Note
Swift is changing rapidly, this question was asked regarding:
Xcode 7, Swift 2.0
Explanation
I'm looking to implement a generic return argument. Quite often, I find it necessary to implement an optional version overload so I can access the underlying type and handle it appropriately. Here's some manufactured functions. The assignment of String is just there as a placeholder for replication:
func ambiguous<T>() -> T {
let thing = "asdf"
return thing as! T
}
func ambiguous<T>() -> T? {
return nil
}
Now, if we look at the implementation:
// Fine
let a: String = ambiguous()
// Ambiguous
let b: String? = ambiguous()
This might seem obvious because you could assign type T to a variable of type T?. So it makes sense that it would have trouble inferring. The problem is, that with a type constraint, it suddenly works. (This can be anything, I'm using Equatable for easy replication.
func nonAmbiguous<T : Equatable>() -> T {
let thing: AnyObject = "asdf"
return thing as! T
}
func nonAmbiguous<T : Equatable>() -> T? {
return nil
}
And now, it functions as expected:
// Fine
let c: String = nonAmbiguous()
// Fine
let d: String? = nonAmbiguous()
Note, this also works with other type:
func nonAmbiguous<T>() -> [T] {
let thing: AnyObject = ["asdf"]
return thing as! [T]
}
func nonAmbiguous<T>() -> [T]? {
return nil
}
// Fine
let e: [String] = nonAmbiguous()
// Fine
let d: [String]? = nonAmbiguous()
Question:
Is there a way to have a return generic argument infer the appropriate overload through optionality?
if no
Is this a language feature, or a bug somewhere. If it's a language feature, please explain the underlying issue preventing the possibility of this behavior.
The first example is ambiguous because T can be inferred as both String
and String?.
The second example is not ambiguous because String is Equatable but String? is not, so T : Equatable cannot be inferred as String?.
The third case is not ambiguous because [T] is not
inferred as [String]?.
Remark: Generally, Optional<Wrapped> does not conform to Equatable
even if Wrapped does, in the same way as Array<Element>
does not conform to Equatable even if Element does.
This is a restriction of the current type system in Swift which
might be improved in a future version, compare
[swift-dev] RFC: Adding Optional variants of == for collections to the std lib.
from the Swift development mailing list.
Related
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)")
}
}
}
It seems absurd that this method signature does not compile in Swift 4:
class Bar<ValueType> {
func version() throws -> String where ValueType == [String: Any] { ... }
}
(Error: where clause cannot be attached to a non-generic declaration)
but this compiles fine:
class Bar<ValueType> {
func version<T>(_ foo: T? = nil) throws -> String where ValueType == [String: Any] { ... }
}
Anyone have insight as to why this is the case?
Because ValueType has nothing to do with this method (in the first example). It would be wrong to put such a method in a type (class/struct/enum), since it's not really a true member of that type. It's conditionally a member of that type, depending on the truth value of the where clause.
To achieve this, you would want to put this method in an extension of your type, with the where clause you want. E.g.
extension YourType where ValueType == [String: Any] {
func version() throws -> String { ... }
}
That has been finally allowed in Swift 5.3. See Contextual Where Clauses in the official Language Guide.
Citing from there:
You can write a generic where clause as part of a declaration that doesn’t have its own generic type constraints, when you’re already working in the context of generic types. For example, you can write a generic where clause on a subscript of a generic type or on a method in an extension to a generic type.
...
extension Container {
func average() -> Double where Item == Int {
var sum = 0.0
for index in 0..<count {
sum += Double(self[index])
}
return sum / Double(count)
}
That compiles just fine in Xcode 12 beta 4, but won't work in Xcode 11.6 that is shipped with Swift 5.2.4.
I am using Swift 3 with constrained generics (i.e. a where clause). I have a problem when I am trying to do generic type casting. Here is a simplified example of the problem:
func jsonToObj<T:DomainResource>(jsonStr: String) -> [T:DomainResource] {
let json = JSON(parseJSON: jsonStr).dictionaryObject
let bundle = SMART.Bundle(json: json)
let result = bundle.entry?.map() {
return $0.resource as! T
}
return result!
}
My problem is when I return from the method, the compiler complains its cannot convert type [T] to type [T:DomainResource]. If I remove the DomainResource constraint from the generic, it compiles and runs just fine.
That's not what I want, so, I tried this:
let result = bundle.entry?.map() {
return $0.resource as! T:DomainResource
}
Swift doesn't seem to know what that means. Any idea on how to work around this problem? I'd like to not just cast them all to DomainResource objects, if possible.
You wrote this function signature:
func jsonToObj<T:DomainResource>(jsonStr: String) -> [T:DomainResource]
This says that the jsonToObj(jsonStr:) method returns a dictionary whose keys are of type T and whose values are of type DomainResource. It looks like you just want to write this function signature:
func jsonToObj<T:DomainResource>(jsonStr: String) -> [T]
I watched some iOS programming tutorial and have a question on "AnyObject".
The bridging does not work.
I have the following code:
import Foundation
class CalculatorBrain
{
private var accumulator = 0.0
var internalProgram = [AnyObject]()
func setOperand (operand: Double) {
accumulator = operand
internalProgram.append(operand)
}
// ...
}
Same for string...
func performOperation (symbol: String) {
internalProgram.append(symbol)
}
I know about NSString and the reference type vs. struct thing, and that double and string are no reference types.
But anyway in the tutorial the bridging worked fine...
What could be the reason for my failure?
As you correctly say, Double and String are not reference types in Swift – they are structs. Therefore you cannot directly store them in an [AnyObject], you would first need to bridge them to Objective-C in order to do so.
Although bridging to Objective-C in this case is unnecessary – if you want an array of heterogenous types, including value types, then you can use an [Any]:
var internalProgram = [Any]()
However, from what I can tell, you don't want a array of anything (it's very rarely a good idea) – you just want an array that can contain a Double or a String.
You can describe this with an enum:
enum OperationArgument { // feel free to give me a better name
case operand(Double)
case symbol(String)
}
And now you can define an array of it:
var internalProgram = [OperationArgument]()
func setOperand (operand: Double) {
internalProgram.append(.operand(operand))
}
func performOperation (symbol: String) {
internalProgram.append(.symbol(symbol))
}
Now instead of conditional type-casting to get back the type of each element, you can just use a switch or an if case:
for element in internalProgram {
switch element {
case let .operand(operand):
print(operand)
case let .symbol(symbol):
print(symbol)
}
if case let .operand(operand) = element {
print(operand)
}
}
To use a Swift value type as an AnyObject, in Swift 3 you need to explicitly cast it to the old Objective-C object type.
So this
internalProgram.append(operand)
becomes this
internalProgram.append(operand as NSNumber)
And this
internalProgram.append(symbol)
becomes this
internalProgram.append(symbol as NSString)
The tutorial you are talking about has probably been written with Swift 2 where you just needed to import Foundation.
var internalProgram = [AnyObject]()
func setOperand (operand: Double) {
// you have cast operand as AnyObject !!
internalProgram.append(operand as AnyObject)
}
setOperand(operand: 10)
print(internalProgram, type(of: internalProgram)) // [10] Array<AnyObject>
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.