It is legal to say this (arr is an Array):
let arrenum = Array(arr.enumerated())
So why isn't it legal to say this?
extension Array {
func f() {
let arrenum = Array(self.enumerated())
// error: type of expression is ambiguous without more context
}
}
EDIT It seems this is a workaround:
extension Array {
func f() {
typealias Tup = (offset:Index, element:Element)
let arrenum = Array<Tup>(self.enumerated())
}
}
But why is that needed? (And is it right?)
This is a known bug (SR-1789). Swift currently has a feature where you can refer to a generic type within its own body without having to repeat its placeholder type(s) – the compiler will infer them for you to be the same as the type of self.
For example:
struct S<T> {
func foo(_ other: S) { // parameter inferred to be `S<T>`.
let x = S() // `x` inferred to be `S<T>`.
}
}
extension S {
func bar(_ other: S) {} // same in extensions too.
}
This is pretty convenient, but the bug you're running into is the fact that Swift will always make this inference, even if it's incorrect.
So, in your example:
extension Array {
func f() {
let arrenum = Array(self.enumerated())
// error: type of expression is ambiguous without more context
}
}
Swift interprets the code as let arrenum = Array<Element>(self.enumerated()), as you're in the body of Array<Element>. This is incorrect, because enumerated() yields a sequence of offset-element tuple pairs – Swift should have inferred Array to be Array<(offset: Int, element: Element)> instead.
One workaround, which you've already discovered, is to explicitly specify the placeholder type in order to prevent the compiler from making this incorrect inference.
extension Array {
func f() {
let arrenum = Array<(offset: Int, element: Element)>(self.enumerated())
}
}
Another possible workaround appears to be using the fully-qualified type, for example:
extension Array {
func f() {
let arrenum = Swift.Array(self.enumerated())
}
}
as it appears Swift doesn't do the same inference for fully-qualified types (I'm not sure if you should rely on this fact though).
Finally it's worth noting that instead of doing a call to Array's initialiser, you could use map(_:) instead to avoid the issue entirely:
extension Array {
func f() {
let arrenum = self.enumerated().map { $0 }
}
}
which, like the initialiser call, will give you back an array of offset-element pairs.
Related
In Swift, you can create a reference to a function in the form of a closure. For example:
func simpleFunc(param: Int) {
}
let simpleFuncReference = simpleFunc(param:) // works just fine
But in one case, I have a function with a generic parameter like this:
func hardFunc<T: StringProtocol>(param: T) {
}
let hardFuncReference = hardFunc(param:) // "Generic parameter 'T' could not be inferred"
To try to remove that error, I attempted to explicitly specify the type, but immediately another error comes up.
func hardFunc<T: StringProtocol>(param: T) {
}
let hardFuncReference = hardFunc(param:) // "Cannot explicitly specialize a generic function"
Is there a way I can get a reference to hardFunc as a closure?
As you already guessed, you have to help type inference out a little:
func hardFunc<T: StringProtocol>(param: T) {
}
let hardFuncReference:(String) -> Void = hardFunc(param:)
Note that you do have to specify the particular type that you're specializing on, but in this case you do it by specifying the type of the variable you're assigning the closure to.
You can't keep it generic unless you're in a generic context specifying that it's generic on the same type. So this will work too
struct Foo<T: StringProtocol> {
let hardFuncReference:(T) -> Void = hardFunc(param:)
}
I am using Xcode Version 11.3.1 (11C504)
I am trying to create a generic function in Swift that will reject its
parameter unless such a parameter is Optional.
In the following code, I was expecting the system to report errors in all calls to onlyCallableByAnOptable() made inside test(), because none of them provide an optional value as a parameter.
However, the system only reports non-protocol conformance if I remove the Optional extension that conforms to Optable!
Which to me, it means that the system is regarding any and all values as Optional, regardless!
Am I doing something wrong?
