I have the next example where I have a problem to initialize the variable tmpClass, I tried Any, AnyObject, AnyClass ... but without success. I need to change and call the class dynamically from one variable
class One{
func doSomething()->Int{ return 100 }
}
class Two{
func doSomething()->Int{ return 200 }
}
class begin{
var tmpClass = ???
var type = "One"
switch (type){
case "One":
tmpClass = One()
case "Two":
tmpClass = Two()
}
println(tmpClass.doSomething())
}
the question is what type need to be tmpClass or some method to do that ?
You can do it in two different ways, see which fits you better:
Protocols
Create a protocol for doSomething
protocol DoSomethingProtocol {
func doSomething()->Int
}
Make your classes conform to it
class One : DoSomethingProtocol {
func doSomething()->Int{ return 100 }
}
class Two : DoSomethingProtocol{
func doSomething()->Int{ return 200 }
}
Set your tempClass variable to be of a protocol compliant class
var tmpClass: DoSomethingProtocol?
Inheritance
Create a base class with the doSomething method
class DoSomethingBaseClass {
func doSomething()->Int{ return 0 }
}
Override doSomething in your classes
class One : DoSomethingBaseClass {
override func doSomething()->Int{ return 100 }
}
class Two : DoSomethingBaseClass{
override func doSomething()->Int{ return 200 }
}
Set your tempClass variable to be of the super class (polymorphism!)
var tmpClass: DoSomethingBaseClass?
General tips
In Swift switches must be exhaustive (all possible options must have a case), to fix this in your example you can simply add
default:
tmpClass = nil
Do not forget to unwrap your optional. For safety you can use the if let syntax for optional binding:
if let tmpUnwrappedClass = tmpClass {
println(tmpUnwrappedClass.doSomething())
}
You could just use the ! unwrap operator as in println(tmpClass!.doSomething()) but this will fail when tmpClass is nil.
Your begin class has two properties and no initializers. Classes with properties must have a designated initializer to set them, that's why I removed the whole class.
Both versions are tested and working in the latest Swift using Playground.
Related
I expected the following code to print "extension" in both cases. But the type constraint on the extension does not take effect on the contained generic type. I see the same behavior when constraining on protocols too.
class Generic1<T1> {
func doSomething() {
print("base")
}
}
extension Generic1 where T1 == String {
func doSomething() {
print("extension")
}
}
class Generic2<T2> {
private let generic1 = Generic1<T2>()
func doSomething() {
generic1.doSomething()
}
}
Generic1<String>().doSomething() // prints extension
Generic2<String>().doSomething() // prints base
The only workaround I currently have is to constrain the outer generic as well like so:
extension Generic2 where T2 == String {
func doSomething() {
generic1.doSomething()
}
}
Why does this happen? Are there better solutions?
Edit: Just for completeness, the workaround that suited my case was the following:
class Generic1<T1> {
func doSomething() {
print("base")
}
}
class StringGeneric1: Generic1<String> {
override func doSomething() {
print("extension")
}
}
class Generic2<T2> {
private let generic1: Generic1<T2>
init (_ generic1: Generic1<T2>) {
self.generic1 = generic1
}
func doSomething() {
generic1.doSomething()
}
}
Generic1<String>().doSomething() // prints "base"
Generic2<String>(StringGeneric1()).doSomething() // prints "extension"
The problem is that methods defined in extensions are statically dispatched. So when you have:
class Generic2<T2> {
private let generic1 = Generic1<T2>()
func doSomething() {
generic1.doSomething()
}
}
The compiler cannot know here whether T2 is going to be a String or not, so it generates a call to the method in the base class. When you explicitly specify that T2 is String, then with that information, the compiler can generate a call to the extension's method here. Otherwise, though, the type of T2 isn't known until runtime, so you can't reach the extension method via static dispatch.
This may be solved when conditional conformances are added with Swift 4.2
I am trying to achieve a design where I can have a base class that has a generic property that I can change values on by conforming to a protocol.
protocol EnumProtocol {
static var startValue: Self { get }
func nextValue() -> Self
}
enum FooState: EnumProtocol {
case foo1, foo2
static var startValue: FooState { return .foo1 }
func nextValue() -> FooState {
switch self {
case .foo1:
return .foo2
case .foo2:
return .foo1
}
}
}
enum BarState: EnumProtocol {
case bar
static var startValue: BarState { return .bar }
func nextValue() -> BarState {
return .bar
}
}
class BaseClass<T: EnumProtocol> {
var state = T.startValue
}
class FooClass: BaseClass<FooState> {
}
class BarClass: BaseClass<BarState> {
}
Is it possible to end up with a solution similar to this where the element type is unknown and the value relies on the nextValue() method.
let foo = FooClass()
let bar = BarClass()
if let test = bar as? BaseClass {
test.state = test.state.nextValue()
}
This works but BarState will be unknown in my case and a lot of classes will be subclasses of BaseClass and have different state types.
let bar = BarClass()
if let test = bar as? BaseClass<BarState> {
test.state = test.state.nextValue()
}
This is a simplified example. In my case I will get a SKNode subclass that has a state property that is an enum with a nextvalue method that have defined rules to decide what the next value will be. I am trying to have a generic implementation of this that only relies on what is returned from the nextValue method. Is there a better pattern to achieve this?
