ios - 如何在 Swift 中通过泛型类型构造属性？

Question

我在 swift 中创建了一个泛型类，我想用“AdaptorType”类型初始化一个适配器，但我收到了一个编译错误

class Sample<ItemType, AdaptorType> {
    var items = Array<ItemType>() 
    let adaptor = AdaptorType()   //Error: 'AdaptorType' is not constructible with '()'
}

我也试过在 init() 中初始化它，但同样的问题发生了

class SampleGnericClass<ItemType, AdaptorType> {
    var items = Array<ItemType>() 
    let adaptor : AdaptorType

    init() {
        adaptor = AdaptorType()      //It doesn't work, same error occors here
    }
}

使用通用类型 AdaptorType 初始化适配器属性的正确方法是什么？非常感谢!

编辑:关于这个问题的完整代码

import UIKit

protocol XYManagedObject {

}

protocol XYNetworkProtocol {

    typealias ManagedObjectType : XYManagedObject

    func object(fromItemDictionary dictionary: NSDictionary?) -> ManagedObjectType?
    func objects(fromEntryDictionary dictionary: NSDictionary?) -> Array<ManagedObjectType>?
    func actionParameters(page: UInt?) -> (action: String, parameters: Dictionary<String, String>?)
    func pageSize() -> UInt? // nil for unlimited
}

class ProductAdaptor : XYNetworkProtocol {

    var filter = Dictionary<String, String>()

    typealias ManagedObjectType = XYProductObject

    func object(fromItemDictionary dictionary: NSDictionary?) -> ManagedObjectType? {
        //... Will return an object here
        return nil
    }

    func objects(fromEntryDictionary dictionary: NSDictionary?) -> Array<ManagedObjectType>? {
        //... will return an array of objects here
        return nil
    }

    func actionParameters(page: UInt?) -> (action: String, parameters: Dictionary<String, String>?) {
        var parameters = filter
        if page {
            parameters["page"] = "\(page!)"
        }

        return ("product_list", parameters)
    }

    func pageSize() -> UInt? {
        return 100
    }
}

class XYDataManager<ItemType: XYManagedObject, AdaptorType: XYNetworkProtocol> {

    var objects = Array<ItemType>()
    let adaptor = ProductAdaptor() //work around, adaptor should be any type conform to XYNetworkProtocol

    func loadPage(page: UInt?) -> (hasNextPage: Bool, items: Array<ItemType>?) {
        let fetcher = XYFetchController()

        let ap = adaptor.actionParameters(page)
        if let fetchedResult = fetcher.JSONObjectForAPI(ap.action, parameters: ap.parameters) {

            let npage = fetchedResult["npage"].integerValue
            var hasNextPage = true
            if !npage || !page {
                hasNextPage = false
            }
            else if npage <= page {
                hasNextPage = false
            }

            let objects = adaptor.objects(fromEntryDictionary: fetchedResult)
            return (hasNextPage, (objects as Array<ItemType>?))
        }
        else {
            return (false, nil)
        }
    }
}

Answer 1

并非所有对象都可以像您尝试的那样在没有参数的情况下进行初始化。你必须将一个初始化的实例带到你类的初始化器中

你也可以定义你自己的协议(protocol)，它需要一个空的初始值设定项(通过向它添加 init)并限制 AdapterType 为该协议(protocol)