Swift 类型删除/具体化行为？

Question

Swift 目前支持类型具体化吗？我无法让这段代码正常工作:

import Swift

protocol P { var name:String { get set } }

class A : P { var name:String = "A" }
class B : P { var name:String = "B" }
class C : P { var name:String = "C" }

func inspect(var p:P) { println("should only be C: \(p.name)") }
func inspect(var p:A) { println("a: \(p.name)") }
func inspect(var p:B) { println("b: \(p.name)") }

func failsToRetainType<T:P>(t:T) {
    print("inspecting \(t.name): ")

    //  let t = t.dynamicType
    inspect(t)
}

inspect(A())
inspect(B())
inspect(C())

failsToRetainType(A()) // expect A, get C
failsToRetainType(B()) // expect B, get C
failsToRetainType(C()) // expect C, get C

我错过了什么吗？

尽管如此，我已经看到另外两篇关于同一主题的帖子，但它们似乎更侧重于构建，并且可以通过创建相同类型的新实例来解决:

• Swift Generics will not instantiate generic when using inheritance
• Swift generics not preserving type

Answer 1

据我所知，您需要主动将泛型函数中的实例转换回特定类型:

func failsToRetainType<T:P>(t:T) {
    print("inspecting \(t.name): ")

    switch t {
    case let t as A: inspect(t)
    case let t as B: inspect(t)
    default: inspect(t)
    }
}

// inspecting A: a: A
// inspecting B: b: B
// inspecting C: should only be C: C