I tried to add an override for enumerate that handles the case of an optional sequence (without crashing). The idea was that if the sequence is valid aka .Some, it would enumerate that sequence, otherwise enumerate an empty sequence:
func enumerate2<Seq : SequenceType>(base: Seq?) -> EnumerateSequence<Seq> {
// if optional sequence is specified
if let b = base { return enumerate(b) }
// enumerate empty sequence
let a = Array<Seq.Generator.Element>()
return Swift.enumerate(a)
}
func enumerate3<Seq : SequenceType, T where T == Seq.Generator.Element>(base: Seq?) -> EnumerateSequence<Seq> {
// if optional sequence is specified
if let b = base { return enumerate(b) }
// enumerate empty sequence
let a = Array<T>()
return Swift.enumerate(a)
}
I am seeing the error:
'Array<Seq.Generator.Element>' does not conform to protocol 'GeneratorType'
on the last return lines: return Swift.enumerate(a)
This confuses me as EnumerateSequence<Seq> does not appear to conform to GeneratorType. This seems like a simple enough exercise, what could I be missing?
Note that the above code is split apart for illustration, and the suffixes 2 and 3 are to keep remove ambiguity.
Edit:
One work around is to return an Optional sequence instead of an empty sequence.
func enumerate<Seq : SequenceType>(base: Seq?) -> EnumerateSequence<Seq>? {
return base != nil ? enumerate(base) : nil
}
The problem then shifts to safely enumerating optionals:
public func each<S:SequenceType, T where T == S.Generator.Element>
(seq: S, with fn:(T)->()) {
for s in seq { fn(s) }
}
public func each<S:SequenceType, T where T == S.Generator.Element>
(seq: S?, with fn:(T)->()) {
if let some = seq {
for s in some { fn(s) }
}
}
let es = enumerate(["a", "b", "c", "d"])
each(es) { p in
println("\(p.0), \(p.1)")
}
let b:[Int]? = nil
let nes = enumerate(b)
each(nes) { p in
println("\(p.0), \(p.1)")
}
println("done")
which produces:
0, a
1, b
2, c
3, d
done
Perhaps though, the explicit use of if let... is better just for being more obvious, but I am still curious about the initial compilation error.
This isn't working because your function declaration says you're returning an EnumerateSequence<Seq>, but that last line would return a EnumerateSequence <Array<Seq.Generator.Element>>—those aren't the same, so the compiler won't allow it.
You need to be able to create an empty instance of Seq type, but the SequenceType protocol doesn't specify an initializer - you need to go down the chain to ExtensibleCollectionType to find a protocol with an initializer, so change your generic constraint to that. Then you can do this:
func enumerate3<Seq : ExtensibleCollectionType>(base: Seq?) -> EnumerateSequence<Seq> {
// if optional sequence is specified
if let b = base { return enumerate(b) }
// enumerate empty sequence
let a = Seq()
return enumerate(a)
}
Note: if you look through the Swift headers, it won't show you that Array conforms to ExtensibleCollectionProtocol, but it actually does through a "hidden" ArrayType protocol.
There was a good solution to this posted on dev forurms.
The solution requires altering the return type and and using SequenceOf(EmptyCollection>(...)), but that does not appear to have any negative impact since the compiler will already differentiate based on the optional argument type. The code is:
func enumerate<Seq: SequenceType>(base: Seq?) -> SequenceOf<(Int, Seq.Generator.Element)> {
if let base = base {
return SequenceOf(Swift.enumerate(base))
} else {
return SequenceOf(EmptyCollection<(Int, Seq.Generator.Element)>())
}
}
Related
A block is defined like below
// Declare block ( optional )
typealias sorting = (([Schedule], [String]) -> [Schedule])?
var sortSchedule: sorting = { (schedules, sortDescription) in
var array = [Schedule]()
for string in sortDescription
{
for (index, schedule) in schedules.enumerate()
{
if string == schedule.startTime
{
array.append(schedule)
break
}
}
}
return array
}
At some points, I am invoking a block by doing
let allSchedules = sortSchedule?(result, sortDescription())
for schedule in allSchedules // Xcode complains at here
{
..........
