Swift5 #dynamicMemberLookup function - swift

#dynamicMemberLookup subscript function.
no use parameterName is Error.
why?
#dynamicCallable is OK.
#dynamicMemberLookup
struct DynamicMember {
// OK.
subscript(dynamicMember string: String) -> String {
return string
}
// NG.
// none parameterName func. Why error?
// #dynamicMemberLookup attribute requires 'Dynamic' to have a 'subscript(dynamicMember:)' method that accepts either 'ExpressibleByStringLiteral' or a keypath
/*
subscript(dynamicMember: String) -> String {
return dynamicMember
}
*/
}
let dm = DynamicMember()
print(dm.dynamicProperty)
#dynamicCallable no parameter name OK.
#dynamicCallable
struct DynamicCall {
// OK.
func dynamicallyCall(withArguments list: [Int]) -> [Int] {
return list
}
// OK.
// none parameterName func.
// dynamicCallable is OK.
/*
func dynamicallyCall(withArguments: [Int]) -> [Int] {
return withArguments
}
*/
}
let dc = DynamicCall()
print(dc(1, 2, 3))
Compiler bug?
Please tell me who you know.

dynamicallyCall is a function, but dynamicMemberLookup is a subscript, which has variations. It is done intentionally
a) to allow key path access (for your example)
print(dm[dynamicMember: "dynamicProperty"])
b) but to disallow ambiguous, something like
dm["dynamicProperty"]
See official details on Swift Evolution

Related

What aren't these two ways of expressing map function equivalent?

I got a surprise today while looking at another SO question:
let s = "1,a"
let arr = s.split(separator: ",")
let result = arr.compactMap{Int($0)} // ok
let result2 = arr.compactMap(Int.init) // error
Why is line 3 legal but line 4 is not? I would have thought these two ways of saying "coerce the incoming parameter to Int if possible" would be completely equivalent.
I understand that line 4 is choking on the Subsequence, and I see how to get out of the difficulty:
let result2 = arr.map(String.init).compactMap(Int.init) // ok
What I don't understand is why they both don't choke in the same way.
Looks like the Int.init overload that accepts a Substring has the following signature:
public init?<S>(_ text: S, radix: Int = 10) where S : StringProtocol
So, Int($0) works because it uses the default radix, but there isn't an Int.init(_:) that accepts a Substring - there's only Int.init(_:radix:) that does - and so it fails.
But if there was one:
extension Int {
public init?<S>(_ text: S) where S : StringProtocol {
self.init(text, radix: 10)
}
}
then this would work:
let result1 = arr.compactMap(Int.init)
In fact the first version (Int($0)) calls this initializer, which has two parameters (one of them has a default value):
#inlinable public init?<S>(_ text: S, radix: Int = 10) where S : StringProtocol
If I define a custom initializer like so, then the second example works too.
extension Int {
init?<S>(_ string: S) where S: StringProtocol {
// convert somehow, e.g: self.init(string, radix: 10)
return nil
}
}
let result2 = arr.compactMap(Int.init)
It seems to me that if I write Int.init in the compactMap, it can call only the exact initializer (or function), and the second parameter of the first called initializer cannot be inferred.
Another example:
func test1<S>(param1: S) -> String where S: StringProtocol {
return ""
}
func test2<S>(param1: S, defaultParam: String = "") -> String where S: StringProtocol {
return ""
}
extension Sequence {
func customCompactMap<ElementOfResult>(_ transform: (Element) -> ElementOfResult?) -> [ElementOfResult] {
compactMap(transform)
}
}
arr.customCompactMap(test1)
arr.customCompactMap(test2) // error
I think the function references cannot hold any default values. Unfortunately I didn't find any official reference to this, but seems interesting.
Proof, last example:
func test3(param1: String, defaultParam: String = "") { }
let functionReference = test3
functionReference("", "")
functionReference("") // error
Here the functionReference's type is (String, String) -> (), even though the test3 function has a default value for the second parameter. As you can see functionReference cannot be called with only one value.
I tried looking for the Swift forum post where someone on the core team explained this, but sorry, I couldn't find it. You can go asking there and get clarification on this point:
Default arguments don't actually produce overloads.
Instead, using default arguments at call site is syntactic sugar for using all arguments. The compiler inserts the defaults for the ones you don't use.
A few results of that…
You cannot use functions with default arguments as closures with simplified signatures. You have to wrap them in new closures, as you demonstrated in your question.
func ƒ(_: Int = 0) { }
let intToVoid: (Int) -> Void = ƒ // compiles
// Cannot convert value of type '(Int) -> ()' to specified type '() -> Void'
let voidToVoid: () -> Void = ƒ
Methods with different default argument patterns, that look the same at call site, are not considered overrides.
class Base {
func ƒ(_: Any? = nil) -> String { "Base" }
}
final class Derived: Base {
// No `override` required.
func ƒ() -> String { "Derived" }
}
Base().ƒ() // "Base"
Derived().ƒ() // "Derived"
(Derived().ƒ as (Any?) -> String)("argument") // "Base"
Default arguments do not allow for satisfaction of protocol requirements.
protocol Protocol {
func ƒ() -> String
}
// Type 'Base' does not conform to protocol 'Protocol'
extension Base: Protocol { }

