Related
I want to create a list of levels with a score required to reach that level. But I'm not sure how best to do it.
For example I tried an enum like this:
enum xpLevels {
case 1 = 0...50
case 2 = 51...100
case 3 = 101...200
}
But this gives me errors saying: "Consecutive declarations on a line must be separated by ';'"
And I'm quite sure I haven't structured this right. Is an enum even the right thing to be using for this? I'd appreciate any help to point me in the right direction.
If you really wanted enums, you might do something like this:
enum xpLevels {
case low = 50
case medium = 100
case high = 200
}
Then, in your code, check if the value is <= low, else <= medium, else it must be "high" ;)
Enums are very unsuitable for this. Try an array instead. For example, you can store:
let levelXps = [0, 50, 100, 200]
Or if you want to write out the level numbers as well:
let levelXps = [0:0, 1:50, 2:100, 3:200].values.sorted()
Then you can have a function that tells you what level it is, given a score:
func level(forScore score: Int) -> Int? {
levelXps.lastIndex(where: { score > $0 })?.advanced(by: 1)
}
This finds the index of the last item in the array that is smaller than the given score, then adds 1 to it. If the score is negative, the function returns nil.
You can create an enumeration with the maximum score for each level as suggested by paulsm4 and declare your enumeration as CaseIterable. Then you can create an initializer to get the corresponding case for the score:
enum Level: Int, CaseIterable {
case one = 50, two = 100, three = 200
}
extension Level {
init(rawValue: Int) {
self = Level.allCases.first { 0...$0.rawValue ~= rawValue } ?? .one
}
}
extension Int {
var level: Level { .init(rawValue: self) }
}
let score = 165
let level = score.level // three
You can also create a custom enumeration of closed ranges. You just need to create a computed property to return each case range:
enum Level: CaseIterable {
case one, two, three, four
}
extension Level: RawRepresentable {
typealias RawValue = ClosedRange<Int>
var rawValue: RawValue {
let range: RawValue
switch self {
case .one: range = 0 ... 50
case .two: range = 51 ... 100
case .three: range = 101 ... 200
case .four: range = 201 ... .max
}
return range
}
init?(rawValue: RawValue) {
guard let level = Level.allCases.first(where: { $0.rawValue == rawValue }) else { return nil }
self = level
}
init(score: Int) {
self = Level.allCases.first { $0.rawValue ~= score } ?? .one
}
}
extension Int {
var level: Level { .init(score: self) }
}
let score = 76
let level = score.level // two
I'm trying to use nested enums to describe my model, in a way that makes illegal states impossible and provides categorisation at the top level. Right now, my (simplified) code is:
enum SportsEvent {
enum RunningEvent {
case sprint
case marathon
}
enum ThrowingEvent {
case hammer
case javelin
case discus
}
case running(event: RunningEvent)
case throwing(event: ThrowingEvent)
func isSameCategory(as other: SportsEvent) -> Bool {
return false
}
}
let sprint = SportsEvent.running(event: .sprint)
let javelin = SportsEvent.throwing(event: .javelin)
let hammer = SportsEvent.throwing(event: .hammer)
sprint.isSameCategory(as: javelin) // False
hammer.isSameCategory(as: javelin) // True
It feels like it should be trivial with an if case ... and a wildcard, but I'm not seeing how to achieve that. I'm hoping a giant switch statement isn't necessary, as my actual model is more complex.
I think you need a switch-statement, with a “compound case” listing all
possible “same value combinations” of the outer enumeration,
plus a default case:
func isSameCategory(as other: SportsEvent) -> Bool {
switch (self, other) {
case (.running, .running),
(.throwing, .throwing):
return true
default:
return false
}
}
Or (attribution goes to #Hamish):
func isSameCategory(as other: SportsEvent) -> Bool {
switch (self, other) {
case (.running, .running),
(.throwing, .throwing):
return true
case (.running, _),
(.throwing, _):
return false
}
}
which has the advantage that the compiler checks that all cases are covered. For an enumeration with n cases that makes 2 * n
cases in the switch statement (which is better than n * n if you checked all possible combinations).
Depending on your use case, you might be able to convert SportEvent to a protocol:
enum RunningEvent {
case sprint
case marathon
}
enum ThrowingEvent {
case hammer
case javelin
case discus
}
enum SportEventCategory {
case running
case throwing
}
protocol SportEvent {
var category: SportEventCategory { get }
}
extension RunningEvent: SportEvent {
var category: SportEventCategory {
return .running
}
}
extension ThrowingEvent: SportEvent {
var category: SportEventCategory {
return .throwing
}
}
let sportEvent1: SportEvent = RunningEvent.sprint
let sportEvent2: SportEvent = ThrowingEvent.hammer
print(sportEvent1.category == sportEvent2.category)
or even as one flat enum:
enum SportEvent {
enum Category {
case running
case throwing
}
case sprint
case marathon
case hammer
case javelin
case discus
var category: Category {
switch self {
case .sprint, .marathon, .hammer:
return .running
case .javelin, .discus:
return .throwing
}
}
}
let sportEvent1: SportEvent = .sprint
let sportEvent2: SportEvent = .marathon
print(sportEvent1.category == sportEvent2.category)
I'd like to have an enum with two cases with a Hashable associated type each conform to Hashable, like so:
enum WordOrNumber {
case Word(String)
case Number(Int)
}
The first idea I had for hashing was the following:
extension WordOrNumber: Hashable {
var hashValue: Int {
switch self {
case let .Word(word):
return word.hashValue & ~1 // ends in 0
case let .Number(number):
return number.hashValue | 1 // ends in 1
}
}
}
What I'm not sure about, is how this will interact with buckets in Swift's Dictionary and Set implementations.
