Enum Associated Value Lookup - swift

I have an enum with different associated value custom classes. Then trying to create a calculated value to determine if one particular type and return the value if so. The below works using a switch in the calculated value but seems a bit clumsy. Are there any cleaner approaches? Thanks.
private enum Situation: Equatable {
case inLocation(LocationMO)
case heldByBeing(BeingMO)
case inContainer(PhysicalObjectMO)
}
private var isInSituation: Situation
var isInLocation: LocationMO? {
switch isInSituation {
case .inLocation(let validLocation): return validLocation
default: return nil
}
}

Related

Sub enumeration rawValue

Consider the following code, where I declared an enum with sub enums inside of it.
enum LocalizeKey {
case message(Messages)
case buttons(Buttons)
enum Buttons: String {
case remove = "Remove"
case add = "Add"
}
enum Messages: String {
case success = "Success"
case failure = "Failure"
}
}
In a normal enum with no sub enums we can easily access .rawValue property and get the raw value of whatever case we picked.
For this case, i created a function like this just to check out what am i getting .
func keyString(for type: LocalizeKey) {
print(type)
}
keyString(for: .message(.failure)) // usage
Problem : there are no other properties than .self to access for this LocalizeKey enum .
What I am trying to achieve: perhaps you can relate by the naming, i am trying to wrap my localized keys, so i can access them easily based on the key type etc, and the rawValue that is refering to the actual key will go into the getLocalizedValue function .
Playground Output : using the function above the playground output was
message(__lldb_expr_21.LocalizeKey.Messages.failure)
Edit: without having to create a variable that switches self on every case, imagine if we had +400 key that would be a huge mess probably.
You need to switch on the type parameter and do pattern matching:
switch type {
case .message(let messages): return messages.rawValue
case .buttons(let buttons): return buttons.rawValue
}
You can also make this an extension of LocalizeKey:
extension LocalizeKey {
var keyString: String {
switch self {
case .message(let messages): return messages.rawValue
case .buttons(let buttons): return buttons.rawValue
}
}
}
You are going to have to switch somewhere. If there are only a handful of "sub-enums", it is probably the easiest to just write a switch manually:
func keyString(for type: LocalizeKey) {
switch type {
case .message(let message):
print(message.rawValue)
case .buttons(let button):
print(button.rawValue)
}
}
If you don't want to write this manually, you either have to change your data structure so it is not needed, or use a code generation tool that generates the boilerplate for you.
Although The mentioned answers do provide the solution, I'd mention the issue of the approach itself:
At this point, each new case (key) has to be added in your switch statement with an associated value, which seems to be undesired boilerplate coding; I assume that you could imagine how it will look like when having many cases in the enums.
Therefore, I'd recommend to follow an approach to be more dynamic instead of adding the value of each case manually in a switch statement. Example:
protocol Localizable {
var value: String { get }
}
extension RawRepresentable where Self: Localizable, Self.RawValue == String {
var value: String { return rawValue }
}
extension CustomStringConvertible where Self: RawRepresentable, Self.RawValue == String {
var description: String { return rawValue }
}
struct LocalizeKey {
enum Buttons: String, Localizable, CustomStringConvertible {
case remove = "Remove"
case add = "Add"
}
enum Messages: String, Localizable, CustomStringConvertible {
case success = "Success"
case failure = "Failure"
}
}
We are applying the same logic for your code, with some improvements to make it easier to maintain.
Based on that, you still able to implement your function as:
func keyString(for type: Localizable) {
print(type)
}
Usage:
keyString(for: LocalizeKey.Buttons.add) // Add
keyString(for: LocalizeKey.Messages.success) // Success
IMO, I find calling it this way seems to be more readable, straightforward rather than the proposed approach (keyString(for: .message(.failure))).

Associating a type with a Swift enum case?

