This is a question that aims merely at elegance, but is there a way to make something to the following code work in Swift? I know the code does not work, what I want is that the result of the code within the closure is stored in a constant. And the underlying theoretical issue is whether or not it is possible to retrieve the returned value from the closure with type Int and not with type () -> Int.
Thanks a lot for any help or comment!
let tableWithBooleans: [Bool] = Array(repeating: false, count: 10)
tableWithBooleans[0] = true
tableWithBooleans[5] = true
let numberOfTrue: Int = {
var result: Int = 0
for i in 0...9 {
if tableWithBooleans[i] {
result += 1
}
}
return result
}
// I want the code to compile and numberOfTrue to be a constant equal to 2
Use a high-order function instead
let numberOfTrue = tableWithBooleans.reduce(0) { $1 ? $0 + 1 : $0 }
Now if you still want to use your closure code then you should add a () after the closing } since you are calling the code inside {} as a function
let numberOfTrue: Int = {
var result: Int = 0
for i in 0...9 {
if tableWithBooleans[i] {
result += 1
}
}
return result
}()
Related
I know that closures capture the values in a given environment. That is not my question. My question is how do I capture the return value of a closure. For example if I use a closure as a parameter like:
func myClosure(yourAge x: Int, completion: (Int) -> String) {
if x == 4 {
completion(x)
}
}
then say:
let answer = myClosure(yourAge: 4) { x in
return "You're just a baby"
}
The warning is:
Constant 'answer' inferred to have type '()', which may be unexpected
And I understand that warning. Pretty much answer will not be an answer at all. It will be Void or ()
Now if I make the entire function return a String such as:
func myClosure(yourAge x: Int, completion: (Int) -> String) -> String {
completion(x)
}
Then I can of course capture the result in a property:
let answer = myClosure(yourAge: 4) { x in
if x < 10 { return "You're just a baby"}
else {
return "You can play the game"
}
}
And maybe I just answered my own question here but is there no simple way to place the return value of a closure into a property or I am using this in an unintended way?
If you wanted to do this purely with a closure (and not a function that accepts a closure), you could do something like:
let myClosure: (Int) -> String = { age in
if age < 10 {
return "You're just a baby"
}
else {
return "You can play the game"
}
}
let answer = myClosure(4) // "You're just a baby"
i am currently trying to do some self learning in swift just for my own interest. in the course i bought it says that we should create a function similar to this one in order to solve my problem. but I'm blankly staring trying to figure out what this function actually does?
func unknown() -> () -> Int {
var x = 0
let z: () -> Int = {
x += 1
return x
}
return z
}
It is a function that returns another function which will return an integer that will be increased everytime you call it:
let afunc = unknown()
let value1 = afunc() // 1
let value2 = afunc() // 2
let value3 = afunc() // 3
The interesting part of this is the return type. () -> Int is a function that returns an Int, which means that unknown returns a function rather than something simple, like a number.
z is then a variable of that same type and is assigned a function definition to be returned.
If you assign the result of unknown to a variable, you can then invoke the returned function.
This implementation of a high order function is an interesting way of defining generators. An infinite sequence-like class would've achieve the same thing, but with more verbosity:
class MySequence {
private var x = 0
func unknown() -> Int {
x += 1
return x
}
}
var seq = MySequence()
let unknown = seq.unknown
print(unknown()) // 1
print(unknown()) // 2
print(unknown()) // 3
// ... and so on
The main difference between the class and the anonymous closure is the storage for x: the closure captures in due to using the variables within its body, while the class declares explicit storage for the property.