Swift syntax explanation with compact map

I did find following code while examine code:
override func layoutAttributesForElements(in rect: CGRect) -> [UICollectionViewLayoutAttributes]? {
return super.layoutAttributesForElements(in: rect)?
.compactMap { $0.copy() as? ParallaxLayoutAttributes }
.compactMap(prepareAttributes)
}
private func prepareAttributes(attributes: ParallaxLayoutAttributes) -> ParallaxLayoutAttributes {
// Lot of code doing stuff with attributes
return attributes
}
So, actually what i want to ask is, that compact is function declared as following:
#inlinable public func compactMap<ElementOfResult>(_ transform: (Element) throws -> ElementOfResult?) rethrows -> [ElementOfResult]
Here in example, we pass just function, without arguments:
.compactMap(prepareAttributes)
That completely bend my mind, because, well, prepareAttributes function declared like this (with argument you have to pass):
private func prepareAttributes(attributes: ParallaxLayoutAttributes) -> ParallaxLayoutAttributes
So, why code above compiles and what how exactly .compactMap(prepareAttributes)
runs when you did not pass an argument for prepareAttributes function?
In the call .compactMap(prepareAttributes), you pass in the function, prepareAttributes to compactMap as a closure. Since prepareAttributes takes a single input argument whose type matches the closure variable of compactMap, the compiler can automatically infer that it needs to pass $0 to prepareAttributes.
So essentially, .compactMap(prepareAttributes) is shorthand for
.compactMap {prepareAttributes(attributes: $0) }
A simple example of the same behaviour with map that is quite often used is to map over a type that you then pass into an init, which you could write as .map { MyType(input: $0) } or simplify to .map(MyType.init).
struct MyInt {
let value: Int
init(value: Int) {
self.value = value
}
}
let ints = [1,2,3]
let myInts = ints.map(MyInt.init) // same as `ints.map { MyInt(value: $0) }

