I'm trying to do create some closure definitions which I'm gonna use a lot in my iOS app. So I thought to use a typealias as it seemed the most promising ...
I did a small Playground example which shows my issue in detail
// Here are two tries for the Closure I need
typealias AnonymousCheck = (Int) -> Bool
typealias NamedCheck = (number: Int) -> Bool
// This works fine
var var1: AnonymousCheck = {
return $0 > 0
}
var1(-2)
var1(3343)
// This works fine
var var2: NamedCheck = {
return $0 > 0
}
var2(number: -2)
var2(number: 12)
// But I want to use the typealias mainly as function parameter!
// So:
// Use typealias as function parameter
func NamedFunction(closure: NamedCheck) {
closure(number: 3)
}
func AnonymousFunction(closure: AnonymousCheck) {
closure(3)
}
// This works as well
// But why write again the typealias declaration?
AnonymousFunction({(another: Int) -> Bool in return another < 0})
NamedFunction({(another: Int) -> Bool in return another < 0})
// This is what I want... which doesn't work
// ERROR: Use of unresolved identifier 'number'
NamedFunction({NamedCheck in return number < 0})
// Not even these work
// ERROR for both: Anonymous closure arguments cannot be used inside a closure that has exlicit arguments
NamedFunction({NamedCheck in return $0 < 0})
AnonymousFunction({AnonymousCheck in return $0 < 0})
Am I missing something or is it just not supported in Swift?
Thanks
EDIT/ADDITION:
The above is just a simple example. In real life my typealias is more complicated. Something like:
typealias RealLifeClosure = (number: Int, factor: NSDecimalNumber!, key: String, upperCase: Bool) -> NSAttributedString
I basically want to use a typealias as a shortcut so I don't have to type that much. Maybe typealias isn't the right choice... Is there another?
You aren't rewriting the typealias declaration in this code, you're declaring the parameters and return type:
AnonymousFunction({(another: Int) -> Bool in return another < 0})
Happily, Swift's type inference lets you use any of the following - pick the style that feels best to you:
AnonymousFunction( { (number: Int) -> Bool in number < 0 } )
AnonymousFunction { (number: Int) -> Bool in number < 0 }
AnonymousFunction { (number) -> Bool in number < 0 }
AnonymousFunction { number -> Bool in number < 0 }
AnonymousFunction { number in number < 0 }
AnonymousFunction { $0 < 0 }
I don't think you'll be able to do what you want. To simplify your example slightly, you can do this:
typealias NamedCheck = (number: Int) -> Bool
let f: NamedCheck = { $0 < 5 }
f(number: 1)
NamedFunction(f)
NamedFunction( { $0 < 5 } as NamedCheck)
But you can't do what you want, which is to rely on the fact that the tuple arg is called number to refer to it inside the closure without giving it as part of the closure:
// compiler error, no idea what "number" is
let g: NamedCheck = { number < 5 }
Bear in mind that you can name the parameter without giving it a type (which is inferred from the type of g):
let g: NamedCheck = { number in number < 5 }
but also, you can name it whatever you want:
let h: NamedCheck = { whatevs in whatevs < 5 }
NamedFunction(h)
Here's what I think is happening (this is partly guesswork). Remember how functions can have external and internal argument names:
func takesNamedArgument(#namedArg: Int) { etc... }
Or, to write it longhand:
func takesNamedArgument(namedArg namedArg: Int) { etc... }
But you can also give as the second, internal, name whatever you like:
func takesNamedArgument(namedArg whatevs: Int) { etc... }
I think this is what is happening with the closures with named tuples. The "external" name is "number", but you must give it an "internal" name too, which is what you must use in the function body. You can't make use of the external argument within your function. In case of closure expressions, if you don't give an internal name, you can use $0 etc, but you can't just skip it, any more than you can skip the internal name altogether and just rely on the external name when defining a regular function.
I was hoping that I could prove this theory by the following:
let f = { (#a: Int, #b: Int)->Bool in a < b }
resulting in f being of type (a: Int, b: Int)->Bool). This compiles, as does:
let g = { (number1 a: Int, number2 b: Int)->Bool in a < b }
but it doesn't look like the external names for the argument make it out to the type of f or g.
