Swift5.2のcallAsFunctionを使ってUseCaseの必須なメソッドとしてみる

usecase_sample.swift

import Foundation

protocol UseCase where Failure: Error {
    associatedtype Parameters
    associatedtype Success
    associatedtype Failure

    func callAsFunction(_ parameters: Parameters, completion: ((Result<Success, Failure>) -> ())?)
}

struct User {
    let id: UUID
}

class GetUserInteractor: UseCase {
    func callAsFunction(_ parameters: UUID, completion: ((Result<User, Never>) -> ())?) {
        // 本来ならなんやかんや色々あるけど今回はとりあえず返すだけ
        completion?(.success(User(id: parameters)))
    }
}

let getUser = GetUserInteractor()
getUser(UUID()) {
    switch $0 {
    case .success(let user):
        print(user)
    case .failure:
        break
    }
}