我很難實作依賴倒置。環顧四周，找到一篇精彩的文章Swift Type Erasure。我不知道如何才能在我的情況下獲得優勢。這就是我想要實作的目標。
網路協議
protocol Networkable {
associatedtype Response: Decodable
func request(handler: @escaping((Result<Response, Error>) -> ()))
}
Networkable協議的具體實作
final class Networker<Response: Decodable>: Networkable {
private let session: URLSession
private let url: URL
init(session: URLSession, url: URL) {
self.session = session
self.url = url
}
func request(handler: @escaping ((Result<Response, Error>) -> ())) {
session.dataTask(with: url) { data, response, error in
handler(.success(try! JSONDecoder().decode(Response.self, from: data!)))
}
}
}
AViewModel依賴于Networkable協議
class ViewModel {
let networker: Networkable
// Error-> Protocol 'Networkable' can only be used as a generic constraint because it has Self or associated type requirements
init(netwrorker: Networkable) {
self.networker = netwrorker
}
func execute() {
networker.request { _ in
print("Responsed")
}
}
}
用法：
class View {
let viewModel: ViewModel
init() {
let networker = Networker<ResponseModel>(session: .shared, url: URL(string: "SOME URL")!)
self.viewModel = ViewModel(netwrorker: networker)
}
func load() {
viewModel.execute()
}
}
struct ResponseModel: Decodable {}
問題：
- 這是實作依賴倒置的正確方法嗎？
- 如何在 Network 類上實作通用行為并仍然開放進行測驗？
uj5u.com熱心網友回復：
你的ViewModel班級格式不正確：
class ViewModel<T: Decodable> {
let networker: Networker<T>
…
}
否則，您可能會創建一個型別擦除的AnyNetworkable具體型別，您可以將其用于：
class ViewModel<T: Decodable> {
let networker: AnyNetworkable<T>
…
}
或者您可以采用更通用的方法，例如：
class ViewModel<R, T: Networkable> where T.Response == R {
let networker: T
}
顯然，這同樣反映在您的View型別上，您還需要對特定的ViewModel.
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