CODEWITHSUNDEEP
swift
Farhad-Taran
Farhad-Taran

Reputation: 6512

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
    } 
}

Upvotes: 106

Views: 84215

Answers (5)

Saurabh Sharma
Saurabh Sharma

Reputation: 207

How to create and use generic Protocol:

protocol Generic {
    
    associatedtype T
    associatedtype U

    func operation(_ t: T) -> U
}


// Use Generic Protocol

struct Test: Generic {

    typealias T = UserModel
    typealias U = Any
    
    func operation(_ t: UserModel) -> Any {
        let dict = ["name":"saurabh"]
        return dict
    }
}

Upvotes: 2

denis_lor
denis_lor

Reputation: 6547

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: ...)

Upvotes: 5

Heath Borders
Heath Borders

Reputation: 32107

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 SourceTypes compatible with the given ApiMapperProtocol:

let dictionary: NSDictionary = ...
let uses = UsesApiMapperProtocol()
let userModel: UserModel = uses.usesApiMapperProtocol(UserMapper()
    source: dictionary)

Upvotes: 29

Lou Franco
Lou Franco

Reputation: 89142

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
    }
}

Upvotes: 172

Dsjove
Dsjove

Reputation: 37

You can use templates methods with type-erasure...

protocol HeavyDelegate : class {
  func heavy<P, R>(heavy: Heavy<P, R>, shouldReturn: P) -> R
}  

class Heavy<P, R> {
    typealias Param = P
    typealias Return = R
    weak var delegate : HeavyDelegate?  
    func inject(p : P) -> R? {  
        if delegate != nil {
            return delegate?.heavy(self, shouldReturn: p)
        }  
        return nil  
    }
    func callMe(r : Return) {
    }
}
class Delegate : HeavyDelegate {
    typealias H = Heavy<(Int, String), String>

    func heavy<P, R>(heavy: Heavy<P, R>, shouldReturn: P) -> R {
        let h = heavy as! H
        h.callMe("Hello")
        print("Invoked")
        return "Hello" as! R
    }  
}

let heavy = Heavy<(Int, String), String>()
let delegate = Delegate()
heavy.delegate = delegate
heavy.inject((5, "alive"))

Upvotes: -3

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