}
Im using ? because I want to make sure that if the block exists, then do something. However, Xcode complains for the for loop
value of optional type [Schedule]? not upwrapped, did you mean to use '!' or '?'?
Im not sure why because the return type of a block is an array which can have 0 or more than one items.
Does anyone know why xcode is complaining.
You are use ? in line let allSchedules = sortSchedule?(result, sortDescription()) not "for sure that if the block exists", but just for note, that you understand that it can be nil. Behind scene allSchedules have type Array<Schedule>?. And you can not use for in cycle for nil. You better use optional binding:
if let allSchedules = sortSchedule?(result, sortDescription())
{
for schedule in allSchedules
{
//..........
}
}
Quite often, you need to write code such as the following:
if someOptional != nil {
// do something with the unwrapped someOptional e.g.
someFunction(someOptional!)
}
This seems a bit verbose, and also I hear that using the ! force unwrap operator can be unsafe and best avoided. Is there a better way to handle this?
It is almost always unnecessary to check if an optional is not nil. Pretty much the only time you need to do this is if its nil-ness is the only thing you want to know about – you don’t care what’s in the value, just that it’s not nil.
Under most other circumstances, there is a bit of Swift shorthand that can more safely and concisely do the task inside the if for you.
Using the value if it isn’t nil
Instead of:
let s = "1"
let i = Int(s)
if i != nil {
print(i! + 1)
}
you can use if let:
if let i = Int(s) {
print(i + 1)
}
You can also use var:
if var i = Int(s) {
print(++i) // prints 2
}
but note that i will be a local copy - any changes to i will not affect the value inside the original optional.
You can unwrap multiple optionals within a single if let, and later ones can depend on earlier ones:
if let url = NSURL(string: urlString),
data = NSData(contentsOfURL: url),
image = UIImage(data: data)
{
let view = UIImageView(image: image)
// etc.
}
You can also add where clauses to the unwrapped values:
if let url = NSURL(string: urlString) where url.pathExtension == "png",
let data = NSData(contentsOfURL: url), image = UIImage(data: data)
{ etc. }
Replacing nil with a default
Instead of:
let j: Int
if i != nil {
j = i
}
else {
j = 0
}
or:
let j = i != nil ? i! : 0
you can use the nil-coalescing operator, ??:
// j will be the unwrapped value of i,
// or 0 if i is nil
let j = i ?? 0
Equating an optional with a non-optional
Instead of:
if i != nil && i! == 2 {
print("i is two and not nil")
}
you can check if optionals are equal to non-optional values:
if i == 2 {
print("i is two and not nil")
}
This also works with comparisons:
if i < 5 { }
nil is always equal to other nils, and is less than any non-nil value.
Be careful! There can be gotchas here:
let a: Any = "hello"
let b: Any = "goodbye"
if (a as? Double) == (b as? Double) {
print("these will be equal because both nil...")
}
Calling a method (or reading a property) on an optional
Instead of:
let j: Int
if i != nil {
j = i.successor()
}
else {
// no reasonable action to take at this point
fatalError("no idea what to do now...")
}
you can use optional chaining, ?.:
let j = i?.successor()
Note, j will also now be optional, to account for the fatalError scenario. Later, you can use one of the other techniques in this answer to handle j’s optionality, but you can often defer actually unwrapping your optionals until much later, or sometimes not at all.