Is it be better to differentiate the two cases by the LSB or MSB, or by a bit somewhere in the middle?
I assume it doesn't make any difference to shift the hashValues left by one and then adding 1, or not, but I would be interested to know if this isn't the case.
EDIT:
I just tested this, and it looks like the hashValue of an Int is the value itself. This is obviously a problem, since you often get ints in succession, like in an enum Bla: Int {}. So my new and improved(?) hashValue is:
extension WordOrNumber: Hashable {
var hashValue: Int {
switch self {
case let .Word(word): // ends in 0
return word.hashValue << 1
case let .Number(number): // ends in 1
return number.hashValue << 1 &+ 1
}
}
}
The above question about LSB and MSB still stands.
Having something like:
extension WordOrNumber: Hashable {
var hashValue: Int {
switch self {
case .Word(let value):
return value.hashValue
case .Number(let value):
return value.hashValue
}
}
static func ==(lhs: WordOrNumber, rhs: WordOrNumber) -> Bool {
switch (lhs, rhs) {
case (.Word(let lhsValue), .Word(let rhsValue)):
return lhsValue == rhsValue
case (.Number(let lhsValue), .Number(let rhsValue)):
return lhsValue == rhsValue
default:
return false
}
}
}
... should be enough.
How can I determine the number of cases in a Swift enum?
(I would like to avoid manually enumerating through all the values, or using the old "enum_count trick" if possible.)
As of Swift 4.2 (Xcode 10) you can declare
conformance to the CaseIterable protocol, this works for all
enumerations without associated values:
enum Stuff: CaseIterable {
case first
case second
case third
case forth
}
The number of cases is now simply obtained with
print(Stuff.allCases.count) // 4
For more information, see
SE-0194 Derived Collection of Enum Cases
I have a blog post that goes into more detail on this, but as long as your enum's raw type is an integer, you can add a count this way:
enum Reindeer: Int {
case Dasher, Dancer, Prancer, Vixen, Comet, Cupid, Donner, Blitzen
case Rudolph
static let count: Int = {
var max: Int = 0
while let _ = Reindeer(rawValue: max) { max += 1 }
return max
}()
}
Xcode 10 update
Adopt the CaseIterable protocol in the enum, it provides a static allCases property which contains all enum cases as a Collection . Just use of its count property to know how many cases the enum has.
See Martin's answer for an example (and upvote his answers rather than mine)
Warning: the method below doesn't seem to work anymore.
I'm not aware of any generic method to count the number of enum cases. I've noticed however that the hashValue property of the enum cases is incremental, starting from zero, and with the order determined by the order in which the cases are declared. So, the hash of the last enum plus one corresponds to the number of cases.
For example with this enum:
enum Test {
case ONE
case TWO
case THREE
case FOUR
static var count: Int { return Test.FOUR.hashValue + 1}
}
count returns 4.
I cannot say if that's a rule or if it will ever change in the future, so use at your own risk :)
I define a reusable protocol which automatically performs the case count based on the approach posted by Nate Cook.
protocol CaseCountable {
static var caseCount: Int { get }
}
extension CaseCountable where Self: RawRepresentable, Self.RawValue == Int {
internal static var caseCount: Int {
var count = 0
while let _ = Self(rawValue: count) {
count += 1
}
return count
}
}
Then I can reuse this protocol for example as follows:
enum Planet : Int, CaseCountable {
case Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune
}
//..
print(Planet.caseCount)
Create static allValues array as shown in this answer
enum ProductCategory : String {
case Washers = "washers", Dryers = "dryers", Toasters = "toasters"
static let allValues = [Washers, Dryers, Toasters]
}
...
let count = ProductCategory.allValues.count
This is also helpful when you want to enumerate the values, and works for all Enum types
If the implementation doesn't have anything against using integer enums, you could add an extra member value called Count to represent the number of members in the enum - see example below:
enum TableViewSections : Int {
case Watchlist
case AddButton
case Count
}
Now you can get the number of members in the enum by calling, TableViewSections.Count.rawValue which will return 2 for the example above.
