I'm in a situation where I have a custom type that contains an associatedtype. In the case where this is equal to Void, I would like to have some default behaviour (to make the call-site more convenient). I tried to boil the example down to:
protocol FooType {
associatedtype T: Any
var bar: (String) -> T { get }
}
struct Foo<T>: FooType {
let bar: (String) -> T
}
extension Foo where T == Void { // Compile error: "Same-type requirement makes generic parameter 'T' non-generic".
init() {
self.bar = { _ in return }
}
}
The idea is that, in the cases where the generic type is Void, it doesn't make sense (in my scenario) to pass in a function (named bar in the example). Therefore, I just want a default implementation for this function in this specific context.
When trying to do the above I get the Same-type requirement makes generic parameter 'T' non-generic which sounds very similar to what happens when one tries to restrict e.g. the Array type when containing specific types. A workaround for this is to introduce a protocol, but I cannot do that for Void. Is it possible to do what I want or is this currently a limitation in Swift 3?
As of Swift 3.1, the code posted in the question now works. That is, the following now works as wanted:
protocol FooType {
associatedtype T: Any
var bar: (String) -> T { get }
}
struct Foo<T>: FooType {
let bar: (String) -> T
}
extension Foo where T == Void {
init() {
self.bar = { _ in return }
}
}
let foo = Foo<String>(bar: { (t: String) in return "" })
let zoo = Foo<Void>()
Related
I have some protocol hierarchies on my code where I have protocols defining the objects I use and protocols defining functions to use with this objects.
The object protocols are inherited by other object protocols that add more functionality to the original protocols and so are the functions that use them. The problem is that I can't find a way to specialize the function to take only the inherited parameter.
Here's some code to clarify what I'm trying to do:
protocol A {
var foo: String { get set }
}
protocol B: A {
var bar: String { get set }
}
struct Test: B {
var foo: String = "foo"
var bar: String = "bar"
}
protocol UseAProtocol {
static func use<T: A>(_ obj: T)
}
protocol UseBProtocol: UseAProtocol {
}
extension UseBProtocol {
//If I change the requirement to <T: B> this won't conform to `UseAProtocol`.
static func use<T: A>(_ obj: T) {
print(obj.foo)
// print(obj.bar) - Since obj does not conform to `B` I can't access ".bar" here without a forced casting.
}
}
struct Manager: UseBProtocol {
}
Manager.use(Test())
What I want to do is make the use function on the UseBProtocol only accept objects that conform to B. B inherits from A, but when I change from <T:A> to <T:B> I got an error saying that Manager does not conform to UseAProtocol and I have to change it back to <T:A>.
I know I can do this using associatedtype and where clauses on the inherit protocols - that's what I use today - but I wanted to move the generic requirement to the method so I could group all of them together under the same struct (I have a lot of this hierarchies and by using associatedtype I must use one struct by hierarchy). When the Conditional Conformances came to Swift this would be possible with associatedtype, but until them...
I could also use as! to force the casting from A to B on the UseBProtocol implementation, but that's a really bad solution and the error would be throw only at runtime.
Is there any way to achieve what I'm looking for?
It seems like what you are actually looking for is an associatedType in UseAProtocol rather than making the use function generic.
By declaring an associated type in UseAProtocol and changing the function signature of use to static func use(_ obj: ProtocolType) your code compiles fine and you can access both foo and bar from Manager.
protocol AProtocol {
var foo: String { get set }
}
protocol BProtocol: AProtocol {
var bar: String { get set }
}
struct Test: BProtocol {
var foo: String = "foo"
var bar: String = "bar"
}
protocol UseAProtocol {
associatedtype ProtocolType
static func use(_ obj: ProtocolType)
}
protocol UseBProtocol: UseAProtocol {
}
extension UseBProtocol {
static func use(_ obj: BProtocol) {
print(obj.foo)
print(obj.bar)
}
}
struct Manager: UseBProtocol {
}
Manager.use(Test()) //prints both "foo" and "bar"
As an exercise in learning I'm rewriting my validation library in Swift.
I have a ValidationRule protocol that defines what individual rules should look like:
protocol ValidationRule {
typealias InputType
func validateInput(input: InputType) -> Bool
//...
}
The associated type InputType defines the type of input to be validated (e.g String). It can be explicit or generic.
Here are two rules:
struct ValidationRuleLength: ValidationRule {
typealias InputType = String
//...
}
struct ValidationRuleCondition<T>: ValidationRule {
typealias InputType = T
// ...
}
Elsewhere, I have a function that validates an input with a collection of ValidationRules:
static func validate<R: ValidationRule>(input i: R.InputType, rules rs: [R]) -> ValidationResult {
let errors = rs.filter { !$0.validateInput(i) }.map { $0.failureMessage }
return errors.isEmpty ? .Valid : .Invalid(errors)
}
I thought this was going to work but the compiler disagrees.
