How do I get the count of a Swift enum?

Question

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.)

Answer 1

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

Answer 2

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
    }()
}

Answer 3

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 :)

Answer 4

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)

Answer 5

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

Answer 6

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!")
  }
}

Answer 7

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
   }

Answer 8

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 :)

Answer 9

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.

Answer 10

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

Answer 11

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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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
}

Answer 17

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.

Answer 18

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
}

Answer 19

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)")

Answer 20

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   
}

Answer 21

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)

Answer 22

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.

Answer 23

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

Answer 24

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()
        }
    }
}

Answer 25

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.