Using generic methods in protocols Swift

I believe I have some misunderstanding of how generics work. I have the protocol:
protocol CommandProtocol {
func execute<T>() -> T
func unExecute<T>() -> T
}
And a class that conforms to it:
class CalculatorCommand: CommandProtocol {
...
func execute<String>() -> String {
return calculator.performOperation(operator: `operator`, with: operand) as! String
}
func unExecute<Double>() -> Double {
return calculator.performOperation(operator: undo(operator: `operator`), with: operand) as! Double
}
...
}
The calculator.performOperation() method actually returns Double, but here I just try to play with generics so I replace return type from Double to String.
After that, I have a class which invokes these methods:
class Sender {
...
// MARK: - Public methods
func undo() -> Double {
if current > 0 {
current -= 1
let command = commands[current]
return command.unExecute()
}
return 0
}
func redo() -> Double? {
if current < commands.count {
let command = commands[current]
current += 1
let value: Double = command.execute()
print(type(of: value))
return command.execute()
}
return nil
}
...
}
In the undo() method everything works as expected (one thing that I did not understand fully is how Swift really knows whether the unExecute value will return Double or not, or compiler infers it based on the undo() return type?)
But in the redo() method, I am calling the execute() method which returns String, but the method expects Double, so I thought that my program would crash, but not, it works totally fine as if execute() method returns Double.
Please, could someone explain to me what exactly happens under the cover of this code? Thank you in advance.
You are correct that you misunderstand generics. First, let's look at this protocol:
protocol CommandProtocol {
func execute<T>() -> T
func unExecute<T>() -> T
}
This says "no matter what type the caller requests, this function will return that type." That's impossible to successfully implement (by "successfully" I mean "correctly returns a value in all cases without crashing"). According this protocol, I'm allowed to write the following code:
func run(command: CommandProtocol) -> MyCustomType {
let result: MyCustomType = command.execute()
return result
}
There's no way to write an execute that will actually do that, no matter what MyCustomType is.
Your confusion is compounded by a subtle syntax mistake:
func execute<String>() -> String {
This does not mean "T = String," which is what I think you expect it to mean. It creates a type variable called String (that has nothing to do with Swift's String type), and it promises to return it. when you later write as! String, that means "if this values isn't compatible with the type requested (not "a string" but whatever was requested by the caller), then crash.
The tool that behaves closer to what you want here is an associated type. You meant to write this:
protocol CommandProtocol {
associatedType T
func execute() -> T
func unExecute() -> T
}
But this almost certainly won't do what you want. For example, with that, it's impossible to have an array of commands.
Instead what you probably want is a struct:
struct Command {
let execute: () -> Void
let undo: () -> Void
}
You then make Commands by passing closures that do what you want:
let command = Command(execute: { self.value += 1 },
undo: { self.value -= 1 })
Alternately, since this is a calculator, you could do it this way:
struct Command {
let execute: (Double) -> Double
let undo: (Double) -> Double
}
let command = Command(execute: { $0 + 1 }, undo: { $0 - 1 })
Then your caller would look like:
value = command.execute(value)
value = command.undo(value)
You think this returns a Swift.Double, but no. This code is no different than using T instead of Double. Swift does not require the names of generic placeholders to match what you put in a protocol.
func unExecute<Double>() -> Double {
return calculator.performOperation(operator: undo(operator: `operator`), with: operand) as! Double
}
You're not actually looking for generic methods. You want this, instead.
protocol CommandProtocol {
associatedtype ExecuteValue
associatedtype UnExecuteValue
func execute() -> ExecuteValue
func unExecute() -> UnExecuteValue
}

Swift: Optional chaining for optional subscripts

I have a let map : [String: String] and a let key: String?.
What is the most concise way to access map[key] (and get back a String? if I had a key and None if I did not)?
let value = key.flatMap { map[$0] }
would to the trick, using the
/// Returns `nil` if `self` is nil, `f(self!)` otherwise.
#warn_unused_result
public func flatMap<U>(#noescape f: (Wrapped) throws -> U?) rethrows -> U?
method from struct Optional.
Alternatively, you can wrap that into a custom subscript method
extension Dictionary {
subscript(optKey : Key?) -> Value? {
return optKey.flatMap { self[$0] }
}
}
and the simply write
let value = map[key]
To avoid confusion with the "normal" subscript method, and to make
the intention more clear to the reader of your code, you can define
the subscript method with an external parameter name:
extension Dictionary {
subscript(optional optKey : Key?) -> Value? {
return optKey.flatMap { self[$0] }
}
}
let value = map[optional: key]
I think I am going with the decidedly unfancy
key == nil ? nil : map[key!]