As the name implies, you can chain them, so you can write:
let j = s.toInt()?.successor()?.successor()
Optional chaining also works with subscripts:
let dictOfArrays: ["nine": [0,1,2,3,4,5,6,7]]
let sevenOfNine = dictOfArrays["nine"]?[7] // returns {Some 7}
and functions:
let dictOfFuncs: [String:(Int,Int)->Int] = [
"add":(+),
"subtract":(-)
]
dictOfFuncs["add"]?(1,1) // returns {Some 2}
Assigning to a property on an optional
Instead of:
if splitViewController != nil {
splitViewController!.delegate = self
}
you can assign through an optional chain:
splitViewController?.delegate = self
Only if splitViewController is non-nil will the assignment happen.
Using the value if it isn’t nil, or bailing (new in Swift 2.0)
Sometimes in a function, there’s a short bit of code you want to write to check an optional, and if it’s nil, exit the function early, otherwise keep going.
You might write this like this:
func f(s: String) {
let i = Int(s)
if i == nil { fatalError("Input must be a number") }
print(i! + 1)
}
or to avoid the force unwrap, like this:
func f(s: String) {
if let i = Int(s) {
print(i! + 1)
}
else {
fatalErrr("Input must be a number")
}
}
but it’s much nicer to keep the error-handling code at the top by the check. This can also lead to unpleasant nesting (the "pyramid of doom").
Instead you can use guard, which is like an if not let:
func f(s: String) {
guard let i = Int(s)
else { fatalError("Input must be a number") }
// i will be an non-optional Int
print(i+1)
}
The else part must exit the scope of the guarded value, e.g. a return or fatalError, to guarantee that the guarded value will be valid for the remainder of the scope.
guard isn’t limited to function scope. For example the following:
var a = ["0","1","foo","2"]
while !a.isEmpty {
guard let i = Int(a.removeLast())
else { continue }
print(i+1, appendNewline: false)
}
prints 321.
Looping over non-nil items in a sequence (new in Swift 2.0)
If you have a sequence of optionals, you can use for case let _? to iterate over all the non-optional elements:
let a = ["0","1","foo","2"]
for case let i? in a.map({ Int($0)}) {
print(i+1, appendNewline: false)
}
prints 321. This is using the pattern-matching syntax for an optional, which is a variable name followed by ?.
You can also use this pattern matching in switch statements:
func add(i: Int?, _ j: Int?) -> Int? {
switch (i,j) {
case (nil,nil), (_?,nil), (nil,_?):
return nil
case let (x?,y?):
return x + y
}
}
add(1,2) // 3
add(nil, 1) // nil
Looping until a function returns nil
Much like if let, you can also write while let and loop until nil:
while let line = readLine() {
print(line)
}
You can also write while var (similar caveats to if var apply).
where clauses also work here (and terminate the loop, rather than skipping):
while let line = readLine()
where !line.isEmpty {
print(line)
}
Passing an optional into a function that takes a non-optional and returns a result
Instead of:
let j: Int
if i != nil {
j = abs(i!)
}
else {
// no reasonable action to take at this point
fatalError("no idea what to do now...")
}
you can use optional’s map operator:
let j = i.map { abs($0) }
This is very similar to optional chaining, but for when you need to pass the non-optional value into the function as an argument. As with optional chaining, the result will be optional.
This is nice when you want an optional anyway. For example, reduce1 is like reduce, but uses the first value as the seed, returning an optional in case the array is empty. You might write it like this (using the guard keyword from earlier):
extension Array {
func reduce1(combine: (T,T)->T)->T? {
guard let head = self.first
else { return nil }
return dropFirst(self).reduce(head, combine: combine)
}
}
[1,2,3].reduce1(+) // returns 6
But instead you could map the .first property, and return that:
extension Array {
func reduce1(combine: (T,T)->T)->T? {
return self.first.map {
dropFirst(self).reduce($0, combine: combine)
}
}
}
Passing an optional into a function that takes an optional and returns a result, avoiding annoying double-optionals
Sometimes, you want something similar to map, but the function you want to call itself returns an optional. For example:
// an array of arrays
let arr = [[1,2,3],[4,5,6]]
// .first returns an optional of the first element of the array
// (optional because the array could be empty, in which case it's nil)
let fst = arr.first // fst is now [Int]?, an optional array of ints
// now, if we want to find the index of the value 2, we could use map and find
let idx = fst.map { find($0, 2) }
But now idx is of type Int??, a double-optional. Instead, you can use flatMap, which “flattens” the result into a single optional:
let idx = fst.flatMap { find($0, 2) }
// idx will be of type Int?
