The new SwiftUI tutorial has the following code:
struct ContentView: View {
var body: some View {
Text("Hello World")
}
}
The second line the word some, and on their site is highlighted as if it were a keyword.
Swift 5.1 does not appear to have some as a keyword, and I don't see what else the word some could be doing there, since it goes where the type usually goes. Is there a new, unannounced version of Swift? Is it a function that's being used on a type in a way I didn't know about?
What does the keyword some do?
some View is an opaque result type as introduced by SE-0244 and is available in Swift 5.1 with Xcode 11. You can think of this as being a "reverse" generic placeholder.
Unlike a regular generic placeholder which is satisfied by the caller:
protocol P {}
struct S1 : P {}
struct S2 : P {}
func foo<T : P>(_ x: T) {}
foo(S1()) // Caller chooses T == S1.
foo(S2()) // Caller chooses T == S2.
An opaque result type is an implicit generic placeholder satisfied by the implementation, so you can think of this:
func bar() -> some P {
return S1() // Implementation chooses S1 for the opaque result.
}
as looking like this:
func bar() -> <Output : P> Output {
return S1() // Implementation chooses Output == S1.
}
In fact, the eventual goal with this feature is to allow reverse generics in this more explicit form, which would also let you add constraints, e.g -> <T : Collection> T where T.Element == Int. See this post for more info.
The main thing to take away from this is that a function returning some P is one that returns a value of a specific single concrete type that conforms to P. Attempting to return different conforming types within the function yields a compiler error:
// error: Function declares an opaque return type, but the return
// statements in its body do not have matching underlying types.
func bar(_ x: Int) -> some P {
if x > 10 {
return S1()
} else {
return S2()
}
}
As the implicit generic placeholder cannot be satisfied by multiple types.
This is in contrast to a function returning P, which can be used to represent both S1 and S2 because it represents an arbitrary P conforming value:
func baz(_ x: Int) -> P {
if x > 10 {
return S1()
} else {
return S2()
}
}
Okay, so what benefits do opaque result types -> some P have over protocol return types -> P?
A major current limitation of protocols is that PATs (protocols with associated types) cannot be used as actual types. Although this is a restriction that will likely be lifted in a future version of the language, because opaque result types are effectively just generic placeholders, they can be used with PATs today.
This means you can do things like:
func giveMeACollection() -> some Collection {
return [1, 2, 3]
}
let collection = giveMeACollection()
print(collection.count) // 3
Because opaque result types enforce a single concrete type is returned, the compiler knows that two calls to the same function must return two values of the same type.
This means you can do things like:
// foo() -> <Output : Equatable> Output {
func foo() -> some Equatable {
return 5 // The opaque result type is inferred to be Int.
}
let x = foo()
let y = foo()
print(x == y) // Legal both x and y have the return type of foo.
This is legal because the compiler knows that both x and y have the same concrete type. This is an important requirement for ==, where both parameters of type Self.
protocol Equatable {
static func == (lhs: Self, rhs: Self) -> Bool
}
This means that it expects two values that are both the same type as the concrete conforming type. Even if Equatable were usable as a type, you wouldn't be able to compare two arbitrary Equatable conforming values with each other, for example:
func foo(_ x: Int) -> Equatable { // Assume this is legal.
if x > 10 {
return 0
} else {
return "hello world"
}
}
let x = foo(20)
let y = foo(5)
print(x == y) // Illegal.
As the compiler cannot prove that two arbitrary Equatable values have the same underlying concrete type.
In a similar manner, if we introduced another opaque type returning function:
// foo() -> <Output1 : Equatable> Output1 {
func foo() -> some Equatable {
return 5 // The opaque result type is inferred to be Int.
}
// bar() -> <Output2 : Equatable> Output2 {
func bar() -> some Equatable {
return "" // The opaque result type is inferred to be String.
}
let x = foo()
let y = bar()
print(x == y) // Illegal, the return type of foo != return type of bar.
The example becomes illegal because although both foo and bar return some Equatable, their "reverse" generic placeholders Output1 and Output2 could be satisfied by different types.
Unlike regular protocol-typed values, opaque result types compose well with regular generic placeholders, for example:
protocol P {
var i: Int { get }
}
struct S : P {
var i: Int
}
func makeP() -> some P { // Opaque result type inferred to be S.
return S(i: .random(in: 0 ..< 10))
}
func bar<T : P>(_ x: T, _ y: T) -> T {
return x.i < y.i ? x : y
}
let p1 = makeP()
let p2 = makeP()
print(bar(p1, p2)) // Legal, T is inferred to be the return type of makeP.
This wouldn't have worked if makeP had just returned P, as two P values may have different underlying concrete types, for example:
struct T : P {
var i: Int
}
func makeP() -> P {
if .random() { // 50:50 chance of picking each branch.
return S(i: 0)
} else {
return T(i: 1)
}
}
let p1 = makeP()
let p2 = makeP()
print(bar(p1, p2)) // Illegal.
At this point you may be thinking to yourself, why not just write the code as:
func makeP() -> S {
return S(i: 0)
}
Well, the use of an opaque result type allows you to make the type S an implementation detail by exposing only the interface provided by P, giving you flexibility of changing the concrete type later down the line without breaking any code that depends on the function.
For example, you could replace:
func makeP() -> some P {
return S(i: 0)
}
with:
func makeP() -> some P {
return T(i: 1)
}
without breaking any code that calls makeP().
See the Opaque Types section of the language guide and the Swift evolution proposal for further information on this feature.