(By the way, the following code used to be working as expected in earlier versions of Swift. I just recently found out that it stopped working, for it was letting a non-Optional go through.)
protocol Optable {
func opt()
}
func onlyCallableByAnOptable<T>( _ value: T) -> T where T: Optable {
return value
}
// Comment the following line to get the errors
extension Optional: Optable { func opt() {} }
class TestOptable {
static func test()
{
let c = UIColor.blue
let s = "hi"
let i = Int(1)
if let o = onlyCallableByAnOptable(c) { print("color \(o)") }
//^ expected ERROR: Argument type 'UIColor' does not conform to expected type 'Optable'
if let o = onlyCallableByAnOptable(s) { print("string \(o)") }
//^ expected ERROR: Argument type 'String' does not conform to expected type 'Optable'
if let o = onlyCallableByAnOptable(i) { print("integer \(o)") }
//^ expected ERROR: Argument type 'Int' does not conform to expected type 'Optable'
}
}
Since you've made all Optionals conform to Optable and you are using the if let syntax to unwrap the result of the call to onlyCallableByAnOptable (which means the return type must be some kind of Optional, which means the parameter must also be that same type of Optional because both the parameter and the return type are of type T in your generic method), Swift is inferring the types being passed in as UIColor?, String?, and Int? (implicitly wrapping them in Optionals) instead of UIColor, String and Int.
I am the one who posted this question.
I was trying to create a generic function in Swift that would reject
its parameter unless such parameter is an Optional.
As #TylerTheCompiler pointed out, using my original implementation (in the question), Swift was inferring type T (used in onlyCallableByAnOptable()), based on the full context of the call, not solely on the type of the value provided as parameter to it, therefore inferring T to be an Optional.
For the sake of helping others who might be trying to achieve the same as I was, the following is my solution to the problem I had.
All calls to onlyCallableByAnOptable(...) now correctly yield errors due to non-protocol conformance.
Errors like: Argument type 'UIColor' does not conform to expected type 'Optable'
If anyone knows of a simpler solution, please do post it as an answer
to: How to create a generic function in Swift that will reject the given parameter unless it is an Optional?.
protocol Optable {
associatedtype OptableType
func optionalOptable() -> OptableType?
func opt()
}
func onlyCallableByAnOptable<T>( _ value: T) -> T.OptableType? where T: Optable {
return value.optionalOptable()
}
extension Optional: Optable {
typealias OptableType = Wrapped //: Wrapped is the type of the element, as defined in Optional
func opt() {}
func optionalOptable() -> OptableType? {
return self
}
}
class TestOptable {
static func test()
{
let c = UIColor.blue
let s = "hi"
let i = Int(1)
if let o = onlyCallableByAnOptable(c) { // ERROR, as was desired.
print("color \(o)")
}
if let o = onlyCallableByAnOptable(s) { // ERROR, as was desired.
print("string \(o)")
}
if let o = onlyCallableByAnOptable(i) { // ERROR, as was desired.
print("integer \(o)")
}
}
}
I am attempting to create a function that can return any type. I do not want it to return an object of type Any, but of other types, i.e. String, Bool, Int, etc. You get the idea.
You can easily do this using generics in this fashion:
func example<T>(_ arg: T) -> T {
// Stuff here
}
But is it possible to do it without passing in any arguments of the same type? Here is what I am thinking of:
func example<T>() -> T {
// Stuff here
}
When I try to do this, everything works until I call the function, then I get this error:
generic parameter 'T' could not be inferred
is it possible to do it without passing in any arguments of the same type?
The answer is yes, but there needs to be a way for the compiler to infer the correct version of the generic function. If it knows what it is assigning the result to, it will work. So for instance, you could explicitly type a let or var declaration. The below works in a playground on Swift 3.
protocol Fooable
{
init()
}
extension Int: Fooable {}
extension String: Fooable {}
func foo<T: Fooable>() -> T
{
return T()
}
let x: String = foo() // x is assigned the empty string
let y: Int = foo() // y is assigned 0
In C#, it's possible to call a generic method by specifying the type:
public T f<T>()
{
return something as T
}
var x = f<string>()
Swift doesn't allow you to specialize a generic method when calling it. The compiler wants to rely on type inference, so this is not possible:
func f<T>() -> T? {
return something as T?
}
let x = f<String>() // not allowed in Swift
What I need is a way to pass a type to a function and that function returning an object of that type, using generics
This works, but it's not a good fit for what I want to do:
let x = f() as String?
EDIT (CLARIFICATION)
I've probably not been very clear about what the question actually is, it's all about a simpler syntax for calling a function that returns a given type (any type).
As a simple example, let's say you have an array of Any and you create a function that returns the first element of a given type:
// returns the first element in the array of that type
func findFirst<T>(array: [Any]) -> T? {
return array.filter() { $0 is T }.first as? T
}
You can call this function like this:
let array = [something,something,something,...]
let x = findFirst(array) as String?
