ios - 使用可选的初始化程序扩展 CollectionType

Question

protocol Decodable {
    init?(data: [String: AnyObject])
}

struct A: Decodable {

    var data: [String: AnyObject]!

    init?(data: [String: AnyObject]) {
        self.data = data
    }

}

当我想创建一个对象时这有效

let d = ["name":"Rahul"]
let a = A(data: d)

我正在尝试实现以下目标，但它在编译时出现错误。

let dArray = [["name":"Rahul"],["name":"Rahul"],["name":"Rahul"]]
let aArray = [A](data: dArray)

以下代码给出错误“nil 是初始化程序中允许的唯一返回值”。

public extension CollectionType where Generator.Element: Decodable {

    init?(data: [[String: AnyObject]]) {
        var elements: [Generator.Element] = []

        for d in data {
            let element = Generator.Element(data: d)
            if let element = element {
                element.append(element)
            }
        }

        return elements

    }

}

==================================== 答案:-

public extension Array where Element: Decodable {

    init?(data: [String: AnyObject]) {
        var elements: [Element] = []
        for d in data {
            let element = Element(data: d)
            if let element = element {
                element.append(element)
            }
        }
        self = elements
    }

}

这将允许您使用以下代码进行初始化

let dArray = [["name":"Rahul"],["name":"Rahul"],["name":"Rahul"]]
let aArray = [A](data: dArray)

Answer 1

您的错误是因为CollectionType是一个协议(protocol)，无法初始化。 您是否尝试过创建一个辅助类方法来返回集合，例如

func collectionWithData(data: [[String: AnyObject]]) -> [Generator.Element] {