// and not Int?? unlike if `map` was used
I think you should go back to the Swift programming book and learn what these things are for. ! is used when you are absolutely sure that the optional isn't nil. Since you declared that you are absolutely sure, it crashes if you're wrong. Which is entirely intentional. It is "unsafe and best avoided" in the sense that asserts in your code are "unsafe and best avoided". For example:
if someOptional != nil {
someFunction(someOptional!)
}
The ! is absolutely safe. Unless there is a big blunder in your code, like writing by mistake (I hope you spot the bug)
if someOptional != nil {
someFunction(SomeOptional!)
}
in which case your app may crash, you investigate why it crashes, and you fix the bug - which is exactly what the crash is there for. One goal in Swift is that obviously your app should work correctly, but since Swift cannot enforce this, it enforces that your app either works correctly or crashes if possible, so bugs get removed before the app ships.
You there is one way. It is called Optional Chaining. From documentation:
Optional chaining is a process for querying and calling properties,
methods, and subscripts on an optional that might currently be nil. If
the optional contains a value, the property, method, or subscript call
succeeds; if the optional is nil, the property, method, or subscript
call returns nil. Multiple queries can be chained together, and the
entire chain fails gracefully if any link in the chain is nil.
Here is some example
class Person {
var residence: Residence?
}
class Residence {
var numberOfRooms = 1
}
let john = Person()
if let roomCount = john.residence?.numberOfRooms {
println("John's residence has \(roomCount) room(s).")
} else {
println("Unable to retrieve the number of rooms.")
}
// prints "Unable to retrieve the number of rooms."
You can check the full article here.
We can use optional binding.
var x:Int?
if let y = x {
// x was not nil, and its value is now stored in y
}
else {
// x was nil
}
After lot of thinking and researching i have came up with the easiest way to unwrap an optional :
Create a new Swift File and name it UnwrapOperator.swift
Paste the following code in the file :
import Foundation
import UIKit
protocol OptionalType { init() }
extension String: OptionalType {}
extension Int: OptionalType {}
extension Int64: OptionalType {}
extension Float: OptionalType {}
extension Double: OptionalType {}
extension CGFloat: OptionalType {}
extension Bool: OptionalType {}
extension UIImage : OptionalType {}
extension IndexPath : OptionalType {}
extension NSNumber : OptionalType {}
extension Date : OptionalType {}
extension UIViewController : OptionalType {}
postfix operator *?
postfix func *?<T: OptionalType>( lhs: T?) -> T {
guard let validLhs = lhs else { return T() }
return validLhs
}
prefix operator /
prefix func /<T: OptionalType>( rhs: T?) -> T {
guard let validRhs = rhs else { return T() }
return validRhs
}
Now the above code has created 2 operator [One prefix and one postfix].
At the time of unwrapping you can use either of these operator before or after the optionals
The explanation is simple, the operators returns the constructor value if they get nil in variable else the contained value inside the variable.
Below is the example of usage :
var a_optional : String? = "abc"
var b_optional : Int? = 123
// before the usage of Operators
print(a_optional) --> Optional("abc")
print(b_optional) --> Optional(123)
// Prefix Operator Usage
print(/a_optional) --> "abc"
print(/b_optional) --> 123
// Postfix Operator Usage
print(a_optional*?) --> "abc"
print(b_optional*?) --> 123
Below is the example when variable contains nil :
var a_optional : String? = nil
var b_optional : Int? = nil
// before the usage of Operators
print(a_optional) --> nil
print(b_optional) --> nil
// Prefix Operator Usage
print(/a_optional) --> ""
print(/b_optional) --> 0
// Postfix Operator Usage
print(a_optional*?) --> ""
print(b_optional*?) --> 0
Now it is your choice which operator you use, both serve the same purpose.