This will not work for this exact scenario because EnumProtocol can not be used as concrete type since it has a Self type requirement, however, in order to achieve this type of behavior in other cases you can create a protocol that the base class conforms to and try to cast objects to that type when you are trying to determine if an object is some subclass of that type.
Consider the following example
class Bitcoin { }
class Ethereum { }
class Wallet<T> {
var usdValue: Double = 0
}
class BitcoinWallet: Wallet<Bitcoin> {
}
class EthereumWallet: Wallet<Ethereum> {
}
let bitcoinWallet = BitcoinWallet() as Any
if let wallet = bitcoinWallet as? Wallet {
print(wallet.usdValue)
}
This will not work, due to the same error that you are referring to:
error: generic parameter 'T' could not be inferred in cast to 'Wallet<_>'
However, if you add the following protocol
protocol WalletType {
var usdValue: Double { get set }
}
and make Wallet conform to that
class Wallet<T>: WalletType
then you can cast values to that protocol and use it as expected:
if let wallet = bitcoinWallet as? WalletType {
print(wallet.usdValue)
}
Is it possible to get the object type from an optional?
For example, if I have a class that has a property that is an optional string, can I somehow just get back the string type?
The exact use case I have is I have many custom classes all of which have a property that is storing another custom class as an optional value. I would like to write a generic function that will create an instance of the object class stored in the optional.
Here is an example of what I am looking for, although .dynamicType does not work since it is an optional:
class Class1 {
}
class Class2 {
var myOp: Class1?
}
var c = Class2()
c.myOp = c.myOp.dynamicType()
Since you wanted to use this with Generics I tried it for you. It works, but it may not be so useful.
First some setup:
This is a helper protocol to make sure our Generic type will have a known init method.
protocol ZeroParameterInit {
init()
}
This is an extension to get the type from an optional:
extension Optional {
var dynamicWrappedType : Wrapped.Type {
return Wrapped.self
}
}
Implemented in your code:
class Class1 : ZeroParameterInit {
required init() {}
}
class Class2 {
var myOp: Class1?
}
var c = Class2()
c.myOp = c.myOp.dynamicWrappedType.init()
Generic implementation:
class Class1 : ZeroParameterInit {
required init() {}
}
class Class2<T where T : ZeroParameterInit> {
var attribute: Optional<T>// used long syntax to remind you of : Optional<Wrapped>
init(attr:T) {
attribute = attr
attribute = nil
}
}
The function to create the instance:
func myFunc<T>(instance: Class2<T>) -> T {
return instance.attribute.dynamicWrappedType.init()
}
Some tests:
let alpha = Class1()
let beta = Class2(attr: alpha)
beta.attribute = myFunc(beta)
The issue:
You can't create an instance of Class2 without informing it about the type of it's generic attribute. So you need to pass it some object/type and that complicates things again.
Some extra methods that might improve how it all works:
init() {
}
let delta = Class2<Class1>()
delta.attribute = myFunc(delta)
init(type:T.Type) {
}
let epsilon = Class2(type: Class1.self)
epsilon.attribute = myFunc(epsilon)
You just need to check if the optional exist:
func myFunc(c: Class2) -> Class1? {
if let c1 = c.myOp{
return c1.dynamicType()
}
return nil
}
OR
func myFunc(c: Class2) -> Class1? {
if c.myOp != nil{
return c.myOp!.dynamicType()
}
return nil
}
Note the your return type need to be optional as well.
Tried this in simulator, seems like doing the right thing, if I understood you
class Class1 {
}
class Class2 {
var myOp: Class1?