When you're handling the enum in a switch statement, make sure to throw an assertion failure when encountering the Count member where you don't expect it:
func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
let currentSection: TableViewSections = TableViewSections.init(rawValue:section)!
switch(currentSection) {
case .Watchlist:
return watchlist.count
case .AddButton:
return 1
case .Count:
assert(false, "Invalid table view section!")
}
}
This kind of function is able to return the count of your enum.
Swift 2:
func enumCount<T: Hashable>(_: T.Type) -> Int {
var i = 1
while (withUnsafePointer(&i) { UnsafePointer<T>($0).memory }).hashValue != 0 {
i += 1
}
return i
}
Swift 3:
func enumCount<T: Hashable>(_: T.Type) -> Int {
var i = 1
while (withUnsafePointer(to: &i, {
return $0.withMemoryRebound(to: T.self, capacity: 1, { return $0.pointee })
}).hashValue != 0) {
i += 1
}
return i
}
String Enum with Index
enum eEventTabType : String {
case Search = "SEARCH"
case Inbox = "INBOX"
case Accepted = "ACCEPTED"
case Saved = "SAVED"
case Declined = "DECLINED"
case Organized = "ORGANIZED"
static let allValues = [Search, Inbox, Accepted, Saved, Declined, Organized]
var index : Int {
return eEventTabType.allValues.indexOf(self)!
}
}
count : eEventTabType.allValues.count
index : objeEventTabType.index
Enjoy :)
Oh hey everybody, what about unit tests?
func testEnumCountIsEqualToNumberOfItemsInEnum() {
var max: Int = 0
while let _ = Test(rawValue: max) { max += 1 }
XCTAssert(max == Test.count)
}
This combined with Antonio's solution:
enum Test {
case one
case two
case three
case four
static var count: Int { return Test.four.hashValue + 1}
}
in the main code gives you O(1) plus you get a failing test if someone adds an enum case five and doesn't update the implementation of count.
This function relies on 2 undocumented current(Swift 1.1) enum behavior:
Memory layout of enum is just a index of case. If case count is from 2 to 256, it's UInt8.
If the enum was bit-casted from invalid case index, its hashValue is 0
So use at your own risk :)
func enumCaseCount<T:Hashable>(t:T.Type) -> Int {
switch sizeof(t) {
case 0:
return 1
case 1:
for i in 2..<256 {
if unsafeBitCast(UInt8(i), t).hashValue == 0 {
return i
}
}
return 256
case 2:
for i in 257..<65536 {
if unsafeBitCast(UInt16(i), t).hashValue == 0 {
return i
}
}
return 65536
default:
fatalError("too many")
}
}
Usage:
enum Foo:String {
case C000 = "foo"
case C001 = "bar"
case C002 = "baz"
}
enumCaseCount(Foo) // -> 3
I wrote a simple extension which gives all enums where raw value is integer a count property:
extension RawRepresentable where RawValue: IntegerType {
static var count: Int {
var i: RawValue = 0
while let _ = Self(rawValue: i) {
i = i.successor()
}
return Int(i.toIntMax())
}
}
Unfortunately it gives the count property to OptionSetType where it won't work properly, so here is another version which requires explicit conformance to CaseCountable protocol for any enum which cases you want to count:
protocol CaseCountable: RawRepresentable {}
extension CaseCountable where RawValue: IntegerType {
static var count: Int {
var i: RawValue = 0
while let _ = Self(rawValue: i) {
i = i.successor()
}
return Int(i.toIntMax())
}
}
It's very similar to the approach posted by Tom Pelaia, but works with all integer types.
enum EnumNameType: Int {
case first
case second
case third
static var count: Int { return EnumNameType.third.rawValue + 1 }
}
print(EnumNameType.count) //3
OR
enum EnumNameType: Int {
case first
case second
case third
case count
}
print(EnumNameType.count.rawValue) //3
*On Swift 4.2 (Xcode 10) can use:
enum EnumNameType: CaseIterable {
case first
case second
case third
}
print(EnumNameType.allCases.count) //3
Of course, it's not dynamic but for many uses you can get by with a static var added to your Enum
static var count: Int{ return 7 }
and then use it as EnumName.count
For my use case, in a codebase where multiple people could be adding keys to an enum, and these cases should all be available in the allKeys property, it's important that allKeys be validated against the keys in the enum. This is to avoid someone forgetting to add their key to the all keys list. Matching the count of the allKeys array(first created as a set to avoid dupes) against the number of keys in the enum ensures that they are all present.
Some of the answers above show the way to achieve this in Swift 2 but none work in Swift 3. Here is the Swift 3 formatted version:
static func enumCount<T: Hashable>(_ t: T.Type) -> Int {
var i = 1
while (withUnsafePointer(to: &i) {
$0.withMemoryRebound(to:t.self, capacity:1) { $0.pointee.hashValue != 0 }
}) {
i += 1
}
return i
}
static var allKeys: [YourEnumTypeHere] {
var enumSize = enumCount(YourEnumTypeHere.self)
let keys: Set<YourEnumTypeHere> = [.all, .your, .cases, .here]
guard keys.count == enumSize else {
fatalError("Missmatch between allKeys(\(keys.count)) and actual keys(\(enumSize)) in enum.")
