我有一个基本协议(模型),一些结构符合.它们也符合Hashable
protocol Model {}
struct Contact: Model,Hashable {
var hashValue: Int { return ... }
static func ==(lhs: Contact,rhs: Contact) -> Bool { return ... }
}
struct Address: Model,Hashable {
var hashValue: Int { return ... }
static func ==(lhs: Address,rhs: Address) -> Bool { return ... }
}
我有一个函数,它接受一个符合Model([Model])的对象数组.
如何将[Model]传递给需要Hashables而不制作Model Hashable的函数？
func complete(with models: [Model]) {
doSomethingWithHashable(models) //can't do this
}
func doSomethingWithHashable <T:Hashable>(_ objects: [T]) {
//
}
我试图避免这种情况
protocol Model: Hashable {}
func complete<T:Model>(with models: [T]) {
runcomparison(models)
}
因为当我这样做时,我得到“模型不能用作通用约束……”
protocol SomethingElse {
var data: [Model] { get }
}
你的代码的问题在于你在谈论Model,它对Hashable一致性没有任何承诺.正如您所指出的那样,告诉编译器关于此问题(即从Hashable派生模型)的问题是,您失去了根据符合模型的异构类型进行交谈的能力.
如果您首先不关心模型一致性,则可以使用标准库的AnyHashable类型擦除包装器来完全任意Hashable一致性实例.
但是,假设您关心模型一致性,则必须为符合Model和Hashable的实例构建自己的type-erased wrapper.在my answer here中,我演示了如何为Equatable标准类型构建类型橡皮擦.可以很容易地为Hashable扩展逻辑 – 我们只需要存储一个额外的函数来返回实例的hashValue.
例如：
struct AnyHashableModel : Model,Hashable {
static func ==(lhs: AnyHashableModel,rhs: AnyHashableModel) -> Bool {
// forward to both lhs's and rhs's _isEqual in order to determine equality.
// the reason that both must be called is to preserve symmetry for when a
// superclass is being compared with a subclass.
// if you kNow you're always working with value types,you can omit one of them.
return lhs._isEqual(rhs) || rhs._isEqual(lhs)
}
private let base: Model
private let _isEqual: (_ to: AnyHashableModel) -> Bool
private let _hashValue: () -> Int
init<T : Model>(_ base: T) where T : Hashable {
self.base = base
_isEqual = {
// attempt to cast the passed instance to the concrete type that
// AnyHashableModel was initialised with,returning the result of that
// type's == implementation,or false otherwise.
if let other = $0.base as? T {
return base == other
} else {
return false
}
}
// simply assign a closure that captures base and returns its hashValue
_hashValue = { base.hashValue }
}
var hashValue: Int { return _hashValue() }
}
然后你会像这样使用它：
func complete(with models: [AnyHashableModel]) {
doSomethingWithHashable(models)
}
func doSomethingWithHashable<T : Hashable>(_ objects: [T]) {
//
}
let models = [AnyHashableModel(Contact()),AnyHashableModel(Address())]
complete(with: models)
在这里,我假设您还希望将其用作Model的要求的包装(假设有一些).或者,您可以公开base属性并从AnyHashableModel本身中删除Model一致性,使调用者访问基础Model符合实例的基础：
struct AnyHashableModel : Hashable {
// ...
let base: Model
// ...
}
但是,您会注意到上述类型擦除的包装器仅适用于Hashable和Model的类型.如果我们想谈论符合实例Hashable的其他协议怎么办？
正如我演示in this Q&A,更通用的解决方案是接受Hashable并且符合其他协议的类型 – 其类型由通用占位符表示.
因为Swift目前还没有办法表达一个通用占位符,它必须符合另一个通用占位符给出的协议;必须由调用者使用转换闭包来定义此关系,以执行必要的向上转换.但是,由于Swift 3.1接受扩展中的具体相同类型要求,我们可以定义一个方便初始化器来删除Model的样板(这可以在其他协议类型中重复).
例如：
/// Type-erased wrapper for a type that conforms to Hashable,/// but inherits from/conforms to a type T that doesn't necessarily require
/// Hashable conformance. In almost all cases,T should be a protocol type.
struct AnySpecificHashable<T> : Hashable {
static func ==(lhs: AnySpecificHashable,rhs: AnySpecificHashable) -> Bool {
return lhs._isEqual(rhs) || rhs._isEqual(lhs)
}
let base: T
private let _isEqual: (_ to: AnySpecificHashable) -> Bool
private let _hashValue: () -> Int
init<U : Hashable>(_ base: U,upcast: (U) -> T) {
self.base = upcast(base)
_isEqual = {
if let other = $0.base as? U {
return base == other
} else {
return false
}
}
_hashValue = { base.hashValue }
}
var hashValue: Int { return _hashValue() }
}
// extension for convenience initialiser for when T is Model.
extension AnySpecificHashable where T == Model {
init<U : Model>(_ base: U) where U : Hashable {
self.init(base,upcast: { $0 })
}
}
您现在想要将您的实例包装在AnySpecificHashable< Model>中：
func complete(with models: [AnySpecificHashable<Model>]) {
doSomethingWithHashable(models)
}
func doSomethingWithHashable<T : Hashable>(_ objects: [T]) {
//
}
let models: [AnySpecificHashable<Model>] = [
AnySpecificHashable(Contact()),AnySpecificHashable(Address())
]
complete(with: models)