I have a generic Swift function like this:
func toNSArray<T>() -> [T] {
...
}
The compiler gives no error but I do not know how to call this function. I tried:
jList.toNSArray<String>()
jList.<String>toNSArray()
but it did not work.
How do I call a Generic function in Swift without input parameters?
You need to tell Swift what the return type needs to be through some calling context:
// either
let a: [Int] = jList.toNSArray()
// or, if you aren’t assigning to a variable
someCall( jList.toNSArray() as [Int] )
Note, in the latter case, this would only be necessary if someCall took a vague type like Any as its argument. If instead, someCall is specified to take an [Int] as an argument, the function itself provides the context and you can just write someCall( jList.toNSArray() )
In fact sometimes the context can be very tenuously inferred! This works, for example:
extension Array {
func asT<T>() -> [T] {
var results: [T] = []
for x in self {
if let y = x as? T {
results.append(y)
}
}
return results
}
}
let a: [Any] = [1,2,3, "heffalump"]
// here, it’s the 0, defaulting to Int, that tells asT what T is...
a.asT().reduce(0, combine: +)
When trying to understand a program, or in some corner-cases, it's useful to find out what type something is. I know the debugger can show you some type information, and you can usually rely on type inference to get away with not specifying the type in those situations, but still, I'd really like to have something like Python's type()
dynamicType (see this question)
Update: this has been changed in a recent version of Swift, obj.dynamicType now gives you a reference to the type and not the instance of the dynamic type.
This one seems the most promising, but I haven't been able to find out the actual type so far.
class MyClass {
var count = 0
}
let mc = MyClass()
# update: this now evaluates as true
mc.dynamicType === MyClass.self
I also tried using a class reference to instantiate a new object, which does work, but oddly gave me an error saying I must add a required initializer:
works:
class MyClass {
var count = 0
required init() {
}
}
let myClass2 = MyClass.self
let mc2 = MyClass2()
Still only a small step toward actually discovering the type of any given object though
edit: I've removed a substantial number of now irrelevant details - look at the edit history if you're interested :)
Swift 3 version:
type(of: yourObject)
Swift 2.0:
The proper way to do this kind of type introspection would be with the Mirror struct,
let stringObject:String = "testing"
let stringArrayObject:[String] = ["one", "two"]
let viewObject = UIView()
let anyObject:Any = "testing"
let stringMirror = Mirror(reflecting: stringObject)
let stringArrayMirror = Mirror(reflecting: stringArrayObject)
let viewMirror = Mirror(reflecting: viewObject)
let anyMirror = Mirror(reflecting: anyObject)
Then to access the type itself from the Mirror struct you would use the property subjectType like so:
// Prints "String"
print(stringMirror.subjectType)
// Prints "Array<String>"
print(stringArrayMirror.subjectType)
// Prints "UIView"
print(viewMirror.subjectType)
// Prints "String"
print(anyMirror.subjectType)
You can then use something like this:
if anyMirror.subjectType == String.self {
print("anyObject is a string!")
} else {
print("anyObject is not a string!")
}
The dynamicType.printClassName code is from an example in the Swift book. There's no way I know of to directly grab a custom class name, but you can check an instances type using the is keyword as shown below. This example also shows how to implement a custom className function, if you really want the class name as a string.
class Shape {
class func className() -> String {
return "Shape"
}
}
class Square: Shape {
override class func className() -> String {
return "Square"
}
}
class Circle: Shape {
override class func className() -> String {
return "Circle"
}
}
func getShape() -> Shape {
return Square() // hardcoded for example
}
let newShape: Shape = getShape()
newShape is Square // true
newShape is Circle // false
newShape.dynamicType.className() // "Square"
newShape.dynamicType.className() == Square.className() // true
Note: that subclasses of NSObject already implement their own className function. If you're working with Cocoa, you can just use this property.