}
return Array(keys)
}
Depending on your use case, you might want to just run the test in development to avoid the overhead of using allKeys on each request
Why do you make it all so complex? The SIMPLEST counter of Int enum is to add:
case Count
In the end. And... viola - now you have the count - fast and simple
enum WeekDays : String , CaseIterable
{
case monday = "Mon"
case tuesday = "Tue"
case wednesday = "Wed"
case thursday = "Thu"
case friday = "Fri"
case saturday = "Sat"
case sunday = "Sun"
}
var weekdays = WeekDays.AllCases()
print("\(weekdays.count)")
If you don't want to base your code in the last enum you can create this function inside your enum.
func getNumberOfItems() -> Int {
var i:Int = 0
var exit:Bool = false
while !exit {
if let menuIndex = MenuIndex(rawValue: i) {
i++
}else{
exit = true
}
}
return i
}
A Swift 3 version working with Int type enums:
protocol CaseCountable: RawRepresentable {}
extension CaseCountable where RawValue == Int {
static var count: RawValue {
var i: RawValue = 0
while let _ = Self(rawValue: i) { i += 1 }
return i
}
}
Credits: Based on the answers by bzz and Nate Cook.
Generic IntegerType (in Swift 3 renamed to Integer) is not supported, as it's a heavily fragmented generic type which lacks a lot of functions. successor is not available with Swift 3 anymore.
Be aware that the comment from Code Commander to Nate Cooks answer is still valid:
While nice because you don't need to hardcode a value, this will
instantiate every enum value each time it is called. That is O(n)
instead of O(1).
As far as I know there is currently no workaround when using this as protocol extension (and not implementing in each enum like Nate Cook did) due to static stored properties not being supported in generic types.
Anyway, for small enums this should be no issue. A typical use case would be the section.count for UITableViews as already mentioned by Zorayr.
Extending Matthieu Riegler answer, this is a solution for Swift 3 that doesn't require the use of generics, and can be easily called using the enum type with EnumType.elementsCount:
extension RawRepresentable where Self: Hashable {
// Returns the number of elements in a RawRepresentable data structure
static var elementsCount: Int {
var i = 1
while (withUnsafePointer(to: &i, {
return $0.withMemoryRebound(to: self, capacity: 1, { return
$0.pointee })
}).hashValue != 0) {
i += 1
}
return i
}
I solved this problem for myself by creating a protocol (EnumIntArray) and a global utility function (enumIntArray) that make it very easy to add an "All" variable to any enum (using swift 1.2). The "all" variable will contain an array of all elements in the enum so you can use all.count for the count
It only works with enums that use raw values of type Int but perhaps it can provide some inspiration for other types.
It also addresses the "gap in numbering" and "excessive time to iterate" issues I've read above and elsewhere.
The idea is to add the EnumIntArray protocol to your enum and then define an "all" static variable by calling the enumIntArray function and provide it with the first element (and the last if there are gaps in the numbering).
Because the static variable is only initialized once, the overhead of going through all raw values only hits your program once.
example (without gaps) :
enum Animals:Int, EnumIntArray
{
case Cat=1, Dog, Rabbit, Chicken, Cow
static var all = enumIntArray(Animals.Cat)
}
example (with gaps) :
enum Animals:Int, EnumIntArray
{
case Cat = 1, Dog,
case Rabbit = 10, Chicken, Cow
static var all = enumIntArray(Animals.Cat, Animals.Cow)
}
Here's the code that implements it:
protocol EnumIntArray
{
init?(rawValue:Int)
var rawValue:Int { get }
}
func enumIntArray<T:EnumIntArray>(firstValue:T, _ lastValue:T? = nil) -> [T]
{
var result:[T] = []
var rawValue = firstValue.rawValue
while true
{
if let enumValue = T(rawValue:rawValue++)
{ result.append(enumValue) }
else if lastValue == nil
{ break }
if lastValue != nil
&& rawValue > lastValue!.rawValue
{ break }
}
return result
}
Or you can just define the _count outside the enum, and attach it statically:
let _count: Int = {
var max: Int = 0
while let _ = EnumName(rawValue: max) { max += 1 }
return max
}()
enum EnumName: Int {
case val0 = 0
case val1
static let count = _count
}
That way no matter how many enums you create, it'll only ever be created once.
(delete this answer if static does that)
The following method comes from CoreKit and is similar to the answers some others have suggested. This works with Swift 4.
public protocol EnumCollection: Hashable {
static func cases() -> AnySequence<Self>
static var allValues: [Self] { get }
}
public extension EnumCollection {
public static func cases() -> AnySequence<Self> {
return AnySequence { () -> AnyIterator<Self> in
var raw = 0
return AnyIterator {
let current: Self = withUnsafePointer(to: &raw) { $0.withMemoryRebound(to: self, capacity: 1) { $0.pointee } }
guard current.hashValue == raw else {
return nil
}
raw += 1
return current
}
}
}
public static var allValues: [Self] {
return Array(self.cases())
}
}
enum Weekdays: String, EnumCollection {
case sunday, monday, tuesday, wednesday, thursday, friday, saturday
}
Then you just need to just call Weekdays.allValues.count.
