How to create generic protocols in Swift?
I'd like to create a protocol with a method that takes a generic input and returns a generic value.
This is what I've tried so far, but it produces the syntax error.
Use of undeclared identifier T.
What am I doing wrong?
protocol ApiMapperProtocol {
func MapFromSource(T) -> U
}
class UserMapper: NSObject, ApiMapperProtocol {
func MapFromSource(data: NSDictionary) -> UserModel {
var user = UserModel() as UserModel
var accountsData:NSArray = data["Accounts"] as NSArray
return user
}
}
Solution 1:
It's a little different for protocols. Look at "Associated Types" in Apple's documentation.
This is how you use it in your example
protocol ApiMapperProtocol {
associatedtype T
associatedtype U
func MapFromSource(_:T) -> U
}
class UserMapper: NSObject, ApiMapperProtocol {
typealias T = NSDictionary
typealias U = UserModel
func MapFromSource(_ data:NSDictionary) -> UserModel {
var user = UserModel()
var accountsData:NSArray = data["Accounts"] as NSArray
// For Swift 1.2, you need this line instead
// var accountsData:NSArray = data["Accounts"] as! NSArray
return user
}
}
Solution 2:
To expound on Lou Franco's answer a bit, If you wanted to create a method that used a particular ApiMapperProtocol
, you do so thusly:
protocol ApiMapperProtocol {
associatedtype T
associatedtype U
func mapFromSource(T) -> U
}
class UserMapper: NSObject, ApiMapperProtocol {
// these typealiases aren't required, but I'm including them for clarity
// Normally, you just allow swift to infer them
typealias T = NSDictionary
typealias U = UserModel
func mapFromSource(data: NSDictionary) -> UserModel {
var user = UserModel()
var accountsData: NSArray = data["Accounts"] as NSArray
// For Swift 1.2, you need this line instead
// var accountsData: NSArray = data["Accounts"] as! NSArray
return user
}
}
class UsesApiMapperProtocol {
func usesApiMapperProtocol<
SourceType,
MappedType,
ApiMapperProtocolType: ApiMapperProtocol where
ApiMapperProtocolType.T == SourceType,
ApiMapperProtocolType.U == MappedType>(
apiMapperProtocol: ApiMapperProtocolType,
source: SourceType) -> MappedType {
return apiMapperProtocol.mapFromSource(source)
}
}
UsesApiMapperProtocol
is now guaranteed to only accept SourceType
s compatible with the given ApiMapperProtocol
:
let dictionary: NSDictionary = ...
let uses = UsesApiMapperProtocol()
let userModel: UserModel = uses.usesApiMapperProtocol(UserMapper()
source: dictionary)
Solution 3:
In order to achieve having generics and as well having it declare like this let userMapper: ApiMapperProtocol = UserMapper()
you have to have a Generic Class conforming to the protocol which returns a generic element.
protocol ApiMapperProtocol {
associatedtype I
associatedType O
func MapFromSource(data: I) -> O
}
class ApiMapper<I, O>: ApiMapperProtocol {
func MapFromSource(data: I) -> O {
fatalError() // Should be always overridden by the class
}
}
class UserMapper: NSObject, ApiMapper<NSDictionary, UserModel> {
override func MapFromSource(data: NSDictionary) -> UserModel {
var user = UserModel() as UserModel
var accountsData:NSArray = data["Accounts"] as NSArray
return user
}
}
Now you can also refer to userMapper
as an ApiMapper
which have a specific implementation towards UserMapper
:
let userMapper: ApiMapper = UserMapper()
let userModel: UserModel = userMapper.MapFromSource(data: ...)