枚举方法返回动态类型 [英] enum method returning a dynamic type

问题描述

例如，我有一个字典 [String：Any] 。要处理我使用枚举值创建一个数组的值：

 枚举Foo {
 case option1 
 case option2 
 
 func createKey（） - > [String] {
 switch self {
 case .option1：return [scenario1] 
 case .option2：return [scenario2] 
} 
} 
} 
  

一旦我有这些值，我需要将它们转换成正确的类型能够使用它们。现在我正在使用 if-statements 手动执行，但如果我可以在枚举中创建一个返回正确类型的方法，那么它会减少很多代码。我当前的代码：

  let origin：[String：Any] = [scenario2：someText] 
 let选项：Foo = .option2 
 
 option.createKey（）。forEach {
 guard let rawValue = origin [$ 0] else {return} 
 
 switch option { 
 case .option1：
 guard let value = rawValue as？ int else {return} 
 print（Value is a Int：，value）
 case .option2：
 guard let value = rawValue as？ String else {return} 
 print（Value is a String：，value）
} 
} 
  

我想要实现的是：

  option.createKey（）。 forEach {
 guard let rawValue = origin [$ 0] as？ option.getType（）else {return} 
} 
  

这是可能吗？ >

解决方案

我认为这里的核心问题是Swift有严格的打字。这意味着在编译时必须知道类型。这显然是合法的：

  let s：Any =howdy
 if let ss = s as？ String {
 print（ss）
} 
  

但这不合法：

  let s：Any =howdy
 let someType = String.self 
 if let ss =作为？ someType {// * 
 print（ss）
} 
  

someType 必须是类型;它不能是在其本身内隐藏一个类型的变量。但这正是您实际要求做的事情。

I have an enum and I'd like to create a method to return a different type for every case.

For example, I have a dictionary [String: Any]. To process the values I'm using the enum to create an array of keys:

enum Foo {
    case option1
    case option2

    func createKey() -> [String] {
        switch self {
        case .option1: return ["scenario1"]
        case .option2: return ["scenario2"]
        }
    }
}

Once I have the values, I need to cast them to a the proper type to be able to use them. Right now I'm doing it manually using if-statements but it would reduce a lot of code if I can somehow create a method in the enum to return the proper type. My current code:

let origin: [String: Any] = ["scenario2": "someText"]
let option: Foo = .option2

option.createKey().forEach {
    guard let rawValue = origin[$0] else { return }

    switch option {
    case .option1:
        guard let value = rawValue as? Int else { return }
        print("Value is an Int:", value)
    case .option2:
        guard let value = rawValue as? String else { return }
        print("Value is a String:", value)
    }
}

What I would like to achieve is something like:

option.createKey().forEach {
    guard let rawValue = origin[$0] as? option.getType() else { return }
}

Is this possible?

解决方案

I think the core of the problem here is that Swift has strict typing. That means types must be known at compile time. This, obviously, is legal:

let s : Any = "howdy"
if let ss = s as? String {
    print(ss)
}

But this is not legal:

let s : Any = "howdy"
let someType = String.self
if let ss = s as? someType { // *
    print(ss)
}

someType must be a type; it cannot be a variable hiding a type inside itself. But that is precisely what, in effect, you are asking to do.

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