Just want to share a solution when you have an enum with associated values.
enum SomeEnum {
case one
case two(String)
case three(String, Int)
}
CaseIterable doesn't provide allCases automatically.
We can't provide a raw type like Int for your enum to calculate cases count somehow.
What we can do is to use power of switch and fallthrough keyword.
extension SomeEnum {
static var casesCount: Int {
var sum = 0
switch Self.one { // Potential problem
case one:
sum += 1
fallthrough
case two:
sum += 1
fallthrough
case three:
sum += 1
}
return sum
}
}
So now you can say SomeEnum.casesCount.
Remarks:
We still have a problem with switch Self.one {..., we hardcoded the first case. You can easily hack this solution. But I used it just for unit tests so that was not a problem.
If you often need to get cases count in enums with associated values, think about code generation.
struct HashableSequence<T: Hashable>: SequenceType {
func generate() -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
if next.hashValue == i {
i += 1
return next
}
return nil
}
}
}
extension Hashable {
static func enumCases() -> Array<Self> {
return Array(HashableSequence())
}
static var enumCount: Int {
return enumCases().enumCount
}
}
enum E {
case A
case B
case C
}
E.enumCases() // [A, B, C]
E.enumCount // 3
but be careful with usage on non-enum types. Some workaround could be:
struct HashableSequence<T: Hashable>: SequenceType {
func generate() -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
guard sizeof(T) == 1 else {
return nil
}
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
if next.hashValue == i {
i += 1
return next
}
return nil
}
}
}
extension Hashable {
static func enumCases() -> Array<Self> {
return Array(HashableSequence())
}
static var enumCount: Int {
return enumCases().count
}
}
enum E {
case A
case B
case C
}
Bool.enumCases() // [false, true]
Bool.enumCount // 2
String.enumCases() // []
String.enumCount // 0
Int.enumCases() // []
Int.enumCount // 0
E.enumCases() // [A, B, C]
E.enumCount // 4
It can use a static constant which contains the last value of the enumeration plus one.
enum Color : Int {
case Red, Orange, Yellow, Green, Cyan, Blue, Purple
static let count: Int = Color.Purple.rawValue + 1
func toUIColor() -> UIColor{
switch self {
case .Red:
return UIColor.redColor()
case .Orange:
return UIColor.orangeColor()
case .Yellow:
return UIColor.yellowColor()
case .Green:
return UIColor.greenColor()
case .Cyan:
return UIColor.cyanColor()
case .Blue:
return UIColor.blueColor()
case .Purple:
return UIColor.redColor()
}
}
}
This is minor, but I think a better O(1) solution would be the following (ONLY if your enum is Int starting at x, etc.):
enum Test : Int {
case ONE = 1
case TWO
case THREE
case FOUR // if you later need to add additional enums add above COUNT so COUNT is always the last enum value
case COUNT
static var count: Int { return Test.COUNT.rawValue } // note if your enum starts at 0, some other number, etc. you'll need to add on to the raw value the differential
}
The current selected answer I still believe is the best answer for all enums, unless you are working with Int then I recommend this solution.
How can I determine the number of cases in a Swift enum?
(I would like to avoid manually enumerating through all the values, or using the old "enum_count trick" if possible.)
As of Swift 4.2 (Xcode 10) you can declare
conformance to the CaseIterable protocol, this works for all
enumerations without associated values:
enum Stuff: CaseIterable {
case first
case second
case third
case forth
}
The number of cases is now simply obtained with
print(Stuff.allCases.count) // 4
For more information, see
SE-0194 Derived Collection of Enum Cases
I have a blog post that goes into more detail on this, but as long as your enum's raw type is an integer, you can add a count this way:
enum Reindeer: Int {
case Dasher, Dancer, Prancer, Vixen, Comet, Cupid, Donner, Blitzen
case Rudolph
static let count: Int = {
var max: Int = 0
while let _ = Reindeer(rawValue: max) { max += 1 }
return max
}()
}
Xcode 10 update
Adopt the CaseIterable protocol in the enum, it provides a static allCases property which contains all enum cases as a Collection . Just use of its count property to know how many cases the enum has.
See Martin's answer for an example (and upvote his answers rather than mine)
Warning: the method below doesn't seem to work anymore.
I'm not aware of any generic method to count the number of enum cases. I've noticed however that the hashValue property of the enum cases is incremental, starting from zero, and with the order determined by the order in which the cases are declared. So, the hash of the last enum plus one corresponds to the number of cases.
For example with this enum:
enum Test {
case ONE
case TWO
case THREE
case FOUR
static var count: Int { return Test.FOUR.hashValue + 1}
}
count returns 4.