The syntax to create a closure is:
{ (parameters) -> return type in
statements
}
What's at the left of in is the closure signature (parameters and return value). In some cases the signature can be omitted or simplified when type inference is able to determine the number of parameters and their type, and the return value.
In your case it doesn't work because you are passing a type alias, but it is interpreted as a parameter name. The 3 lines work if either you:
name the parameter properly
NamedFunction({number in return number < 0})
AnonymousFunction({number in return number < 0})
use shorthand arguments:
NamedFunction({ return $0 < 0})
AnonymousFunction({ return $0 < 0})
use shorthand arguments and implicit return:
NamedFunction({ $0 < 0})
AnonymousFunction({ $0 < 0})
Related
I've read this post, but need a little additional help.
I would like to construct a Closure which takes in a variable amount of Doubles, compares them to a threshold (which is also a Double) and returns a Bool checking if ALL entries were greater than the threshold. The return type should be (Double...) -> Bool
Here is what I have so far:
func isAllAbove(lower: Double) -> (Double...) -> Bool {
return {
var conditions: [Bool] = []
for i in 0...$0.count {
conditions.append(lower < $0[i])
}
return !conditions.contains(false)
}
}
However, the compiler complains that
Cannot convert return expression of type '(_) -> _' to return type '(Double...) -> Bool'
Why is this happening and how can I fix this? Thanks!
Try to specify parameter type and return type in closure to helps compiler to understand what value it should take and return. Also, you have a mistake in for loop. The interval should be like this 0 ..< values.count:
func isAllAbove(lower: Double) -> (Double...) -> Bool {
return { (values: Double...) -> Bool in
var conditions: [Bool] = []
for i in 0 ..< values.count {
conditions.append(lower < values[i])
}
return !conditions.contains(false)
}
}
let allAbove = isAllAbove(lower: 2)
print(allAbove(1, 2, 3)) // false
Also, you can write it almost in 1 line of code:
let lower = 2
let isAllAbove = ![1, 2, 3].contains { $0 < lower }
print(isAllAbove1) // false
Is there a more elegant way to filter with an additional parameter (or map, reduce).
When I filter with a single parameter, we get a beautiful easy to ready syntax
let numbers = Array(1...10)
func isGreaterThan5(number:Int) -> Bool {
return number > 5
}
numbers.filter(isGreaterThan5)
However, if I need to pass an additional parameter to my function it turns out ugly
func isGreaterThanX(number:Int,x:Int) -> Bool {
return number > x
}
numbers.filter { (number) -> Bool in
isGreaterThanX(number: number, x: 8)
}
I would like to use something like
numbers.filter(isGreaterThanX(number: $0, x: 3))
but this gives a compile error annonymous closure argument not contained in a closure
You could change your function to return a closure which serves
as predicate for the filter method:
func isGreaterThan(_ lowerBound: Int) -> (Int) -> Bool {
return { $0 > lowerBound }
}
let filtered = numbers.filter(isGreaterThan(5))
isGreaterThan is a function taking an Int argument and returning
a closure of type (Int) -> Bool. The returned closure "captures"
the value of the given lower bound.
If you make the function generic then it can be used with
other comparable types as well:
func isGreaterThan<T: Comparable>(_ lowerBound: T) -> (T) -> Bool {
return { $0 > lowerBound }
}
print(["D", "C", "B", "A"].filter(isGreaterThan("B")))
In this particular case however, a literal closure is also easy to read:
let filtered = numbers.filter( { $0 > 5 })
And just for the sake of completeness: Using the fact that
Instance Methods are Curried Functions in Swift, this would work as well:
extension Comparable {
func greaterThanFilter(value: Self) -> Bool {
return value > self
}
}
let filtered = numbers.filter(5.greaterThanFilter)
but the "reversed logic" might be confusing.
