如何将 AnyKeyPath 转换为 WritableKeyPath？

Question

我有一个枚举案例数组，其中每个案例都有一个keyPath属性，该属性返回AnyKeyPath与枚举案例同名的匹配类属性：

protocol PathAccessor: CodingKey {
    var keyPath: AnyKeyPath { get }
    static var allCases: [Self] { get }

    init?(rawValue: Int)
}

extension PathAccessor {
    static var allCases: [Self] {
        var cases: [Self] = []
        var index: Int = 0
        while let element = Self.init(rawValue: index) {
            cases.append(element)
            index += 1
        }

        return cases
    }
}

class Robot {

    let name: String
    var age: Int
    var powered: Bool
    var hasItch: Bool?

    enum CodingKeys: Int, PathAccessor {
        case name
        case age
        case powered
        case hasItch

        var keyPath: AnyKeyPath {
            switch self {
            case .name: return \Robot.name
            case .age: return \Robot.age
            case .powered: return \Robot.powered
            case .hasItch: return \Robot.hasItch
            }
        }
    }

    init(name: String, age: Int, powered: Bool) {
        self.name = name
        self.age = age
        self.powered = powered
    }
}

for element in Robot.CodingKeys.allCases {
    // Trying to implement
}

在上面的循环中，我想检查keyPathcase 的属性以查看它是否是 a WritableKeyPath，如果是，则创建一个闭包来修改键路径访问的属性。

问题在于 aWritableKeyPath是泛型类型。我知道该Root类型，但该Value类型几乎可以是现有的任何类型。我可以为每种最可能的类型创建一堆案例：

if let path = element.keyPath as? WritableKeyPath<Robot, Int> {

} else if let path = element.keyPath as? WritableKeyPath<Robot, String> {

} // So on and so forth

但这非常耗时、丑陋且难以维护。

我确实尝试转换为动态类型，但这会产生编译器错误（Use of undeclared type 'valueType'）：

let valueType = type(of: element.keyPath).valueType
guard let path = element.keyPath as? WritableKeyPath<Self, valueType> else {
    continue
}

我可以使用类型已经符合的协议，但由于某种原因，这也失败了：

guard let path = element.keyPath as? WritableKeyPath<Robot, NodeInitializable> else {
    print("bad")
    continue
}
print("good")

// Output:
// bad
// bad
// bad
// bad

那么，是否有可能将 an 转换AnyKeyPath为 a WritableKeyPath，而不需要大量的展开语句或不应该在生产中使用的奇怪的 hack？

Answer 1

经过几个小时的代码研究，我得到的最好的结果是：

struct Person {
var name: String
var collection: [String]
var optionalCollection: [String]?
let birthdate: Date

fileprivate func keyPath<V>(from label: String, type: V.Type) -> WritableKeyPath<Person, V> {
    print("type: \(type)")
    let valuePair: [String: PartialKeyPath<Person>] = ["name": \Person.name,
                                                       "birthdate": \Person.birthdate,
                                                       "collection": \Person.collection,
                                                       "optionalCollection": \Person.optionalCollection]
    return valuePair[label] as! WritableKeyPath<Person, V>
}


}



var person = Person(name: "john",
                    collection: [],
                    optionalCollection: ["gotcha"],
                    birthdate: Date(timeIntervalSince1970: 0))


let name = person[keyPath: person.keyPath(from: "name", type: String.self)] // john
let birthdate = person[keyPath: person.keyPath(from: "birthdate", type: Date.self)] // Jan 1, 1970
let collection = person[keyPath: person.keyPath(from: "collection", type: Array<String>.self)] // []
let optionalCollection = person[keyPath: person.keyPath(from: "optionalCollection", type: Optional<Array<String>>.self)] // ["gotcha"]

但是，您必须始终将类型作为参数传递。如果只有 Mirror 类允许我们从每个属性获取实际类型就好了。