In the following example, even though the input is a String, rule1's InputType is a String, and rule2s InputType is a String...
func testThatItCanEvaluateMultipleRules() {
let rule1 = ValidationRuleCondition<String>(failureMessage: "message1") { $0.characters.count > 0 }
let rule2 = ValidationRuleLength(min: 1, failureMessage: "message2")
let invalid = Validator.validate(input: "", rules: [rule1, rule2])
XCTAssertEqual(invalid, .Invalid(["message1", "message2"]))
}
... I'm getting extremely helpful error message:
_ is not convertible to ValidationRuleLength
which is cryptic but suggests that the types should be exactly equal?
So my question is... how do I append different types that all conform to a protocol with an associated type into a collection?
Unsure how to achieve what I'm attempting, or if it's even possible?
EDIT
Here's it is without context:
protocol Foo {
typealias FooType
func doSomething(thing: FooType)
}
class Bar<T>: Foo {
typealias FooType = T
func doSomething(thing: T) {
print(thing)
}
}
class Baz: Foo {
typealias FooType = String
func doSomething(thing: String) {
print(thing)
}
}
func doSomethingWithFoos<F: Foo>(thing: [F]) {
print(thing)
}
let bar = Bar<String>()
let baz = Baz()
let foos: [Foo] = [bar, baz]
doSomethingWithFoos(foos)
Here we get:
Protocol Foo can only be used as a generic constraint because it has
Self or associated type requirements.
I understand that. What I need to say is something like:
doSomethingWithFoos<F: Foo where F.FooType == F.FooType>(thing: [F]) {
}
Protocols with type aliases cannot be used this way. Swift doesn't have a way to talk directly about meta-types like ValidationRule or Array. You can only deal with instantiations like ValidationRule where... or Array<String>. With typealiases, there's no way to get there directly. So we have to get there indirectly with type erasure.
Swift has several type-erasers. AnySequence, AnyGenerator, AnyForwardIndex, etc. These are generic versions of protocols. We can build our own AnyValidationRule:
struct AnyValidationRule<InputType>: ValidationRule {
private let validator: (InputType) -> Bool
init<Base: ValidationRule where Base.InputType == InputType>(_ base: Base) {
validator = base.validate
}
func validate(input: InputType) -> Bool { return validator(input) }
}
The deep magic here is validator. It's possible that there's some other way to do type erasure without a closure, but that's the best way I know. (I also hate the fact that Swift cannot handle validate being a closure property. In Swift, property getters aren't proper methods. So you need the extra indirection layer of validator.)
With that in place, you can make the kinds of arrays you wanted:
let len = ValidationRuleLength()
len.validate("stuff")
let cond = ValidationRuleCondition<String>()
cond.validate("otherstuff")
let rules = [AnyValidationRule(len), AnyValidationRule(cond)]
let passed = rules.reduce(true) { $0 && $1.validate("combined") }
Note that type erasure doesn't throw away type safety. It just "erases" a layer of implementation detail. AnyValidationRule<String> is still different from AnyValidationRule<Int>, so this will fail:
let len = ValidationRuleLength()
let condInt = ValidationRuleCondition<Int>()
let badRules = [AnyValidationRule(len), AnyValidationRule(condInt)]
// error: type of expression is ambiguous without more context
Consider the following:
protocol Foo {
typealias A
func hello() -> A
}
protocol FooBar: Foo {
func hi() -> A
}
extension FooBar {
func hello() -> A {
return hi()
}
}
class FooBarClass: FooBar {
typealias A = String
func hi() -> String {
return "hello world"
}
}
This code compiles. But if I comment out explicit definition of associated type typealias A = String, then for some reason, swiftc fails to infer the type.
I'm sensing this has to do with two protocols sharing the same associated type but without a direct assertion through, for example, type parameterization (maybe associated type is not powerful/mature enough?), which makes it ambiguous for type inference.
I'm not sure if this is a bug / immaturity of the language, or maybe, I'm missing some nuances in protocol extension which rightfully lead to this behaviour.
Can someone shed some light on this?
look at this example
protocol Foo {
typealias A
func hello() -> A
}
protocol FooBar: Foo {
typealias B
func hi() -> B
}
extension FooBar {
func hello() -> B {
return hi()
}
}
class FooBarClass: FooBar {
//typealias A = String
func hi() -> String {
return "hello world"
}
}
with generics
class FooBarClass<T>: FooBar {
var t: T?
func hi() -> T? {
return t
}
}
let fbc: FooBarClass<Int> = FooBarClass()
fbc.t = 10
fbc.hello() // 10
fbc.hi() // 10
Providing explicit values for associated types in a protocol is required for conformance to said protocol. This can be accomplished by hard coding a type, as you've done with typealias A = String, or using a parameterized type as you mentioned, such as below:
class FooBarClass<T>: FooBar {
typealias A = T
...