Remark: In earlier Swift versions you could use a curried function
syntax:
func isGreaterThan(lowerBound: Int)(value: Int) -> Bool {
return value > lowerBound
}
but this feature has been removed in Swift 3.
I'm confused on how getFunctionNeededForReference is running. There is no call for it and where are the functions returned to? where are they going? I know they are being referenced but where are the functions going to, there is not call for getFunctionNeededForReference in the beginning? there is no call sending the argument flag anyway?
func add ( a: Int , b : Int)-> Int {
//returing a result and not a variable
return a + b
}
func multiply ( a: Int, b: Int) -> Int{
return a * b
}
// declaring a function as a variable, it takes in 2 Ints and returns an Int
var f1 : (Int, Int)-> Int
f1 = add
f1 = multiply
// Function as a parameter
func arrayOperation (f: (Int, Int) -> Int , arr1: [Int] , arr2: [Int]) -> [Int]
{
// Declaring and initializing an empty array to return
var returningArray = [Int]()
for (i, val) in enumerate(arr1)
{
returningArray.append(f(arr1 [i], arr2 [i]))
}
return returningArray
}
arrayOperation(add, [2,3,4], [4,5,6])
arrayOperation(multiply, [2,3,4], [4,5,6])
//Function as a return value
func getFunctionNeededForReference (flag : Int) -> (Int,Int) ->Int
{
if flag == 0 {
return add
}else {
return multiply
}
}
What you've posted is just some example code showing things that Swift supports. It's not code that's useful for anything. It's just demonstrating Swift's syntax for first-class functions.
If you don't understand what “first-class functions” means, you can look up the term in your favorite search engine and find many explanations.
As an exercise, I'm trying to extend Array in Swift to add a sum() member function. This should be type safe in a way that I want a call to sum() to compile only if the array holds elements that can be added up.
I tried a few variants of something like this:
extension Array {
func sum<U : _IntegerArithmeticType where U == T>() -> Int {
var acc = 0
for elem in self {
acc += elem as Int
}
return acc
}
}
The idea was to say, “OK, this is a generic function, the generic type must be something like an Int, and must also be the same as T, the type of the elements of the array”. But the compiler complains: “Same-type requirement make generic parameters U and T equivalent”. That's right, and they should be, with the additional contraint T : _IntegerArithmeticType.
Why isn't the compiler letting me do this? How can I do it?
(I know that I should later fix how things are added up and what the return type exactly is, but I'm stuck at the type constraint for now.)
As per Martin R's comment, this is not currently possible. The thing I'm tempted to use in this particular situation would be an explicit passing of a T -> Int conversion function:
extension Array {
func sum(toInt: T -> Int?) -> Int {
var acc = 0
for elem in self {
if let i = toInt(elem) {
acc += i
}
}
return acc
}
}
Then I can write stuff like this:
func itself<T>(t: T) -> T {
return t
}
let ss = ["1", "2", "3", "4", "five"].sum { $0.toInt() }
let si = [1, 2, 3, 4].sum(itself)
An explicit function has to be passed, though. The (itself) part can of course be replaced by { $0 }. (Others have called the itself function identity.)
Note that an A -> B function can be passed when A -> B? is needed.
This piece of code comes from Swift documentation https://developer.apple.com/library/prerelease/mac/documentation/Swift/Conceptual/Swift_Programming_Language/Extensions.html
extension Int {
subscript(var digitIndex: Int) -> Int {
var decimalBase = 1
while digitIndex > 0 {
decimalBase *= 10
--digitIndex
}
return (self / decimalBase) % 10
}
}
Apparently var is a reserved word, so why it is legal to declare: subscript(var digitIndex: Int) -> Int?
If I change the signature to subscript(#digitIndex: Int) -> Int, I will get this compiler error:
My questions are:
1) why the signature is valid?
2) why my change causes an exception?
Declaring a function argument with var means that it can be modified, is not a constant. In your case, without using var, you had a constant argument but you attempted to decrement it. Thus the error.
Your two cases are:
func foo (x: int) { /* x is a constant, like `let x: int` */ }
func foo (var x: int) { /* x is not a constant */ }