Some fancy stuff can result by using high order functions, like a generator for the Fibonacci numbers:
func fibonnaciSequence() -> () -> Int? {
var a = 0, b = 1
return { let c = a; a += b; b = c; return c }
}
let fibo = fibonnaciSequence()
while let f = fibo() {
// this will print forever
// actually not forever, it will stop at some point due to += overflowing
print(f)
}
Another question asked, essentially, how to implement a take function which would return the first n elements of a sequence. My answer was:
struct TakeFromSequenceSequence<S:SequenceType> : SequenceType {
var limit : Int
var sequence : S
func generate() -> AnyGenerator<S.Generator.Element> {
var generator = sequence.generate()
var limit = self.limit
return anyGenerator {
guard limit > 0 else {
return nil
}
limit = limit - 1
return generator.next()
}
}
}
extension SequenceType {
func take(count:Int) -> TakeFromSequenceSequence<Self> {
return TakeFromSequenceSequence(limit: count, sequence: self)
}
}
but it seems like I ought to be able to use AnySequence and anyGenerator to do it all inline in my take function:
extension SequenceType {
func take(count:Int) -> AnySequence<Self.Generator.Element> {
// cannot invoke initializer for type 'AnySequence<_>' with an argument list of type '(() -> _)'
return AnySequence({
var generator = self.generate()
var limit = count
// cannot invoke 'anyGenerator' with an argument list of type '(() -> _)'
return anyGenerator({
guard limit > 0 else {
return nil
}
limit = limit - 1
return generator.next()
})
})
}
}
Unfortunately, this yields multiple typing errors, mostly (I think) because type inference is failing.
Anybody have any suggestions on how to get this (using AnySequence and anyGenerator inline) to work?
(The answer is now based on Swift 2.2/Xcode 7.3. A solution for Swift 2.1 can be found in the edit history.)
The type of the closure passed to the AnySequence init method
must be specified explicitly:
extension SequenceType {
func take(count:Int) -> AnySequence<Generator.Element> {
return AnySequence { () -> AnyGenerator<Generator.Element> in
var generator = self.generate()
var limit = count
return AnyGenerator {
guard limit > 0 else {
return nil
}
limit = limit - 1
return generator.next()
}
}
}
}
Note that the (redundant) Self. in Self.Generator.Element is omitted, otherwise it does not compile.
Example:
let sequence = [1,2,3,4,5].take(2)
print(Array(sequence)) // [1, 2]
print(Array(sequence)) // [1, 2]
Alternatively, the method can be defined as
extension SequenceType {
func take(count:Int) -> AnySequence<Generator.Element> {
var generator = self.generate()
var limit = count
return AnySequence {
return AnyGenerator {
guard limit > 0 else {
return nil
}
limit = limit - 1
return generator.next()
}
}
}
}
Now the closure passed to the AnySequence init method is a "single-expression closure" and the type is inferred by the compiler.
But – as David Berry noted – the created sequence then behaves differently, the generate() method cannot be called repeatedly
with identical results:
let sequenceX = [1,2,3,4,5].take(2)
print(Array(sequenceX)) // [1, 2]
print(Array(sequenceX)) // []
This is permitted behavior, as stated in the SequenceType protocol reference:
... It is not correct to assume that a sequence will either be
"consumable" and will resume iteration, or that a sequence is a
collection and will restart iteration from the first element. A
conforming sequence that is not a collection is allowed to produce an
arbitrary sequence of elements from the second generator.
So one can choose among these implementations, dependent on the desired behavior.
I want to initialize a dictionary with a dictionary nested inside like this:
var a = [Int:[Int:Float]]()
a[1][2] = 12
But I get an error:
(Int:[Int:Float]) does not have a member named 'subscript'
I've hacked at a variety of other approaches, all of them running into some kind of issue.
Any idea why this doesn't work?
You can create your own 2D dictionary like this:
struct Dict2D<X:Hashable,Y:Hashable,V> {
var values = [X:[Y:V]]()
subscript (x:X, y:Y)->V? {
get { return values[x]?[y] }
set {
if values[x] == nil {
values[x] = [Y:V]()
}
values[x]![y] = newValue
}
}
}
var a = Dict2D<Int,Int,Float>()
a[1,2] = 12
println(a[1,2]) // Optional(12.0)
println(a[0,2]) // nil
The point is you access the element via a[x,y] instead of a[x][y] or a[x]?[y].
