如何接受两种类型之一作为 Swift/SwiftUI 中函数的参数？

Question

我想编写一个函数，仅接受

Int

或

Double

并根据它们返回不同的值。实际用例的特点是 SwiftUI

struct: View

具有更复杂的

init()

。如果可能的话，我想避免使用不同的输入类型重复它。这怎么办？

理想情况下，我想要一个如下所示的解决方案：

struct MyStruct<T>: View where T is (Int | Double) {  // Specifying type here is the problematic part
var value: T
var description: String

init(_ value: T) {
   self.value = value
   switch value {
      case is Int:
         description = "Integer"
      case is Double:
         description = "Double"
   }
}

var body: some View {
   Text(description)
}

Answer 1

至少有两个选择：

1. Int

   和

   Double

   在扩展中采用的自定义虚拟协议

   protocol IntOrDouble {}
   extension Int : IntOrDouble {}
   extension Double : IntOrDouble {}
   
   struct MyStruct<T>: View where T : IntOrDouble {
       var value: T
       let description: String
   
       init(_ value: T) {
          self.value = value
           description = value is Int ? "Integer" : "Double"
       }
   
       var body: some View {
           Text(description)
       }
   }
   

2. 具有关联值的枚举

   enum IntOrDouble {
       case integer(Int)
       case double(Double)
   }
   
   struct MyStruct : View {
       var value: IntOrDouble
       let description: String
   
       init(value: IntOrDouble) {
           self.value = value
           switch value {
               case .integer:
                   description = "Integer"
               case .double:
                   description = "Double"
           }
       }
   
       var body: some View {
           Text(description)
       }
   }