When a function's return value is another function,there's no way to get the returned function's argument names.Is this a pitfall of swift language?
For example:
func makeTownGrand(budget:Int,condition: (Int)->Bool) -> ((Int,Int)->Int)?
{
guard condition(budget) else {
return nil;
}
func buildRoads(lightsToAdd: Int, toLights: Int) -> Int
{
return toLights+lightsToAdd
}
return buildRoads
}
func evaluateBudget(budget:Int) -> Bool
{
return budget > 10000
}
var stopLights = 0
if let townPlan = makeTownGrand(budget: 30000, condition: evaluateBudget)
{
stopLights = townPlan(3, 8)
}
Be mindful of townPlan,townPlan(lightsToAdd: 3, toLights: 8) would be much more sensible to townPlan(3, 8), right?
You're correct. From the Swift 3 release notes:
Argument labels have been removed from Swift function types... Unapplied references to functions or initializers no longer carry argument labels.
Thus, the type of townPlan, i.e. the type returned from calling makeTownGrand, is (Int,Int) -> Int — and carries no external argument label information.
For a full discussion of the rationale, see https://github.com/apple/swift-evolution/blob/545e7bea606f87a7ff4decf656954b0219e037d3/proposals/0111-remove-arg-label-type-significance.md
Related
I want to save a function that takes two parameters as a function that only takes one parameter. I know I learned this with functional programming but I can't remember the methodology name or how to implement it.
Example: a methods like this:
func add (a: Int, b: Int) {
return a + b
}
And you can manipulate and save a new method that let’s say only increments a by 1:
let increment = add(b:1)
print(increment(a: 4))
// prints 5
Can you do this in swift?
It seems you're looking for function currying. This was a part of swift in earlier versions but was removed because it added too much complexity inside the compiler. (Like seen here: https://github.com/apple/swift-evolution/blob/master/proposals/0002-remove-currying.md)
I guess the closest you can get to a curried function is if you do something like this:
func add(_ x: Int) -> (Int) -> Int {
return { y in
y + x
}
}
With this you can say:
let add2 = add(2)
print(add2(3)) // prints 5
You are looking for Default Parameter Values
You can assign default values to the parameters. When calling this function if you pass the value to the parameters that passed value will be used else the default value will be used. To use the default value you can avoid the parameter in the function calling
func add(a: Int = 1, b: Int = 1) -> Int {
return a + b
}
print(add(a: 5, b: 5))//prints 10
print(add(a: 4))//prints 5
print(add(b: 4))//prints 5
print(add())//prints 2
In The Swift Programming Language, it says:
Functions can also take a variable number of arguments, collecting them into an array.
func sumOf(numbers: Int...) -> Int {
...
}
When I call such a function with a comma-separated list of numbers (`sumOf(1, 2, 3, 4), they are made available as an array inside the function.
Question: what if I already have an array of numbers that I want to pass to this function?
let numbers = [1, 2, 3, 4]
sumOf(numbers)
This fails with a compiler error, “Could not find an overload for '__conversion' that accepts the supplied arguments”. Is there a way to turn an existing array into a list of elements that I can pass to a variadic function?
Splatting is not in the language yet, as confirmed by the devs. Workaround for now is to use an overload or wait if you cannot add overloads.
Here's a work around that I found. I know it's not exactly what you want, but it seems to be working.
Step 1: Declare the function you'd like with an array instead of variadic arguments:
func sumOf(numbers: [Int]) -> Int {
var total = 0
for i in numbers {
total += i
}
return total
}
Step 2: Call this from within your variadic function:
func sumOf(numbers: Int...) -> Int {
return sumOf(numbers)
}
Step 3: Call Either Way:
var variadicSum = sumOf(1, 2, 3, 4, 5)
var arraySum = sumOf([1, 2, 3, 4, 5])
It seems strange, but it is working in my tests. Let me know if this causes unforeseen problems for anyone. Swift seems to be able to separate the difference between the two calls with the same function name.
Also, with this method if Apple updates the language as #manojid's answer suggests, you'll only need to update these functions. Otherwise, you'll have to go through and do a lot of renaming.
You can cast the function:
typealias Function = [Int] -> Int
let sumOfArray = unsafeBitCast(sumOf, Function.self)
sumOfArray([1, 2, 3])
You can use a helper function as such:
func sumOf (numbers : [Int]) -> Int { return numbers.reduce(0, combine: +) }
func sumOf (numbers : Int...) -> Int { return sumOf (numbers) }
I did this (Wrapper + Identity Mapping):
func addBarButtonItems(types: REWEBarButtonItemType...) {
addBarButtonItems(types: types.map { $0 })
}
func addBarButtonItems(types: [REWEBarButtonItemType]) {
// actual implementation
}
I know this response does not answer your exact question, but I feel its worth noting. I too was starting to play with Swift and immediately ran into a similar question. Manojlds answer is better for your question, I agree, but again, another workaround I came up with. I do happen to like Logan's better too.
In my case I just wanted to pass an array:
func sumOf(numbers: Array<Int>) -> Int {
var sum = 0
for number in numbers {
sum += number
}
return sum
}
var someNums = [8,7,2,9,12]
sumOf(someNums)
sumOf([10, 15, 20])
Just wanted to share, in case anyone else was thinking like me. Most of the time I would prefer pass the array like this, but I don't think the "Swiftly" yet. :)
Swift 5
This is an approach with #dynamicCallable feature that allows to avoid overloading or unsafeBitCast but you should make a specific struct to call:
#dynamicCallable
struct SumOf {
func dynamicallyCall(withArguments args: [Int]) -> Int {
return args.reduce(0, +)
}
}
let sum = SumOf()
// Use a dynamic method call.
sum(1, 2, 3) // 6
// Call the underlying method directly.
sum.dynamicallyCall(withArguments: [1, 2, 3]) // 6
Swift function parameters not accepted. Missing argument?
Calculate(theA, theB) //error: Missing argument label 'sideB:' in call
func Calculate(sideA: Int, sideB: Int) -> Int {
var ans = sideA + sideB
return ans;
}
You are missing the sideB: in your function call. I didn't want to rewrite your code (since you posted an image) but here's the working function call.
func calcButton(sender: AnyObject) {
let a: Int = 10
let b: Int = 11
calculate(a, sideB: b) //<-- Missing it here
}
func calculate(sideA: Int, sideB: Int) -> Int {
let a = sideA + sideB
return a
}
you might also want to have both variables in the function call so you can do this instead:
func calcButton(sender: AnyObject) {
let a: Int = 10
let b: Int = 11
calculate(sideA: a, sideB: b)
}
func calculate(sideA A: Int, sideB B: Int) -> Int {
return A + B
}
Just an FYI, use tab completion instead of writing out the function. Xcode will let you know all the function variables with placeholders so you can type them in.
you have missed the sideB param name in swift 3 first parameter is optional but second param is Mandatory that _ in there That’s an underscore. It changes the way the method is called. To illustrate this, here’s a very simple function:
func doStuff(thing: String) {
// do stuff with "thing"
}
It’s empty, because its contents don’t matter. Instead, let’s focus on how it’s called. Right now, it’s called like this:
doStuff(thing: "Hello")
You need to write the name of the thing parameter when you call the doStuff() function. This is a feature of Swift, and helps make your code easier to read. Sometimes, though, it doesn’t really make sense to have a name for the first parameter, usually because it’s built into the method name.
When that happens, you use the underscore character like this:
func doStuff(_ thing: String) {
// do stuff with "thing"
}
That means “when I call this function I don’t want to write thing, but inside the function I want to use thing to refer to the value that was passed in.
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: +)
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)>())
}
}