I’m using Swift 2, and I’d like to associate a struct type with each case in an enum.
At the moment, I’ve solved this by adding a function to the enum called type which returns an instance of the relevant type for each case using a switch statement, but I’m wondering if this is necessary. I know you can associate strings, integers etc. with a Swift enum, but is it possible to associate a type, too? All structs of that type conform to the same protocol, if that helps.
This is what I’m doing now, but I’d love to do away with this function:
public enum Thingy {
case FirstThingy
case SecondThingy
func type() -> ThingyType {
switch self {
case .FirstThingy:
return FirstType()
case .SecondThingy:
return SecondType()
}
}
}
I think you are saying that you want the raw value to be of type ThingyType, but that is not possible.
What you could do is make type a computed property, to remove the () and only needing to access it with thingy.type.
public enum Thingy {
case FirstThingy
case SecondThingy
var type: ThingyType {
switch self {
case .FirstThingy:
return FirstType()
case .SecondThingy:
return SecondType()
}
}
}

In Swift, is there a way of determining whether an enum is based on a certain type (eg. String)?

In order to write generic code for an NSValueTransformer, I need to be able to check that an enum is of type String for example. Ie.:
enum TestEnum: String {
case Tall
case Short
}
I am expecially interested in a test that can be used with the guard statement. Something allong the line of:
guard let e = myEnum as <string based enum test> else {
// throw an error
}
Please note that not all enums have raw values. For eample:
enum Test2Enum {
case Fat
case Slim
}
Hence a check on the raw value type can not be used alone for this purpose.
EDIT
After some further investigation it has become clear that NSValueTransformer can not be used to transform Swift enums. Please see my second comment from matt's answer.
First, it's your enums, so you can't not know what type they are. Second, you're not going to receive an enum type, but an enum instance. Third, even if you insist on pretending not to know what type this enum is, it's easy to make a function that can be called only with an enum that has a raw value and check what type that raw value is:
enum E1 {
case One
case Two
}
enum E2 : String {
case One
case Two
}
enum E3 : Int {
case One
case Two
}
func f<T:RawRepresentable>(t:T) -> Bool {
return T.RawValue.self == String.self
}
f(E3.One) // false
f(E2.One) // true
f(E1.One) // compile error
Generics to the rescue :
func enumRawType<T>(of v:T)-> Any?
{ return nil }
func enumRawType<T:RawRepresentable>(of v:T)-> Any?
{
return type(of:v.rawValue)
}
enumRawType(of:E1.One) // nil
enumRawType(of:E2.One) // String.Type
enumRawType(of:E3.One) // Int.Type

Swift - Nested Enums default values

I have an enum as follows
enum AccountForm: String {
case Profile
enum Content: String {
case Feedback
case Likes
}
enum Actions: String {
case Redeem
case Help
}
}
This represents a form, where profile content and actions are sections and the cases are rows.
These resolve to strings and work as expected
AccountForm.Profile.rawValue returns "Profile"
AccountForm.Content.Feedback.rawValue returns "Feedback"
However, I'd like AccountForm.Content.rawValue to return "Content"
Is this possible? Or is there a better way besides enums to achieve this?
I'm guessing you've got an answer to this by now but just in case you didn't try this:
enum AccountForm : String {
case profile
enum Content: String {
static let rawValue = "Content"
case feedback = "Feedback"
case likes = "Likes"
}
enum Actions : String {
static let rawValue = "Actions"
case redeem = "Redeem"
case help = "Help"
}
}
Static properties on both the Content and Actions enumerations should achieve what you want. A word of warning though. By calling the properties rawValue you're obviously implying the returned values are raw values when technically they aren't. If you can I'd think of a better name (maybe sectionTitle?).
Couple of other things to note.
First, you have to define the properties as static as it sounds like you want to call them on the enumeration type (e.g. AccountForm.Content.rawValue) rather than on an individual enumeration case (e.g. AccountForm.Content.feedback.rawValue). I'll leave you to decide whether that makes sense in your context.
Secondly, when Swift 3.0 arrives, the recommendation for enumeration cases is going to be that case labels follow a lowerCamelCase naming convention rather than the UpperCamelCase convention that was recommended in Swift 2.2.
I've followed the Swift 3.0 recommendation here but the result is that explicit raw-value assignments is needed as you won't be able to rely on using the implicit raw-value assignment mechanism assigning a string with an UpperCamelCase representation which is kind of annoying but those are the implications.
Anyway, hope it helps.
enum Typo {
case Bold
case Normal
case Italic
case All
}
enum Format: CustomStringConvertible {
case Header(Typo)
case Text(Typo)
var description:String {
switch self {
case .Header(let value) where value != .All:
return "Header.\(value)"
case .Header(let value) where value == .All:
return "Header"
case .Text(let value) where value == .All:
return "Text"
case .Text(let value) where value != .All:
return "Text.\(value)"
default:
return ""
}
}
}
let a:Format = .Header(.Bold)
let b:Format = .Text(.Italic)
Format.Header(.All) // Header
Format.Text(.Bold) // Text.Bold
Format.Text(.All) // Text

