字典的元类型

Question

我想编写一个接受任何类型的函数来确定它的工作。但我遇到了dictionary.type 的错误。它为其占位符的键请求特定类型。

有什么办法可以做到吗？

我可能会这样使用：

foo(type: type(of: someDict)) // someDict can be any type of dictionary.
// can be [String: Any] or [Int, Any] or [Double: Any] or [Character: Any], etc. 


func foo(type: Any.Type) {
    switch type {
    case is Int.Type:
        print("is Int")
    //case is Dictionary: // error, Reference to generic type 'Dictionary' requires arguments in <...>
    case is Dictionary<Hashable, Any>: // also error, Using 'Hashable' as a concrete type conforming to protocol 'Hashable' is not supported
        print("is Dictionary")
    default:
        print("don't know")
    }
}

Answer 1

首先，您不应该使用Any 作为函数输入参数的类型。相反，您应该使您的函数通用。

要解决匹配Dictionary 的元类型的问题，您只需将Hashable 更改为AnyHashable，这是Hashable 值的类型擦除容器类型。

func foo<T>(type: T.Type) {
    switch type {
    case is Int.Type:
        print("is Int")
    case is Dictionary<AnyHashable, Any>.Type:
        print("is Dictionary")
    default:
        print("don't know")
    }
}

Answer 2

也许您应该使用Any 而不是Any.Type？这工作正常：

let dict: [Int: Any] = [:]
foo(type: dict) // is Dictionary
foo(type: 12) // is Int
foo(type: "12") // don't know

func foo(type: Any) {
    switch type {
    case is Int:
        print("is Int")
    case is Dictionary<AnyHashable, Any>:
        print("is Dictionary")
    default:
        print("don't know")
    }
}