Swift枚举与不同类型的关联值

Question

我有一个这样的 Swift 枚举：

public enum AnimationType {

    case position(Float)
    case position([Keyframe<Float>])
    case scale(Float)
    case scale([Keyframe<Float>])
    case rect(CGRect)
    case rect([Keyframe<CGRect>])
    case transform(CGAffineTransform)
    case transform([Keyframe<CGAffineTransform>])
    ...
    ...
}

正如我们所见，每种类型都有两个可能的值 - T 类型的固定值或值类型为 T 的关键帧数组（[Keyframe]）。我想知道是否有什么办法可以避免在枚举中重复同名并合并两个枚举案例类型？还是我的建模方式不对？

Answer 1

我假设您在低端使用一些通用代码来利用这两种类型，因此您可以通过将它们分组到某个协议下来受益于多态性：

public enum AnimatationType {
    case position(PositionProtocol)
    case scale(ScaleProtocol)
    case rect(RectProtocol)
    case transform(TransformProtocol)
    ...
    ...
}

然后简单地扩展类型：

extension Float: PositionProtocol {
     func someCommonGround() -> SomeCommonType { ... }
}

extension Keyframe: PositionProtocol where KeyframeGenericArgument == Float {
     func someCommonGround() -> SomeCommonType { ... }
}

Answer 2

我会使用Kind 枚举类型来解决这个问题，适用于每种变化。

public enum AnimationType {
    public enum Kind<Value> {
        case scalar(Value)
        case keyframes([Keyframe<Value>])
    }

    case position(Kind<Float>)
    case scale(Kind<Float>)
    case rect(Kind<CGRect>)
    case transform(Kind<CGAffineTransform>)
}

用法：

let anim1 = AnimationType.position(.scalar(10))
let anim2 = AnimationType.position(.keyframes([Keyframe(10)]))

获取值：

switch anim1 {
case .position(let kind):
    switch kind {
    case .scalar(let value):
        print("value: \(value)")
    case .keyframes(let keyframes):
        print("keyframes: \(keyframes)")
    }

default: // You would implement the rest
    break
}

switch anim1 {
case .position(.scalar(let value)):
    print("value: \(value)")

case .position(.keyframes(let keyframes)):
    print("keyframes: \(keyframes)")

default: // You would implement the rest
    break
}

if case .position(.scalar(let value)) = anim1 {
    print("value: \(value)")
}

---

您还可以添加Codable 一致性：

public struct Keyframe<Value: Codable> {
    let value: Value

    init(_ value: Value) {
        self.value = value
    }
}

extension Keyframe: Codable {
    public func encode(to encoder: Encoder) throws {
        var container = encoder.singleValueContainer()
        try container.encode(value)
    }

    public init(from decoder: Decoder) throws {
        let container = try decoder.singleValueContainer()
        value = try container.decode(Value.self)
    }
}

public enum AnimationType {
    public enum Kind<Value: Codable> {
        case scalar(Value)
        case keyframes([Keyframe<Value>])
    }

    case position(Kind<Float>)
    case scale(Kind<Float>)
    case rect(Kind<CGRect>)
    case transform(Kind<CGAffineTransform>)
}

extension AnimationType.Kind: Codable {
    public func encode(to encoder: Encoder) throws {
        var container = encoder.singleValueContainer()

        switch self {
        case .scalar(let value): try container.encode(value)
        case .keyframes(let keyframes): try container.encode(keyframes)
        }
    }

    public init(from decoder: Decoder) throws {
        let container = try decoder.singleValueContainer()

        if let scalar = try? container.decode(Value.self) {
            self = .scalar(scalar)
            return
        }
        if let keyframes = try? container.decode([Keyframe<Value>].self) {
            self = .keyframes(keyframes)
            return
        }

        // You should throw error here instead
        fatalError("Failed to decode")
    }
}

extension AnimationType: Codable {
    private enum CodingKeys: CodingKey {
        case position
        case scale
        case rect
        case transform
    }

    public func encode(to encoder: Encoder) throws {
        var container = encoder.container(keyedBy: CodingKeys.self)

        switch self {
        case .position(let kind): try container.encode(kind, forKey: .position)
        case .scale(let kind): try container.encode(kind, forKey: .scale)
        case .rect(let kind): try container.encode(kind, forKey: .rect)
        case .transform(let kind): try container.encode(kind, forKey: .transform)
        }
    }

    public init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)

        if let position = try? container.decode(Kind<Float>.self, forKey: .position) {
            self = .position(position)
            return
        }
        if let scale = try? container.decode(Kind<Float>.self, forKey: .scale) {
            self = .scale(scale)
            return
        }
        if let rect = try? container.decode(Kind<CGRect>.self, forKey: .rect) {
            self = .rect(rect)
            return
        }
        if let transform = try? container.decode(Kind<CGAffineTransform>.self, forKey: .transform) {
            self = .transform(transform)
            return
        }

        // You should throw error here instead
        fatalError("Failed to decode")
    }
}

示例编码：

do {
    let data = try JSONEncoder().encode(anim1)
    if let str = String(data: data, encoding: .utf8) {
        print(str)
        // Prints: {"position":10}
    }
} catch {
    print(error)
}

与anim2 相同的事情返回{"position":[10]}。

Answer 3

@George 的回答确实解决了这个问题。

Swift 的解决方案多种多样。

这是我的建议：

public struct Keyframe<T> {
    let v : T
}

public enum AnimaKind{
    case simple
    case series
}

public enum AnimatationType {
    case position
    case scale
    case rect
    case transform
}

extension AnimatationType{
    
    func simple<T>(info: T) -> (type: AnimatationType, kind: AnimaKind, info: T){
        return (self, .simple, info)
    }
    
    func series<T>(info: [T]) -> (type: AnimatationType, kind: AnimaKind, info: [Keyframe<T>]){
        let result = info.map { x in Keyframe(v: x) }
        return (self, .series, result)
    }
}

如您所见，打开包装的方式更容易

func test(){
    let animaTest = AnimatationType.position.simple(info: Float(10))
}

你从animaTest获得价值很方便，

tuple 与 enum nested