It's giving you that error because your first subscript returns an optional so it may return a dictionary or nil. In the case that it returns nil the second subscript would be invalid. You can force it to unwrap the optional value by using an exlamation point.
var a = [1 : [ 2: 3.14]]
a[1]
a[1]![2]
If you aren't positive that a[1] is non-nil you may want to safely unwrap with a question mark instead.
var a = [1 : [ 2: 3.14]]
a[1]
a[1]?[2]
You can also assign using this method. (As of Beta 5)
var a = [Int:[Int:Float]]()
a[1] = [Int: Float]()
a[1]?[2] = 12.0
a[1]?[2] //12.0
Another way to do it is with an extension to the standard dictionary:
extension Dictionary {
mutating func updateValueForKey(key: Key, updater: ((previousValue: Value?) -> Value)) {
let previousValue = self[key]
self[key] = updater(previousValue: previousValue)
}
}
Example:
var a = [Int:[Int:Float]]()
a.updateValueForKey(1) { nestedDict in
var nestedDict = nestedDict ?? [Int:Float]()
nestedDict[2] = 12
return nestedDict
}
I want to have a constant using let that may be one of several values.
For instance:
if condition1 {
constant = "hi"
}
else if condition2 {
constant = "hello"
}
else if condition3 {
constant = "hey"
}
else if condition4 {
constant = "greetings"
}
I'm not sure how to do this with Swift and the let feature. But I'm inclined to believe it's possible, as this is in the Swift book:
Use let to make a constant and var to make a variable. The value of a constant doesn’t need to be known at compile time, but you must assign it a value exactly once.
How would I accomplish this?
As pointed out in the other answers you can't directly do this. But if you're looking to just variably set the initial value of a constant, then yes, that is possible. Here's an example with a computed property.
class MyClass {
let aConstant: String = {
if something == true {
return "something"
} else {
return "something else"
}
}()
}
I think you are looking for variable which will be assigned later inside switch-case:
let constant :String
switch conditions {
case condition1:
constant = "hi"
case condition2:
constant = "hello"
case condition3:
constant = "hey"
case condition4:
constant = "greetings"
default:
constant = "salute"
}
One option would be something like this, using a closure:
let constant: String = ({ value in
if conditionOne {
return "Hi"
} else if conditionTwo {
return "Bye"
}
return "Oops!"
})(myData /*needed for condition*/)
Or, for another twist, using generics:
func fancySwitch<S, T>(val: S, fn: S -> T) -> T {
return fn(val)
}
let x: String = fancySwitch(3) { val in
if val == 2 {
return "Hi"
} else if val < 5 {
return "Bye"
}
return "Oops"
}
let y: String = fancySwitch((3, 4)) { (a, b) in
if a == 2 {
return "Hi"
} else if b < 5 {
return "Bye"
}
return "Oops"
}
I understand what you're looking for. In Scala and some other functional languages this can be done using the match statement (kind of like switch) because the entire statement resolves to a value like this:
val b = true
val num = b match {
case true => 1
case false => 0
}
This is unfortunately not directly possible in Swift because there is no way to get a value from a branch statement. As stated in the Swift book, "Swift has two branch statements: an if statement and a switch statement." Neither of these statements resolve to a value.
The closest code structure I can think of is to first use a variable to retrieve the correct value and then assign it to a constant to be used in any later code:
let b = true
var num_mutable: Int
switch b {
case true:
num_mutable = 1
default:
num_mutable = 0
}
let num = num_mutable
Just add the line let constant: String before your if/else statement.
Below, an excerpt from Swift 1.2 and Xcode 6.3 beta - Swift Blog - Apple Developer elaborates.
let constants are now more powerful and consistent — The new rule is
that a let constant must be initialized before use (like a var), and
that it may only be initialized, not reassigned or mutated after
initialization. This enables patterns like:
let x : SomeThing
if condition {
x = foo()
} else {
x = bar()
}
use(x)
This formerly required the use of a var even though there is no
mutation taking place. Properties have been folded into this model to
simplify their semantics in initializers as well.
I found the Swift blog post above from the article "Let It Go: Late Initialization of Let in Swift", which I found by googling: swift let constant conditional initialize.