我有这样的事情:

class Lumber { }
class Fruit { }

enum Size {
    case small
    case medium
    case large
}

let lumberSize = [
    Size.small: "2x4",
    Size.medium: "4x6",
    Size.large: "6x10"
]

let fruitSize = [
    Size.small: "grape",
    Size.medium: "apple",
    Size.large: "watermelon"
]

let size:[AnyObject.Type:Dictionary] = [
    Lumber.Type: lumberSize,
    Fruit.Type: fruitSize
]

在我的size字典定义中,我从Xcode编辑器得到了这个实时错误:



如何完成对size的尝试?也就是说,如何创建将类型链接到其特定大小字典的字典?

我认为 ObjectIdentifier 可以帮助我,因为它是Hashable,但我不知道如何使用它,或者它是否是正确的选择。

最佳答案

HashableObjectIdentifier实现的协议(protocol)。这意味着ObjectIdentifier(Lumber.Type)是可哈希的,而不是Lumber.Type是可哈希的。您可以尝试将代码更改为使用ObjectIdentifier,如下所示:

class Lumber { }
class Fruit { }

enum Size {
    case small
    case medium
    case large
}

let lumberSize = [
    Size.small: "2x4",
    Size.medium: "4x6",
    Size.large: "6x10"
]

let fruitSize = [
    Size.small: "grape",
    Size.medium: "apple",
    Size.large: "watermelon"
]

let size:[ObjectIdentifier:[Size:String]] = [
    ObjectIdentifier(Lumber.self): lumberSize,
    ObjectIdentifier(Fruit.self): fruitSize
]

let t = size[ObjectIdentifier(Lumber.self)]
let s = t?[.small]
print(s ?? "no s?")

这将编译并打印“2x4”,但是我不确定它是否满足您的特定需求。就个人而言,我只是将类名的字符串版本用作键String(Lumber)。 IE。:
let size:[String:[Size:String]] = [
    String(describing:Lumber.self): lumberSize,
    String(describing:Fruit.self): fruitSize
]

let t = size[String(describing:Lumber.self)]
let s = t?[.small]
print(s ?? "no s?")

10-04 14:29