class MyObj: NSObject {
init() {
super.init()
println("My class is \(self.className)")
}
}
MyObj()
As of Xcode 6.0.1 (at least, not sure when they added it), your original example now works:
class MyClass {
var count = 0
}
let mc = MyClass()
mc.dynamicType === MyClass.self // returns `true`
Update:
To answer the original question, you can actually use the Objective-C runtime with plain Swift objects successfully.
Try the following:
import Foundation
class MyClass { }
class SubClass: MyClass { }
let mc = MyClass()
let m2 = SubClass()
// Both of these return .Some("__lldb_expr_35.SubClass"), which is the fully mangled class name from the playground
String.fromCString(class_getName(m2.dynamicType))
String.fromCString(object_getClassName(m2))
// Returns .Some("__lldb_expr_42.MyClass")
String.fromCString(object_getClassName(mc))
If you simply need to check whether the variable is of type X, or that it conforms to some protocol, then you can use is, or as? as in the following:
var unknownTypeVariable = …
if unknownTypeVariable is <ClassName> {
//the variable is of type <ClassName>
} else {
//variable is not of type <ClassName>
}
This is equivalent of isKindOfClass in Obj-C.
And this is equivalent of conformsToProtocol, or isMemberOfClass
var unknownTypeVariable = …
if let myClass = unknownTypeVariable as? <ClassName or ProtocolName> {
//unknownTypeVarible is of type <ClassName or ProtocolName>
} else {
//unknownTypeVariable is not of type <ClassName or ProtocolName>
}
Swift 3:
if unknownType is MyClass {
//unknownType is of class type MyClass
}
For Swift 3.0
String(describing: <Class-Name>.self)
For Swift 2.0 - 2.3
String(<Class-Name>)
Here is 2 ways I recommend doing it:
if let thisShape = aShape as? Square
Or:
aShape.isKindOfClass(Square)
Here is a detailed example:
class Shape { }
class Square: Shape { }
class Circle: Shape { }
var aShape = Shape()
aShape = Square()
if let thisShape = aShape as? Square {
println("Its a square")
} else {
println("Its not a square")
}
if aShape.isKindOfClass(Square) {
println("Its a square")
} else {
println("Its not a square")
}
Old question, but this works for my need (Swift 5.x):
print(type(of: myObjectName))
Comment: I don't see how #JérémyLapointe answers the question. Using type(of:) only works by checking the compile-time information even if the actual type is a more specific subclass. There is now an easier way to dynamically query the type in Swift 5.1 without resorting to dynamicType like #Dash suggests. For more details on where I got this information, see SE-0068: Expanding Swift Self to class members and value types.
Code
Swift 5.1
// Within an instance method context
Self.self
// Within a static method context
self
This allows the use of Self as shorthand for referring to the containing type (in the case of structs, enums, and final class) or the dynamic type (in the case of non-final classes).
Explanation
The proposal explains well why this approach improves on dynamicType:
Introducing Self addresses the following issues:
dynamicType remains an exception to Swift's lowercased keywords rule. This change eliminates a special case that's out of step with
Swift's new standards. Self is shorter and clearer in its intent. It
mirrors self, which refers to the current instance.
It provides an easier way to access static members. As type names grow large, readability suffers.
MyExtremelyLargeTypeName.staticMember is unwieldy to type and read.
Code using hardwired type names is less portable than code that automatically knows its type.
Renaming a type means updating any TypeName references in code. Using self.dynamicType fights against Swift's goals of concision and
clarity in that it is both noisy and esoteric.
Note that self.dynamicType.classMember and TypeName.classMember
may not be synonyms in class types with non-final members.
