Swift Array Type

Question

I think I should be more clear, here are my classes:

    import Foundation
    import Alamofire

    public class BaseService<T>  : ServiceRequest {

        public var requestType: Alamofire.Method = .GET
        public var path: String = "/"

        var requireLogin:Bool = false

        var consumer:RequestInformer?


        public func requestSuccess<T>(request: Request, response: T) {
            consumer?.requestSuccess(request, response: response)

        }

        public func requestFailed(request: Request, error: NSError) {
            consumer?.requestFailed(request, error: error)
        }
    }

    extension BaseService where T:Sortable {

        func start() {
            if requireLogin {

            }
//problem is here, T is [CampaignModel] but I need CampaignModel class, not array.            NetworkManager.sharedInstance.sendRequest(self,ResponseListModel<T>.self)
        }
    }



    extension BaseService where T:NSObject {

        func start() {
            if requireLogin {

            }
            NetworkManager.sharedInstance.sendRequest(self,ResponseModel<T>.self)
        }
    }

Then I want to use BaseService like below:

import Alamofire

public class ListCampaignService : BaseService<[CampaignModel]> {

    override init() {
        super.init()
        path = "/xxx/yyy.json"
        requestType = .GET
    }
}

That means ListCampaignService returns CampaignModel array. Bu I have to handle object return type rather than array, something like below

import Alamofire

public class SomeOtherService : BaseService<SomeClass> {

    override init() {
        super.init()
        path = "/aaa/bbb.json"
        requestType = .GET
    }
}

what are you going to do with the type once you have it ?

Wain
– Wain

2016-05-17 10:53:51 +00:00
Commented May 17, 2016 at 10:53 — Wain
– Wain, Commented May 17, 2016 at 10:53
will use it for json to object mapping

Kerem
– Kerem

2016-05-17 11:37:08 +00:00
Commented May 17, 2016 at 11:37 — Kerem
– Kerem, Commented May 17, 2016 at 11:37
I think you've lost the question in your "clarification"...

Grimxn
– Grimxn

2016-05-17 15:42:09 +00:00
Commented May 17, 2016 at 15:42 — Grimxn
– Grimxn, Commented May 17, 2016 at 15:42

Grimxn · Accepted Answer · 2016-05-17 11:44:15Z

3

Try this as a start:

func getType<T>(array:[T]) -> T.Type {
    return T.self
}

getType(["1", "2", "3"]) // String.Type
getType([1, 2, 3]) // Int.Type
getType([1, "2", 3.0]) // NSObject.Type
getType([[1], [2], [3]]) // Array<Int>.Type
getType([]) // compile time error "Generic parameter 'T' could not be inferred"

edited May 17, 2016 at 11:44

answered May 17, 2016 at 11:22

Grimxn

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4 Comments

Hamish Over a year ago

You should probably replace Any with T.Type

Grimxn Over a year ago

One doesn't need to - I guess it depends what he wants to do with it afterwards!

Hamish Over a year ago

True, that's why I said "should probably", not "need" ;) Although I would always go with the most descriptive type available – and T.Type is a lot more descriptive than Any.

Grimxn Over a year ago

Agreed. Changed the answer.

Casey Fleser · Accepted Answer · 2016-05-17 11:26:55Z

2

If you're trying to get all the objects of a particular type from a generic Array you could extend Array like this:

extension Array {
    func itemsOfType<T>() -> [T] {
        return self.flatMap { $0 as? T }
    }
}

let x = [1, 2, 3, "foo", "bar", NSDate()]

let ints : [Int] = x.itemsOfType()      // [1, 2, 3]
let altInts = x.itemsOfType() as [Int]  // [1, 2, 3]
let str : [String] = x.itemsOfType()    // ["foo", "bar"]
let dates : [NSDate] = x.itemsOfType()  // ["May 17, 2016, 6:23 AM"]

answered May 17, 2016 at 11:26

Casey Fleser

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Comments

Kametrixom · Accepted Answer · 2016-05-17 12:11:26Z

1

If you use generics in your function, it's really simple:

func test<T>(array: [T]) {
    // array elements are of type T
    print(T.self) 
}

If you don't use generics (outside functions) then you can do this:

let array = [1, 2, 3]
let type = array.dynamicType.Element.self
print(type)     // Int.Type

If you're coming from the future: Swift 3.0 will use Self instead of dynamicType

answered May 17, 2016 at 12:11

Kametrixom

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