如何在 Swift 中不使用任何其他 Swift 类型的情况下构建自定义类型？

Question

What I am trying to find out and learn is that how I could build a custom Type, for example I created this simple enum called Allowed , which has 2 cases Yes and No , and I would like to use it as real Type like Bool Type, and in the other hand I do not want just inherit Bool Type to this custom Type. How I can do that? beside this question I think this custom Type should confirm to "==" and "!=" for if statement, and Size of this type is 1 bit.

enum Allowed { case Yes, No }

---

use case:

let result: Allowed = Allowed.No

My Goal use case:

let result: Allowed = No

Answer 1

// To get == and !=, conform to Hashable
enum Allowed: Hashable {
    case yes, no // I recommend using standard naming conventions
}

let Yes = Allowed.yes  // But as global constants, capitals are probably clearer
let No = Allowed.no

let result = No // No need to include a type

As Sean notes, a two value enum will be one byte.

(I don't recommend this particular set of names, but the general approach is fine.)

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To your question in the comments, if you want init() to initialized this to .no , then you would need to write that initializer:

extension Allowed {
    init() { self = .no }
}

let answer: Allowed = Allowed()

Answer 2

实际上，在 swift 中，您可以省略 enum 类型，因为编译器可以推断出它：

let result: Allowed = .No