协议具有泛型函数和associatedType

Question

我有以下代码：

protocol NextType {
    associatedtype Value
    associatedtype NextResult

    var value: Value? { get }

    func next<U>(param: U) -> NextResult
}

struct Something<Value>: NextType {

    var value: Value?

    func next<U>(param: U) -> Something<Value> {
        return Something()
    }
}

现在，问题在于Something实施next。我想返回Something<U>而不是Something<Value>。

但是，当我这样做时，我得到以下错误。

type 'Something<Value>' does not conform to protocol 'NextType'
protocol requires nested type 'Value'

Answer 1

我测试了以下代码并编译（Xcode 7.3 - Swift 2.2）。在这种状态下它们不是很有用，但我希望它能帮助您找到所需的最终版本。

Version 1

因为，Something是使用V定义的，我认为你不能只返回Something<U>。但你可以使用Something和U重新定义V，如下所示：

protocol NextType {
    associatedtype Value
    associatedtype NextResult

    var value: Value? { get }

    func next<U>(param: U) -> NextResult
}

struct Something<V, U>: NextType {
    typealias Value = V
    typealias NextResult = Something<V, U>

    var value: Value?

    func next<U>(param: U) -> NextResult {
        return NextResult()
    }
}

let x = Something<Int, String>()
let y = x.value
let z = x.next("next")

Version 2

或者只使用Something定义V：

protocol NextType {
    associatedtype Value
    associatedtype NextResult

    var value: Value? { get }

    func next<U>(param: U) -> NextResult
}

struct Something<V>: NextType {
    typealias Value = V
    typealias NextResult = Something<V>

    var value: Value?

    func next<V>(param: V) -> NextResult {
        return NextResult()
    }
}

let x = Something<String>()
let y = x.value
let z = x.next("next")