I cannot say if that's a rule or if it will ever change in the future, so use at your own risk :)
I define a reusable protocol which automatically performs the case count based on the approach posted by Nate Cook.
protocol CaseCountable {
static var caseCount: Int { get }
}
extension CaseCountable where Self: RawRepresentable, Self.RawValue == Int {
internal static var caseCount: Int {
var count = 0
while let _ = Self(rawValue: count) {
count += 1
}
return count
}
}
Then I can reuse this protocol for example as follows:
enum Planet : Int, CaseCountable {
case Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune
}
//..
print(Planet.caseCount)
Create static allValues array as shown in this answer
enum ProductCategory : String {
case Washers = "washers", Dryers = "dryers", Toasters = "toasters"
static let allValues = [Washers, Dryers, Toasters]
}
...
let count = ProductCategory.allValues.count
This is also helpful when you want to enumerate the values, and works for all Enum types
If the implementation doesn't have anything against using integer enums, you could add an extra member value called Count to represent the number of members in the enum - see example below:
enum TableViewSections : Int {
case Watchlist
case AddButton
case Count
}
Now you can get the number of members in the enum by calling, TableViewSections.Count.rawValue which will return 2 for the example above.
When you're handling the enum in a switch statement, make sure to throw an assertion failure when encountering the Count member where you don't expect it:
func tableView(tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
let currentSection: TableViewSections = TableViewSections.init(rawValue:section)!
switch(currentSection) {
case .Watchlist:
return watchlist.count
case .AddButton:
return 1
case .Count:
assert(false, "Invalid table view section!")
}
}
This kind of function is able to return the count of your enum.
Swift 2:
func enumCount<T: Hashable>(_: T.Type) -> Int {
var i = 1
while (withUnsafePointer(&i) { UnsafePointer<T>($0).memory }).hashValue != 0 {
i += 1
}
return i
}
Swift 3:
func enumCount<T: Hashable>(_: T.Type) -> Int {
var i = 1
while (withUnsafePointer(to: &i, {
return $0.withMemoryRebound(to: T.self, capacity: 1, { return $0.pointee })
}).hashValue != 0) {
i += 1
}
return i
}
String Enum with Index
enum eEventTabType : String {
case Search = "SEARCH"
case Inbox = "INBOX"
case Accepted = "ACCEPTED"
case Saved = "SAVED"
case Declined = "DECLINED"
case Organized = "ORGANIZED"
static let allValues = [Search, Inbox, Accepted, Saved, Declined, Organized]
var index : Int {
return eEventTabType.allValues.indexOf(self)!
}
}
count : eEventTabType.allValues.count
index : objeEventTabType.index
Enjoy :)
Oh hey everybody, what about unit tests?
func testEnumCountIsEqualToNumberOfItemsInEnum() {
var max: Int = 0
while let _ = Test(rawValue: max) { max += 1 }
XCTAssert(max == Test.count)
}
This combined with Antonio's solution:
enum Test {
case one
case two
case three
case four
static var count: Int { return Test.four.hashValue + 1}
}
in the main code gives you O(1) plus you get a failing test if someone adds an enum case five and doesn't update the implementation of count.
This function relies on 2 undocumented current(Swift 1.1) enum behavior:
Memory layout of enum is just a index of case. If case count is from 2 to 256, it's UInt8.
If the enum was bit-casted from invalid case index, its hashValue is 0
So use at your own risk :)
func enumCaseCount<T:Hashable>(t:T.Type) -> Int {
switch sizeof(t) {
case 0:
return 1
case 1:
for i in 2..<256 {
if unsafeBitCast(UInt8(i), t).hashValue == 0 {
return i
}
}
return 256
case 2:
for i in 257..<65536 {
if unsafeBitCast(UInt16(i), t).hashValue == 0 {
return i
}
}
return 65536
default:
fatalError("too many")
}
}
Usage:
enum Foo:String {
case C000 = "foo"
case C001 = "bar"
case C002 = "baz"
}
enumCaseCount(Foo) // -> 3
I wrote a simple extension which gives all enums where raw value is integer a count property:
extension RawRepresentable where RawValue: IntegerType {
static var count: Int {
var i: RawValue = 0
while let _ = Self(rawValue: i) {
i = i.successor()
}
return Int(i.toIntMax())
}
}
Unfortunately it gives the count property to OptionSetType where it won't work properly, so here is another version which requires explicit conformance to CaseCountable protocol for any enum which cases you want to count:
protocol CaseCountable: RawRepresentable {}
extension CaseCountable where RawValue: IntegerType {
static var count: Int {
var i: RawValue = 0
while let _ = Self(rawValue: i) {
i = i.successor()
}
return Int(i.toIntMax())
}
}
It's very similar to the approach posted by Tom Pelaia, but works with all integer types.