}
Swift will not infer your associated type from an implemented method of the protocol, as there could be ambiguity with multiple methods with mismatching types. This is why the typealias must be explicitly resolved in your implementing class.
I've been dabbling with Swift recently, and I've hit a weird stumbling block with type constraints not operating as I would expect they should (in comparison to say, Scala).
protocol Foo {
typealias T: Hashable
func somethingWithT() -> T
}
struct Bar: Foo {
typealias T = Int
func somethingWithT() -> T { return 1 }
}
func baz() -> [Foo] {
var myBars = [Foo]()
myBars.append(Bar())
return myBars
}
This throws an error in XCode:
Any ideas what is going on here? I want T to be hashable for use as a key in a dictionary, but from what I've read Hashable has a reference to Self somewhere, so it can only be used as a generic constraint, thus removing my ability to just have a [Foo].
I want to have a list of things that do Foo, but at this rate it seems I'll either have to remove the Hashable constraint or make it less generic..
I've tried cleaning my project and re-making, but no dice :(
The problem here is that the type of T (of the protocol) has to be known at runtime. Due to the lack of generics with type aliases you can work around that by making an AnyFoo type (like in the standard library AnyGenerator and AnySequence):
protocol Foo {
typealias T: Hashable
func somethingWithT() -> T
}
struct AnyFoo<T: Hashable>: Foo {
let function: () -> T
init<F: Foo where F.T == T>(_ foo: F) {
// storing a reference to the function of the original type
function = foo.somethingWithT
}
func somethingWithT() -> T {
return function()
}
}
struct Bar: Foo {
typealias T = Int
func somethingWithT() -> T { return 1 }
}
// instead of returning [Foo] you can return [AnyFoo<Int>]
func baz() -> [AnyFoo<Int>] {
var myBars = [AnyFoo<Int>]()
// converting the type or Bar
myBars.append(AnyFoo(Bar()))
return myBars
}
This is not a generic function but you can convert any Foo type with the same T to the same AnyFoo type
myBars is an array of T: Foo. You could certainly pass in a Bar, because it satisfies the constraint on T. But so could another concrete type. Since T must be a concrete type, you can't arbitrarily put a Bar on there without guaranteeing that the only thing you'll put in there are Bars. And the only way to do that in Swift with this function signature would be to pass in a Bar.
I think maybe what you want is an array of protocol objects, not a generic array.
I have the following protocol and a class that conforms to it:
protocol Foo{
typealias BazType
func bar(x:BazType) ->BazType
}
class Thing: Foo {
func bar(x: Int) -> Int {
return x.successor()
}
}
When I try to create an Array of foos, I get an odd error:
var foos: Array<Foo> = [Thing()]
Protocol Foo can only be used as a generic constraint because it has
Self or associated type requirements.
OK, so it can only be used if it has an associated type requirement (which it does), but for some reason this is an error?? WTF?!
I'm not sure I fully understand what the compiler is trying to tell me...
Let's say, if we could put an instance of Thing into array foos, what will happen?
protocol Foo {
associatedtype BazType
func bar(x:BazType) -> BazType
}
class Thing: Foo {
func bar(x: Int) -> Int {
return x.successor()
}
}
class AnotherThing: Foo {
func bar(x: String) -> String {
return x
}
}
var foos: [Foo] = [Thing()]
Because AnotherThing conforms to Foo too, so we can put it into foos also.
foos.append(AnotherThing())
Now we grab a foo from foos randomly.
let foo = foos[Int(arc4random_uniform(UInt32(foos.count - 1)))]
and I'm going to call method bar, can you tell me that I should send a string or an integer to bar?
foo.bar("foo") or foo.bar(1)
Swift can't.
So it can only be used as a generic constraint.
What scenario requires a protocol like this?
Example:
class MyClass<T: Foo> {
let fooThing: T?
init(fooThing: T? = nil) {
self.fooThing = fooThing
}
func myMethod() {
let thing = fooThing as? Thing // ok
thing?.bar(1) // fine
let anotherThing = fooThing as? AnotherThing // no problem
anotherThing?.bar("foo") // you can do it
// but you can't downcast it to types which doesn't conform to Foo
let string = fooThing as? String // this is an error
}
}
I have been playing with your code trying to understand how to implement the protocol. I found that you can't use Typealias as a generic type because it is just an alias not a type by itself. So if you declare the Typealias outside your protocol and your class you can effectively use it in your code without any problem.
Note: the Typealias has the Int type in its declaration, that way you can always use the alias instead of the Int type and use all of its associated methods and functions.
Here's how I make it work:
typealias BazType = Int
protocol Foo{
func bar(x:BazType) -> BazType
}
class Thing: Foo {
func bar(x: BazType) -> BazType {
return x.successor()
}
}
let elements: Array<Foo> = [Thing(), Thing()]