Why class cannot be initialised in Switch statement in Swift

wondering why Swift switch statement does not allow to instantiate classes like in other languages and how to solve this. Would be glad if anyone can help out on this.
In the example i have created a simple Vehicle class and trying to instantiate its sub classes via switch depending on classSetter value. However the final line of print statement cannot print name property of any of the classes if it is instantiated within switch (or seems to any other kind of conditional) statement.
import UIKit
class Vehicle {
var name: String {return ""}
}
class Car: Vehicle {
override var name: String {return "Car"}
}
class Bike: Vehicle {
override var name: String {return "Bike"}
}
var classSetter:Int = 1
switch classSetter {
case 1:
println("Initializing Car")
var vehicle = Car()
case 2:
println("Initialized Bike")
let vehicle = Bike()
default:
println("just defaulted")
}
println("Name property from initialization is \(vehicle.name)")
Your two vehicles are being declared within the switch’s { }, so they only exist in that block (that is their “scope”). They don’t exist outside it, so you can’t refer to them there, hence the error.
The default solution to this (that other answers are giving) is to declare the vehicle as a var outside the switch, but here’s an alternative: wrap the switch in a closure expression and return a value from it. Why do this? Because then you can use let and not var to declare vehicle:
let vehicle: Vehicle = { // start of a closure expression that returns a Vehicle
switch classSetter {
case 1:
println("Initializing Car")
return Car()
case 2:
println("Initialized Bike")
return Bike()
default:
println("just defaulted")
return Vehicle()
}
}() // note the trailing () because you need to _call_ the expression
println("Name property from initialization is \(vehicle.name)")
It would be nice if if and switch were expressions (i.e. evaluated to a result) so you didn’t need this closure, but for now it’s a reasonable workaround.
Note, several of the answers here that use the var approach suggest making vehicle an Optional value (i.e. Vehicle?). This is not necessary – so long as the code is guaranteed to assign vehicle a value before it is used (and the compiler will check this for you), it doesn’t have to be optional. But I still think the closure expression version is a better way.
By the way, you might want to consider using a protocol for Vehicle instead of a base class, since that way you don’t have to give Vehicle a default but invalid implementation for name:
protocol Vehicle {
var name: String { get }
}
// one of the benefits of this is you could
// make Car and Bike structs if desired
struct Car: Vehicle {
var name: String {return "Car"}
}
struct Bike: Vehicle {
var name: String {return "Bike"}
}
Though this would mean you couldn’t have a default return from the switch statement of a Vehicle(). But chances are that would be bad anyway – an optional Vehicle? with nil representing failure might be a better option:
let vehicle: Vehicle? = {
switch classSetter {
case 1:
println("Initializing Car")
return Car()
case 2:
println("Initialized Bike")
return Bike()
default:
println("no valid value")
return nil
}
}()
// btw since vehicle is a Vehicle? you need to unwrap the optional somehow,
// one way is with optional chaining (will print (nil) here)
println("Name property from initialization is \(vehicle?.name)")
If you didn’t want this to be a possibility at all, you could consider making the indicator for different kinds of vehicles an enum so it could only be one of a valid set of vehicles:
enum VehicleKind {
case Bike, Car
}
let classSetter: VehicleKind = .Car
let vehicle: Vehicle = {
switch classSetter {
case .Car:
println("Initializing Car")
return Car()
case .Bike:
println("Initialized Bike")
return Bike()
// no need for a default clause now, since
// those are the only two possibilities
}
}()
Your assumption is incorrect.
The scope of the variable vehicle is inside the switch. You are then trying to access it outside the switch.
You need to create the variable outside the switch. You can still instantiate it inside.
var vehicle: Vehicle?
Switch... {
Case
vehicle = Car()
}
println(vehicle)
Writing on my phone so couldn't give full proper code but this will give you the idea.
What you're doing doesn't work in other languages either. You need to declare the vehicle variable before you enter the switch so that it's in the same scope as your println. After that, you an assign it whatever you need to inside the switch.