enum EnumNameType: Int {
case first
case second
case third
static var count: Int { return EnumNameType.third.rawValue + 1 }
}
print(EnumNameType.count) //3
OR
enum EnumNameType: Int {
case first
case second
case third
case count
}
print(EnumNameType.count.rawValue) //3
*On Swift 4.2 (Xcode 10) can use:
enum EnumNameType: CaseIterable {
case first
case second
case third
}
print(EnumNameType.allCases.count) //3
Of course, it's not dynamic but for many uses you can get by with a static var added to your Enum
static var count: Int{ return 7 }
and then use it as EnumName.count
For my use case, in a codebase where multiple people could be adding keys to an enum, and these cases should all be available in the allKeys property, it's important that allKeys be validated against the keys in the enum. This is to avoid someone forgetting to add their key to the all keys list. Matching the count of the allKeys array(first created as a set to avoid dupes) against the number of keys in the enum ensures that they are all present.
Some of the answers above show the way to achieve this in Swift 2 but none work in Swift 3. Here is the Swift 3 formatted version:
static func enumCount<T: Hashable>(_ t: T.Type) -> Int {
var i = 1
while (withUnsafePointer(to: &i) {
$0.withMemoryRebound(to:t.self, capacity:1) { $0.pointee.hashValue != 0 }
}) {
i += 1
}
return i
}
static var allKeys: [YourEnumTypeHere] {
var enumSize = enumCount(YourEnumTypeHere.self)
let keys: Set<YourEnumTypeHere> = [.all, .your, .cases, .here]
guard keys.count == enumSize else {
fatalError("Missmatch between allKeys(\(keys.count)) and actual keys(\(enumSize)) in enum.")
}
return Array(keys)
}
Depending on your use case, you might want to just run the test in development to avoid the overhead of using allKeys on each request
Why do you make it all so complex? The SIMPLEST counter of Int enum is to add:
case Count
In the end. And... viola - now you have the count - fast and simple
enum WeekDays : String , CaseIterable
{
case monday = "Mon"
case tuesday = "Tue"
case wednesday = "Wed"
case thursday = "Thu"
case friday = "Fri"
case saturday = "Sat"
case sunday = "Sun"
}
var weekdays = WeekDays.AllCases()
print("\(weekdays.count)")
If you don't want to base your code in the last enum you can create this function inside your enum.
func getNumberOfItems() -> Int {
var i:Int = 0
var exit:Bool = false
while !exit {
if let menuIndex = MenuIndex(rawValue: i) {
i++
}else{
exit = true
}
}
return i
}
A Swift 3 version working with Int type enums:
protocol CaseCountable: RawRepresentable {}
extension CaseCountable where RawValue == Int {
static var count: RawValue {
var i: RawValue = 0
while let _ = Self(rawValue: i) { i += 1 }
return i
}
}
Credits: Based on the answers by bzz and Nate Cook.
Generic IntegerType (in Swift 3 renamed to Integer) is not supported, as it's a heavily fragmented generic type which lacks a lot of functions. successor is not available with Swift 3 anymore.
Be aware that the comment from Code Commander to Nate Cooks answer is still valid:
While nice because you don't need to hardcode a value, this will
instantiate every enum value each time it is called. That is O(n)
instead of O(1).
As far as I know there is currently no workaround when using this as protocol extension (and not implementing in each enum like Nate Cook did) due to static stored properties not being supported in generic types.
Anyway, for small enums this should be no issue. A typical use case would be the section.count for UITableViews as already mentioned by Zorayr.
Extending Matthieu Riegler answer, this is a solution for Swift 3 that doesn't require the use of generics, and can be easily called using the enum type with EnumType.elementsCount:
extension RawRepresentable where Self: Hashable {
// Returns the number of elements in a RawRepresentable data structure
static var elementsCount: Int {
var i = 1
while (withUnsafePointer(to: &i, {
return $0.withMemoryRebound(to: self, capacity: 1, { return
$0.pointee })
}).hashValue != 0) {
i += 1
}
return i
}
I solved this problem for myself by creating a protocol (EnumIntArray) and a global utility function (enumIntArray) that make it very easy to add an "All" variable to any enum (using swift 1.2). The "all" variable will contain an array of all elements in the enum so you can use all.count for the count
It only works with enums that use raw values of type Int but perhaps it can provide some inspiration for other types.
It also addresses the "gap in numbering" and "excessive time to iterate" issues I've read above and elsewhere.
The idea is to add the EnumIntArray protocol to your enum and then define an "all" static variable by calling the enumIntArray function and provide it with the first element (and the last if there are gaps in the numbering).