}
func myFunc(c: Class2) -> AnyObject {
if let c1 = c.myOp{
return c1.self
}
return c
}
var object = Class2()
object.myOp = Class1()
myFunc(object) // Class1
I have some generic type class but no instance of object to test. What I would like to do is to alter the behavior of the function according to the runtime type.
class MyGenericUtility<SomeGenericClass> {
func myFunction() {
// so far I have tested "is", "==" and "==="
if SomeGenericClass is SomeRealClass {
println("some special stuff there")
}
println("some generic stuff as the name tells")
}
}
You can compare the class type, using SomeGenericClass.self == SomeRealClass.self as,
class MyGenericUtility<SomeGenericClass> {
func myFunction() {
if SomeGenericClass.self == SomeRealClass.self {
print("SomeRealClass stuffs")
} else if SomeGenericClass.self == String.self {
print("String stuffs")
}
}
}
let someRealUtility = MyGenericUtility<SomeRealClass>()
someRealUtility.myFunction()
let stringUtility = MyGenericUtility<String>()
stringUtility.myFunction()
Rather than testing at runtime, you should generally handle this at compile time with constrained extensions (this assumes Swift 2). Doing it this way avoids any need to do unsafe as! casting when you need to access type-specific parts of the instance.
class MyGenericUtility<SomeGenericClass> {
}
// Special handling for `SomeRealClass`
extension MyGenericUtility where SomeGenericClass: SomeRealClass {
func myFunction() {
print("SomeRealClass stuffs")
}
}
// Default handling for any unspecified class
extension MyGenericUtility {
func myFunction() {
print("Other stuffs")
}
}
let someRealUtility = MyGenericUtility<SomeRealClass>()
someRealUtility.myFunction()
let stringUtility = MyGenericUtility<String>()
stringUtility.myFunction()
Note that this is based on inheritance, not equality, so any subclass of SomeRealClass would get the SomeRealClass behavior.
You can't use the generic type directly, you need to use a property of that type when comparing with "is".
class MyGenericUtility<T> {
var a: T
func myFunction() {
if a is Int {
println("some special stuff there")
}
println("some generic stuff as the name tells")
}
init(value: T) {
a = value
}
}
let test = MyGenericUtility(value: 5)
test.myFunction()
// Output: some special stuff there
// some generic stuff as the name tells
let test2 = MyGenericUtility(value: "foo")
test2.myFunction()
// Output: some generic stuff as the name tells
From this answer, I know that I can create an instance of a subclass from a superclass. Yet, I can't figure out how to create an array of the subclass from the superclass.
Drawing on the above example, here's my best shot so far:
class Calculator {
func showKind() { println("regular") }
required init() {}
}
class ScientificCalculator: Calculator {
let model: String = "HP-15C"
override func showKind() { println("\(model) - Scientific") }
required init() {
super.init()
}
}
extension Calculator {
class func createMultiple<T:Calculator>(num: Int) -> T {
let subclass: T.Type = T.self
var calculators = [subclass]()
for i in 0..<num {
calculators.append(subclass())
}
return calculators
}
}
let scis: [ScientificCalculator] = ScientificCalculator.createMultiple(2)
for sci in scis {
sci.showKind()
}
With that code, the line var calculators = [subclass]() shows the error Invalid use of '()' to call a value of non-function type '[T.Type]'.
How can I return an array of ScientificCalculators from Calculator.createMultiple?
You were on the right track but you've made some mistakes.
First you need to return a array of T and not just a single element. So you need to change the return type from T to [T]:
class func createMultiple<T:Calculator>(num: Int) -> [T] {
Also you can just use T to initialize new instances of your subclass like that:
var calculators:[T] = [T]()
But the other parts are correct. So you final method would look like that:
extension Calculator {
class func createMultiple<T:Calculator>(num: Int) -> [T] {
let subclass: T.Type = T.self
var calculators = [T]()
for i in 0..<num {
calculators.append(subclass())
}
return calculators
}
}
Edit
If you are using Swift 1.2 you don't have to deal with subclass anymore and you will be able to use T instead like shown in Airspeeds answer.
calculators.append(T())
EDIT: this behaviour appears to have changed in the latest Swift 1.2 beta. You shouldn’t need to use T.self. T is the type you want to create. But if you are using 1.1, it appears not to work (even if T is the subtype, it creates the supertype), and using the metatype to create the type works around this problem. See end of answer for a 1.1 version.
You don’t need to mess with subclass: T.Type = T.self. Just use T – that itself is the type (or rather, a placeholder for whatever type is specified by the caller):
extension Calculator {
// you meant to return an array of T, right?
class func createMultiple<T: Calculator>(num: Int) -> [T] {
// declare an array of T
var calculators = [T]()
for i in 0..<num {
// create new T and append
calculators.append(T())
}
return calculators
}
}
btw, you can replace that for loop with map:
class func createMultiple<T: Calculator>(num: Int) -> [T] {
return map(0..<num) { _ in T() }
}
If you are still on Swift 1.1, you need to use T.self to work around a problem where the subtype is not properly created:
extension Calculator {
// only use this version if you need this to work in Swift 1.1:
class func createMultiple<T: Calculator>(num: Int) -> [T] {
let subclass: T.Type = T.self
return map(0..<num) { _ in subclass() }
}
}