Following code:
protocol SomeProtocol {
typealias SomeType = Int // used typealias-assignment
func someFunc(someVar: SomeType)
}
class SomeClass: SomeProtocol {
func someFunc(someVar: SomeType) {
print(someVar)
}
}
gives compile-time error:
Use of undeclared type 'SomeType'
Adding, say typealias SomeType = Double
, to the SomeClass
resolves the error.
The question is, what's the point of typealias-assignment part (which is optional btw) of protocol associated type declaration though?
In this case the assignment of Int
to the typealias
is equal to no assignment because it gets overridden by your conforming type:
// this declaration is equal since you HAVE TO provide the type for SomeType
protocol SomeProtocol {
typealias SomeType
func someFunc(someVar: SomeType)
}
Such an assignment provides a default type for SomeType
which gets overridden by your implementation in SomeClass
, but it is especially useful for protocol extensions:
protocol Returnable {
typealias T = Int // T is by default of type Int
func returnValue(value: T) -> T
}
extension Returnable {
func returnValue(value: T) -> T {
return value
}
}
struct AStruct: Returnable {}
AStruct().returnValue(3) // default signature: Int -> Int
You get the function for free only by conforming to the protocol without specifying the type of T
. If you want to set your own type write typealias T = String // or any other type
in the struct body.
Some additional notes about the provided code example
You solved the problem because you made it explicit which type the parameter has. Swift also infers your used type:
class SomeClass: SomeProtocol {
func someFunc(someVar: Double) {
print(someVar)
}
}
So SomeType
of the protocol is inferred to be Double
.
Another example where you can see that SomeType
in the class declaration doesn't refer to to the protocol:
class SomeClass: SomeProtocol {
typealias Some = Int
func someFunc(someVar: Some) {
print(someVar)
}
}
// check the type of SomeType of the protocol
// dynamicType returns the current type and SomeType is a property of it
SomeClass().dynamicType.SomeType.self // Int.Type
// SomeType gets inferred form the function signature
However if you do something like that:
protocol SomeProtocol {
typealias SomeType: SomeProtocol
func someFunc(someVar: SomeType)
}
SomeType
has to be of type SomeProtocol
which can be used for more explicit abstraction and more static code whereas this:
protocol SomeProtocol {
func someFunc(someVar: SomeProtocol)
}
would be dynamically dispatched.
There is some great information in the documentation on "associated types" in protocols.
Their use is abundant throughout the standard library, for an example reference the SequenceType protocol, which declares a typealias
for Generator
(and specifies that it conforms to GeneratorType
). This allows the protocol declaration to refer to that aliased type.
In your case, where you used typealias SomeType = Int
, perhaps what you meant was "I want SomeType to be constrained to Integer-like behavior because my protocol methods will depend on that constraint" - in which case, you may want to use typealias SomeType: IntegerType
in your protocol, and then in your class go on to assign a type to that alias which conforms to IntegerType
.
UPDATE
After opening a bug w/ Apple on this and having had extensive discussion around it, I have come to an understanding of what the base issue is at the heart of this:
when conforming to a protocol, you cannot directly refer to an associated type that was declared only within that protocol
(note, however, that when extending a protocol the associated type is available, as you would expect)
So in your initial code example:
protocol SomeProtocol {
typealias SomeType = Int
func someFunc(someVar: SomeType)
}
class SomeClass: SomeProtocol {
func someFunc(someVar: SomeType) { // use of undeclared type "SomeType"
print(someVar)
}
}
...the error re: "use of undeclared type" is correct, your class SomeClass
has not declared the type SomeType
However, an extension to SomeProtocol
has access to the associated type, and can refer to it when providing an implementation:
(note that this requires using a where
clause in order to define the requirement on the associated type)
protocol SomeProtocol {
typealias SomeType = Int
func someFunc(someVar: SomeType)
}
extension SomeProtocol where SomeType == Int {
func someFunc(someVar: SomeType) {
print("1 + \(someVar) = \(1 + someVar)")
}
}
class SomeClass: SomeProtocol {}
SomeClass().someFunc(3) // => "1 + 3 = 4"
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