Because the static variable is only initialized once, the overhead of going through all raw values only hits your program once.
example (without gaps) :
enum Animals:Int, EnumIntArray
{
case Cat=1, Dog, Rabbit, Chicken, Cow
static var all = enumIntArray(Animals.Cat)
}
example (with gaps) :
enum Animals:Int, EnumIntArray
{
case Cat = 1, Dog,
case Rabbit = 10, Chicken, Cow
static var all = enumIntArray(Animals.Cat, Animals.Cow)
}
Here's the code that implements it:
protocol EnumIntArray
{
init?(rawValue:Int)
var rawValue:Int { get }
}
func enumIntArray<T:EnumIntArray>(firstValue:T, _ lastValue:T? = nil) -> [T]
{
var result:[T] = []
var rawValue = firstValue.rawValue
while true
{
if let enumValue = T(rawValue:rawValue++)
{ result.append(enumValue) }
else if lastValue == nil
{ break }
if lastValue != nil
&& rawValue > lastValue!.rawValue
{ break }
}
return result
}
Or you can just define the _count outside the enum, and attach it statically:
let _count: Int = {
var max: Int = 0
while let _ = EnumName(rawValue: max) { max += 1 }
return max
}()
enum EnumName: Int {
case val0 = 0
case val1
static let count = _count
}
That way no matter how many enums you create, it'll only ever be created once.
(delete this answer if static does that)
The following method comes from CoreKit and is similar to the answers some others have suggested. This works with Swift 4.
public protocol EnumCollection: Hashable {
static func cases() -> AnySequence<Self>
static var allValues: [Self] { get }
}
public extension EnumCollection {
public static func cases() -> AnySequence<Self> {
return AnySequence { () -> AnyIterator<Self> in
var raw = 0
return AnyIterator {
let current: Self = withUnsafePointer(to: &raw) { $0.withMemoryRebound(to: self, capacity: 1) { $0.pointee } }
guard current.hashValue == raw else {
return nil
}
raw += 1
return current
}
}
}
public static var allValues: [Self] {
return Array(self.cases())
}
}
enum Weekdays: String, EnumCollection {
case sunday, monday, tuesday, wednesday, thursday, friday, saturday
}
Then you just need to just call Weekdays.allValues.count.
Just want to share a solution when you have an enum with associated values.
enum SomeEnum {
case one
case two(String)
case three(String, Int)
}
CaseIterable doesn't provide allCases automatically.
We can't provide a raw type like Int for your enum to calculate cases count somehow.
What we can do is to use power of switch and fallthrough keyword.
extension SomeEnum {
static var casesCount: Int {
var sum = 0
switch Self.one { // Potential problem
case one:
sum += 1
fallthrough
case two:
sum += 1
fallthrough
case three:
sum += 1
}
return sum
}
}
So now you can say SomeEnum.casesCount.
Remarks:
We still have a problem with switch Self.one {..., we hardcoded the first case. You can easily hack this solution. But I used it just for unit tests so that was not a problem.
If you often need to get cases count in enums with associated values, think about code generation.
struct HashableSequence<T: Hashable>: SequenceType {
func generate() -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
if next.hashValue == i {
i += 1
return next
}
return nil
}
}
}
extension Hashable {
static func enumCases() -> Array<Self> {
return Array(HashableSequence())
}
static var enumCount: Int {
return enumCases().enumCount
}
}
enum E {
case A
case B
case C
}
E.enumCases() // [A, B, C]
E.enumCount // 3
but be careful with usage on non-enum types. Some workaround could be:
struct HashableSequence<T: Hashable>: SequenceType {
func generate() -> AnyGenerator<T> {
var i = 0
return AnyGenerator {
guard sizeof(T) == 1 else {
return nil
}
let next = withUnsafePointer(&i) { UnsafePointer<T>($0).memory }
if next.hashValue == i {
i += 1
return next
}
return nil
}
}
}
extension Hashable {
static func enumCases() -> Array<Self> {
return Array(HashableSequence())
}
static var enumCount: Int {
return enumCases().count
}
}
enum E {
case A
case B
case C
}
Bool.enumCases() // [false, true]
Bool.enumCount // 2
String.enumCases() // []
String.enumCount // 0
Int.enumCases() // []
Int.enumCount // 0
E.enumCases() // [A, B, C]
E.enumCount // 4
It can use a static constant which contains the last value of the enumeration plus one.
enum Color : Int {
case Red, Orange, Yellow, Green, Cyan, Blue, Purple
static let count: Int = Color.Purple.rawValue + 1
func toUIColor() -> UIColor{
switch self {
case .Red:
return UIColor.redColor()
case .Orange:
return UIColor.orangeColor()
case .Yellow:
return UIColor.yellowColor()
case .Green:
return UIColor.greenColor()
case .Cyan:
return UIColor.cyanColor()
case .Blue:
return UIColor.blueColor()
case .Purple:
return UIColor.redColor()
}
}
}
This is minor, but I think a better O(1) solution would be the following (ONLY if your enum is Int starting at x, etc.):
enum Test : Int {
case ONE = 1
case TWO
case THREE
case FOUR // if you later need to add additional enums add above COUNT so COUNT is always the last enum value
case COUNT
static var count: Int { return Test.COUNT.rawValue } // note if your enum starts at 0, some other number, etc. you'll need to add on to the raw value the differential
}
The current selected answer I still believe is the best answer for all enums, unless you are working with Int then I recommend this solution.