That's pretty simple, but what if the type returned is some protocol with a method and you want to call the method on the returned object:
(findFirst(array) as MyProtocol?)?.SomeMethodInMyProtocol()
(findFirst(array) as OtherProtocol?)?.SomeMethodInOtherProtocol()
That syntax is just awkward. In C# (which is just as strongly typed as Swift), you can do this:
findFirst<MyProtocol>(array).SomeMethodInMyProtocol();
Sadly, that's not possible in Swift.
So the question is: is there a way to accomplish this with a cleaner (less awkward) syntax.
Unfortunately, you cannot explicitly define the type of a generic function (by using the <...> syntax on it). However, you can provide a generic metatype (T.Type) as an argument to the function in order to allow Swift to infer the generic type of the function, as Roman has said.
For your specific example, you'll want your function to look something like this:
func findFirst<T>(in array: [Any], ofType _: T.Type) -> T? {
return array.lazy.compactMap { $0 as? T }.first
}
Here we're using compactMap(_:) in order to get a sequence of elements that were successfully cast to T, and then first to get the first element of that sequence. We're also using lazy so that we can stop evaluating elements after finding the first.
Example usage:
protocol SomeProtocol {
func doSomething()
}
protocol AnotherProtocol {
func somethingElse()
}
extension String : SomeProtocol {
func doSomething() {
print("success:", self)
}
}
let a: [Any] = [5, "str", 6.7]
// Outputs "success: str", as the second element is castable to SomeProtocol.
findFirst(in: a, ofType: SomeProtocol.self)?.doSomething()
// Doesn't output anything, as none of the elements conform to AnotherProtocol.
findFirst(in: a, ofType: AnotherProtocol.self)?.somethingElse()
Note that you have to use .self in order to refer to the metatype of a specific type (in this case, SomeProtocol). Perhaps not a slick as the syntax you were aiming for, but I think it's about as good as you're going to get.
Although it's worth noting in this case that the function would be better placed in an extension of Sequence:
extension Sequence {
func first<T>(ofType _: T.Type) -> T? {
// Unfortunately we can't easily use lazy.compactMap { $0 as? T }.first
// here, as LazyMapSequence doesn't have a 'first' property (we'd have to
// get the iterator and call next(), but at that point we might as well
// do a for loop)
for element in self {
if let element = element as? T {
return element
}
}
return nil
}
}
let a: [Any] = [5, "str", 6.7]
print(a.first(ofType: String.self) as Any) // Optional("str")
What you probably need to do is create a protocol that looks something like this:
protocol SomeProtocol {
init()
func someProtocolMethod()
}
And then add T.Type as a parameter in your method:
func f<T: SomeProtocol>(t: T.Type) -> T {
return T()
}
Then assuming you have a type that conforms to SomeProtocol like this:
struct MyType: SomeProtocol {
init() { }
func someProtocolMethod() { }
}
You can then call your function like this:
f(MyType.self).someProtocolMethod()
Like others have noted, this seems like a convoluted way to do what you want. If you know the type, for example, you could just write:
MyType().someProtocolMethod()
There is no need for f.
Is there a keyword to place-hold unused type parameters?
In this example, receiver does not use T of MyGen.
In Java language, it can be written MyGen<?> v.
I could not find a counterpart in swift language documents.
import Foundation
class MyGen<T: Printable> {
var value: T
init(v: T) {
value = v
}
}
func receiver(v: MyGen<WHAT_COMES_HERE>) {
println(v);
}
let s = MyGen<NSString>(v: "hello")
receiver(s)
I know that giving receiver a type parameter solves the problem, but it is not welcome because upper bound Printable is repeated as many as functions and the code has redundant information.
// This works, but not welcome
func receiver<T: Printable>(v: MyGen<T>) {
println(v);
}
In Swift, despite it not being used, you have to declare a type parameters. But, you don't need :Printable in here, You can use just T or whatever as you like.
func receiver<A>(v: MyGen<A>) {
println(v)
}
Use : for the type parameter, only if receiver requires more specialized MyGet.T. For example:
class MyGen<T: Printable> {
var value: T
init(v: T) {
value = v
}
}
// `receiver` accepts `MyGen` only if its `T` is a sequence of `Int`
func receiver<A: SequenceType where A.Generator.Element == Int>(v: MyGen<A>) {
println(reduce(v.value, 0, +))
}
let s = MyGen<[Int]>(v: [1,2,3])
receiver(s) // -> 6