So, need you help, I have loops, where I'd like to find first positive element and it will be text of label and exit from all loops, but every time I get last element:
for j in getArrayOfAllTimes[i].timeForGetDifference()
{
switch j - timeNow() {
case let x where x > 0:
nextTimeLabel.text = String(j - timeNow())
break
default:
break
}
}
How get first element > 0?
The first(where:) method of Array
Instead of explicitly breaking out of a loop when a first element that fulfills som predicate is found, you could simply make use of the Array method first(where:).
Since you haven't provided us with a minimal, complete and verifiable example (which you should) we'll construct such an example:
/* Example setup */
struct Foo: CustomStringConvertible {
private let bar: Int
init(_ bar: Int) { self.bar = bar }
func timeForGetDifference() -> Int {
return bar
}
var description: String {
return String(bar)
}
}
func timeNow() -> Int { return 10 }
let getArrayOfAllTimes = [Foo(6), Foo(2), Foo(9), Foo(4), Foo(11), Foo(3), Foo(13)]
// nextTimeLabel: some UILabel
For the example as per above, we could set the text property of the nextTimeLabel as follows, using first(where:) to find the first element fulfilling our predicate, given that it exists (otherwise; will return nil in which case we will not enter the optional binding block below).
if let firstNonNegativeFoo = getArrayOfAllTimes
.first(where: { $0.timeForGetDifference() - timeNow() > 0 }) {
nextTimeLabel.text = String(describing: firstNonNegativeFoo) // 11
}
As to why your own approach does not work as intended: a break statement within a case of a switch statement will simply end the execution of the switch statement (not the loop which is one level above the switch statement.
From the Language Reference - Statements:
Break Statement
A break statement ends program execution of a loop, an if statement,
or a switch statement.
In you case, you've added the break statements as the last statements of each case: here, particularly, the break has truly no effect (since the switch statement would break out anyway, after exiting the case which it entered).
for i in 1...3 {
switch i {
case is Int: print(i); break // redundant 'break'
case _: ()
}
} // 1 2 3
// ... the same
for i in 1...3 {
switch i {
case is Int: print(i)
case _: ()
}
} // 1 2 3
for j in getArrayOfAllTimes[i].timeForGetDifference() {
bool a = NO;
switch j - timeNow() {
case let x where x > 0:
a = YES
nextTimeLabel.text = String(j - timeNow())
break
default:
break
}
if a == YES {
break;
}
}
Is there a way to make this statement shorter?
let number = 1
if number == 0 || number == 1 {
print("ok")
}
like
let number = 1
if number == (0 || 1) {
print("ok")
}
This one doesn't work but I was wondering if there is a shorter way that I am not aware of.
EDIT:
I am aware of the switch case that might be useful when many case appear, but in my case I was more looking for something to use with ternary operators:
let oldNumber = 1
let newNumber = (oldNumber = 0 || 1) ? 4 : 13
Boolean equation expressions are evaluated always providing left (number) and right (0) side.
Not really shorter but you could write
let number = 1
if (0...1).contains(number) {
print("ok")
}
just for variety:
let c = { (number:Int, args:Int...) -> Bool in return args.contains(number) }
let number = 1;
if c(number, 0,1,9,4) {
print("ok")
}
only shorter if you are going to reuse the closure, has the advantage over the 0...1 range because you can specify a disjoint set of numbers. Can be used for inline if statements etc quite neatly
Little poor but nevertheless.
let number = 1
print((0...1).contains(number) ? "ok" : "bad")
Or:
let a = 1
let b = (0...1).contains(a) ? 4 : 5
When checking for several possible cases the switch statement might be more concise.
let number = 1
switch (number) {
case 0, 1:
print("ok")
default:
print("something else")
}
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.