Swift

Question

我需要像这样声明一个数组：

var cups: [Cup<Drink>] = []

Cup 是一个结构，Drink 是一个协议，但我得到以下错误：

Value of protocol type 'Drink' cannot conform to 'Drink'; only struct/enum/class types can conform to protocols

我知道协议类型 'Drink' 可以被 AnyDrink 结构擦除，流动的代码就是一个例子。

但实际上，Type-Erasure在associatetype、Self和static-method（非final的情况下class采用Drink，比如协议Equatable和静态方法==)在Drink.

中使用

我的问题是：有没有更好的方法来声明cups数组？

或：有没有简单的方法来进行类型擦除？它应该是一个内置功能。

protocol Drink {
    ...
}

struct AnyDrink: Drink {
    let drink: Drink
    ...
}

struct Water: Drink {
    ...
}
struct Coffee: Drink {
    ...
}
struct Tea: Drink {
    ...
}

struct Cup<T: Drink> {
    private(set) var drink: T?
    mutating func bottomUp() {
        drink = nil
    }
}

struct Waiter {
    var cups: [Cup<AnyDrink>] = []
    mutating func makeACupOfSth(_ cup: Cup<AnyDrink>) {
        cups.append(cup)
    }
    mutating func pleaseGiveMeACupOfSthToDrink() -> Cup<AnyDrink> {
        return cups.removeFirst()
    }
    static func excuse(_ customer: Customer) -> Waiter {
        return Waiter()
    }
}

struct Customer {
    var me: Self { self }
    func drink() {
        var waiter = Waiter.excuse(me)
        var cup = waiter.pleaseGiveMeACupOfSthToDrink()
        cup.bottomUp()
    }
}

Answer 1

问题：Is there any better way to declare the cups array

你可以像这样使用另一个名为 DrinkGeneric 的协议，并通过 Cup Struct 实现它：

protocol DrinkGeneric {
    func smaple()
    func typOfDrink() -> Drink.Type
}

struct Cup<T: Drink>: DrinkGeneric {
    public var drink: T?
    mutating func bottomUp() {
        drink = nil
    }
    
    public func typOfDrink() -> Drink.Type {
        return type(of: drink!)
    }
    
    func smaple() {
        print("sample")
    }
    
}

然后创建一个类型为 DrinkGeneric 的数组，如下所示：

var cups: [DrinkGeneric] = [Cup<Water>(drink: Water.init()), Cup<Tea>(drink: Tea.init())]

支票类型：

if type(of: cups[0].typOfDrink()) == Water.self {
    // any work
}

Swift - 如何从类型中删除泛型？

Swift - How to erase generics from a type?

generics

protocols

type-erasure