Using a Type Variable in a Generic

I have this question except for Swift. How do I use a Type variable in a generic?
I tried this:
func intType() -> Int.Type {
return Int.self
}
func test() {
var t = self.intType()
var arr = Array<t>() // Error: "'t' is not a type". Uh... yeah, it is.
}
This didn't work either:
var arr = Array<t.Type>() // Error: "'t' is not a type"
var arr = Array<t.self>() // Swift doesn't seem to even understand this syntax at all.
Is there a way to do this? I get the feeling that Swift just doesn't support it and is giving me somewhat ambiguous error messages.
Edit: Here's a more complex example where the problem can't be circumvented using a generic function header. Of course it doesn't make sense, but I have a sensible use for this kind of functionality somewhere in my code and would rather post a clean example instead of my actual code:
func someTypes() -> [Any.Type] {
var ret = [Any.Type]()
for (var i = 0; i<rand()%10; i++) {
if (rand()%2 == 0){ ret.append(Int.self) }
else {ret.append(String.self) }
}
return ret
}
func test() {
var ts = self.someTypes()
for t in ts {
var arr = Array<t>()
}
}
Swift's static typing means the type of a variable must be known at compile time.
In the context of a generic function func foo<T>() { ... }, T looks like a variable, but its type is actually known at compile time based on where the function is called from. The behavior of Array<T>() depends on T, but this information is known at compile time.
When using protocols, Swift employs dynamic dispatch, so you can write Array<MyProtocol>(), and the array simply stores references to things which implement MyProtocol — so when you get something out of the array, you have access to all functions/variables/typealiases required by MyProtocol.
But if t is actually a variable of kind Any.Type, Array<t>() is meaningless since its type is actually not known at compile time. (Since Array is a generic struct, the compiler needs know which type to use as the generic parameter, but this is not possible.)
I would recommend watching some videos from WWDC this year:
Protocol-Oriented Programming in Swift
Building Better Apps with Value Types in Swift
I found this slide particularly helpful for understanding protocols and dynamic dispatch:
There is a way and it's called generics. You could do something like that.
class func foo() {
test(Int.self)
}
class func test<T>(t: T.Type) {
var arr = Array<T>()
}
You will need to hint the compiler at the type you want to specialize the function with, one way or another. Another way is with return param (discarded in that case):
class func foo() {
let _:Int = test()
}
class func test<T>() -> T {
var arr = Array<T>()
}
And using generics on a class (or struct) you don't need the extra param:
class Whatever<T> {
var array = [T]() // another way to init the array.
}
let we = Whatever<Int>()
jtbandes' answer - that you can't use your current approach because Swift is statically typed - is correct.
However, if you're willing to create a whitelist of allowable types in your array, for example in an enum, you can dynamically initialize different types at runtime.
First, create an enum of allowable types:
enum Types {
case Int
case String
}
Create an Example class. Implement your someTypes() function to use these enum values. (You could easily transform a JSON array of strings into an array of this enum.)
class Example {
func someTypes() -> [Types] {
var ret = [Types]()
for _ in 1...rand()%10 {
if (rand()%2 == 0){ ret.append(.Int) }
else {ret.append(.String) }
}
return ret
}
Now implement your test function, using switch to scope arr for each allowable type:
func test() {
let types = self.someTypes()
for type in types {
switch type {
case .Int:
var arr = [Int]()
arr += [4]
case .String:
var arr = [String]()
arr += ["hi"]
}
}
}
}
As you may know, you could alternatively declare arr as [Any] to mix types (the "heterogenous" case in jtbandes' answer):
var arr = [Any]()
for type in types {
switch type {
case .Int:
arr += [4]
case .String:
arr += ["hi"]
}
}
print(arr)
I would break it down with the things you already learned from the first answer. I took the liberty to refactor some code. Here it is:
func someTypes<T>(t: T.Type) -> [Any.Type] {
var ret = [Any.Type]()
for _ in 0..<rand()%10 {
if (rand()%2 == 0){ ret.append(T.self) }
else {
ret.append(String.self)
}
}
return ret
}
func makeArray<T>(t: T) -> [T] {
return [T]()
}
func test() {
let ts = someTypes(Int.self)
for t in ts {
print(t)
}
}
This is somewhat working but I believe the way of doing this is very unorthodox. Could you use reflection (mirroring) instead?
Its possible so long as you can provide "a hint" to the compiler about the type of... T. So in the example below one must use : String?.
func cast<T>(_ value: Any) -> T? {
return value as? T
}
let inputValue: Any = "this is a test"
let casted: String? = cast(inputValue)
print(casted) // Optional("this is a test")
print(type(of: casted)) // Optional<String>
Why Swift doesn't just allow us to let casted = cast<String>(inputValue) I'll never know.
One annoying scenerio is when your func has no return value. Then its not always straightford to provide the necessary "hint". Lets look at this example...
func asyncCast<T>(_ value: Any, completion: (T?) -> Void) {
completion(value as? T)
}
The following client code DOES NOT COMPILE. It gives a "Generic parameter 'T' could not be inferred" error.
let inputValue: Any = "this is a test"
asyncCast(inputValue) { casted in
print(casted)
print(type(of: casted))
}
But you can solve this by providing a "hint" to compiler as follows:
asyncCast(inputValue) { (casted: String?) in
print(casted) // Optional("this is a test")
print(type(of: casted)) // Optional<String>
}