f2 : foo = foo2()
f2.prt()
var f3 : foo = foo3()
f3.prt()
I was trying to make a Weather abstract class, but using protocols wasn't ideal since I had to write the same init methods over and over again. Extending the protocol and writing an init method had it's issues, especially since I was using NSObject conforming to NSCoding.
So I came up with this for the NSCoding conformance:
required init?(coder aDecoder: NSCoder) {
guard type(of: self) != Weather.self else {
fatalError("<Weather> This is an abstract class. Use a subclass of `Weather`.")
}
// Initialize...
}
As for init:
fileprivate init(param: Any...) {
// Initialize
}
Move all references to abstract properties and methods of Base class to protocol extension implementation, where Self constraint to Base class. You will gain access to all methods and properties of Base class. Additionally compiler check implementation of abstract methods and properties in protocol for derived classes
protocol Commom:class{
var tableView:UITableView {get};
func update();
}
class Base{
var total:Int = 0;
}
extension Common where Self:Base{
func update(){
total += 1;
tableView.reloadData();
}
}
class Derived:Base,Common{
var tableView:UITableView{
return owner.tableView;
}
}
I’ve got a Swift class Parent that has a method doSomething() and I want to detect (in Parent) if a subclass has overridden doSomething(). How do I do that?
class Parent {
func doSomething() {}
func subclassOverridesDoSomething() -> Bool {
// what goes here?
return true
}
}
class Child: Parent {
override func doSomething() {}
}
I know I can do this with NSObject or even the Objective C runtime functions, but how can I do it with Swift classes?
You can use the Objective-C runtime provided you expose the function with dynamic:
class Parent {
func doSomething() {}
func subclassOverridesDoSomething(t:Parent.Type) -> Bool {
let originalMethod = class_getInstanceMethod(t, "doSomething")
return originalMethod != nil
}
}
class Child: Parent {
dynamic override func doSomething() {}
}
Parent().subclassOverridesDoSomething(Child.self)
If you omit dynamic, it won't work because Objective-C can't see the method.
Well Swift doesn't really offer methods to do that.
Also in my opinion, there is no disadvantage in using Objective-c methods like method_getImplementation or the NSObject. You will have to use the methods objective-c offers you to solve your problem.
For example:
let selector = Selector("viewWillAppear:")
let originalMethod = class_getInstanceMethod(YourClass, selector)