If you get an "always true/fails" warning you may need to cast to Any before using is
(foo as Any) is SomeClass
If a parameter is passed as Any to your function, you can test on a special type like so :
func isADate ( aValue : Any?) -> Bool{
if (aValue as? Date) != nil {
print ("a Date")
return true
}
else {
print ("This is not a date ")
return false
}
}
Depends on the use case. But let's assume you want to do something useful with your "variable" types. The Swift switch statement is very powerful and can help you get the results you're looking for...
let dd2 = ["x" : 9, "y" : "home9"]
let dds = dd2.filter {
let eIndex = "x"
let eValue:Any = 9
var r = false
switch eValue {
case let testString as String:
r = $1 == testString
case let testUInt as UInt:
r = $1 == testUInt
case let testInt as Int:
r = $1 == testInt
default:
r = false
}
return r && $0 == eIndex
}
In this case, have a simple dictionary that contains key/value pairs that can be UInt, Int or String. In the .filter() method on the dictionary, I need to make sure I test for the values correctly and only test for a String when it's a string, etc. The switch statement makes this simple and safe!
By assigning 9 to the variable of type Any, it makes the switch for Int execute. Try changing it to:
let eValue:Any = "home9"
..and try it again. This time it executes the as String case.
//: Playground - noun: a place where people can play
import UIKit
class A {
class func a() {
print("yeah")
}
func getInnerValue() {
self.dynamicType.a()
}
}
class B: A {
override class func a() {
print("yeah yeah")
}
}
B.a() // yeah yeah
A.a() // yeah
B().getInnerValue() // yeah yeah
A().getInnerValue() // yeah
I have this for loop, p is a NSManagedObject, fathers is a to-many relationship, so I need to cast NSMutableOrderedSet to [Family] but it does not work, why?
for f in p.fathers as [Family] {
}
You can obtain an array representation of the set via the array property - then you can downcast it to the proper type and assign to a variable:
let families = p.fathers.array as [Family]
but of course you can also use it directly in the loop:
for f in p.fathers.array as [Family] {
....
}
Update
A forced downcast is now required using the ! operator, so the code above should be:
let families = p.fathers.array as! [Family]
The simple solution by Antonio should be used in this case. I'd just like to discuss this a bit more. If we try to enumerate an instance of 'NSMutableOrderedSet' in a 'for' loop, the compiler will complain:
error: type 'NSMutableOrderedSet' does not conform to protocol
'SequenceType'
When we write
for element in sequence {
// do something with element
}
the compiler rewrites it into something like this:
var generator = sequence.generate()
while let element = generator.next() {
// do something with element
}
'NS(Mutable)OrderedSet' doesn't have 'generate()' method i.e. it doesn't conform to the 'SequenceType' protocol. We can change that. First we need a generator:
public struct OrderedSetGenerator : GeneratorType {
private let orderedSet: NSMutableOrderedSet
public init(orderedSet: NSOrderedSet) {
self.orderedSet = NSMutableOrderedSet(orderedSet: orderedSet)
}
mutating public func next() -> AnyObject? {
if orderedSet.count > 0 {
let first: AnyObject = orderedSet.objectAtIndex(0)
orderedSet.removeObjectAtIndex(0)
return first
} else {
return nil
}
}
}
Now we can use that generator to make 'NSOrderedSet' conform to 'SequenceType':
extension NSOrderedSet : SequenceType {
public func generate() -> OrderedSetGenerator {
return OrderedSetGenerator(orderedSet: self)
}
}
'NS(Mutable)OrderedSet' can now be used in a 'for' loop:
let sequence = NSMutableOrderedSet(array: [2, 4, 6])
for element in sequence {
println(element) // 2 4 6
}
We could further implement 'CollectionType' and 'MutableCollectionType' (the latter for 'NSMutableOrderedSet' only) to make 'NS(Mutable)OrderedSet' behave like Swift's standard library collections.
Not sure if the above code follows the best practises as I'm still trying to wrap my head around details of all these protocols.