协议类型的数组

Question

我已经在stackoverflow上检查了关于这个问题的所有答案,但仍然无法弄清楚如何解决这个问题.我的模型看起来像这样

protocol Commandable: Equatable {
    var condition: Condition? {get set}

    func execute() -> SKAction
}

以及实施该协议的3个结构

struct MoveCommand: Commandable {

    var movingVector: CGVector!

    //MARK: - Commandable

    var condition: Condition?
    func execute() -> SKAction {
       ...
    }
}

extension MoveCommand {
    // MARK:- Equatable

    static func ==(lhs: MoveCommand, rhs: MoveCommand) -> Bool {
        return lhs.movingVector == rhs.movingVector && lhs.condition == rhs.condition
    }
}

---

struct RotateCommand: Commandable {
    var side: RotationSide!

    // MARK: - Commandable

    var condition: Condition?
    func execute() -> SKAction {
        ...
    }
}

extension RotateCommand {
    // MARK: - Equatable
    static func ==(lhs: RotateCommand, rhs: RotateCommand) -> Bool {
        return lhs.side == rhs.side && lhs.condition == rhs.condition
    }
}

---

当我尝试创建具有[Commandable]数组的第三个结构时,问题就开始了:

struct FunctionCommand: Commandable {
    var commands = [Commandable]()

编译器输出:Protocol 'Commandable' can only be used as a generic constraint because it has Self or associated type requirements.然后我用这种方式重写了我的结构:

struct FunctionCommand<T : Equatable>: Commandable {
    var commands = [T]()

我解决了这个问题但是出现了新的问题.现在我无法FunctionCommand使用旋转和移动命令的实例创建,只能使用其中一个实例:(:

let f = FunctionCommand(commands: [MoveCommand(movingVector: .zero, condition: nil), 
RotateCommand(side: .left, condition: nil)], condition: nil)

任何帮助,将不胜感激.

更新:那篇文章帮助我弄清楚了 - https://krakendev.io/blog/generic-protocols-and-their-shortcomings

Answer 1

你需要做的是使用类型擦除,就像AnyHashable在Swift标准库中一样.

你做不到:

var a: [Hashable] = [5, "Yo"]
// error: protocol 'Hashable' can only be used as a generic constraint because it has Self or associated type requirements

你要做的是使用类型擦除类型AnyHashable:

var a: [AnyHashable] = [AnyHashable(5), AnyHashable("Yo")]
a[0].hashValue // => shows 5 in a playground

所以,你的解决办法是在更小的部分先被分割的协议,推动Equatable以Hashable(重用AnyHashable)

protocol Conditionable {
    var condition: Condition? { get set }
}

protocol Executable {
    func execute() -> SKAction
}

protocol Commandable: Hashable, Executable, Conditionable {}

然后创建一个AnyCommandable结构,如下所示:

struct AnyCommandable: Commandable, Equatable {
    var exeBase: Executable
    var condBase: Conditionable
    var eqBase: AnyHashable

    init<T: Commandable>(_ commandable: T) where T : Equatable {
        self.condBase = commandable
        self.exeBase = commandable
        self.eqBase = AnyHashable(commandable)
    }

    var condition: Condition? {
        get {
            return condBase.condition
        }
        set {
            condBase.condition = condition
        }
    }

    var hashValue: Int {
        return eqBase.hashValue
    }

    func execute() -> SKAction {
        return exeBase.execute()
    }

    public static func ==(lhs: AnyCommandable, rhs: AnyCommandable) -> Bool {
        return lhs.eqBase == rhs.eqBase
    }
}

然后你可以像这样使用它:

var a = FunctionCommand()
a.commands = [AnyCommandable(MoveCommand()), AnyCommandable(FunctionCommand())]

并且您可以轻松访问属性commands,因为AnyCommandable实现Commandable

a.commands[0].condition

您需要记住现在添加Hashable和Equatable所有命令.我使用这些实现进行测试:

struct MoveCommand: Commandable {

    var movingVector: CGVector!

    var condition: Condition?
    func execute() -> SKAction {
        return SKAction()
    }

    var hashValue: Int {
        return Int(movingVector.dx) * Int(movingVector.dy)
    }

    public static func ==(lhs: MoveCommand, rhs: MoveCommand) -> Bool {
        return lhs.movingVector == rhs.movingVector
    }
}

struct FunctionCommand: Commandable {
    var commands = [AnyCommandable]()

    var condition: Condition?

    func execute() -> SKAction {
        return SKAction.group(commands.map { $0.execute() })
    }

    var hashValue: Int {
        return commands.count
    }

    public static func ==(lhs: FunctionCommand, rhs: FunctionCommand) -> Bool {
        return lhs.commands == rhs.commands
    }
}