Swift协议作为通用参数 [英] Swift Protocol as Generic Parameter

问题描述

提供此类:

class ServiceRegistry {

  var store = [String : AnyObject]()
  var allRegisteredType: [String] {
      return store.map { $0.0 }
  }

  func registerInstance<T>(instance:AnyObject, forType type: T.Type) {
    store[String(type)] = instance
  }

  func instanceForType<T>(type: T.Type) -> T? {
    return store[String(type)] as? T
  }
}

有没有一种方法可以强制T必须是一个协议，而无需使用@obj?

Is there a way I can enforce that T must be a Protocol, without using the @obj?

推荐答案

这是我的类型声明技术的修改版本.我添加了"as?AnyClass" 断言，以便该类型只能是协议类型. 执行此操作的方法可能更优雅，但请仔细阅读我的笔记并研究有关类断言的内容.

This is a modified version of my type assertion technique. I added the "as? AnyClass" assert so that the type can only be of protocol type. There might be a more elegant way of doing this but going through my notes and research about class assertion this is what I came up with.

import Foundation

protocol IDescribable:class{}
class A:IDescribable{}
class B:A{}

let a = A()
let b = B()

func ofType<T>(instance:Any?,_ type:T.Type) -> T?{/*<--we use the ? char so that it can also return a nil*/
    if(instance as? T != nil && type as? AnyClass == nil){return instance as? T}
    return nil
}

Swift.print(ofType(a,A.self))//nil
Swift.print(ofType(a,IDescribable.self))//A
Swift.print(ofType(b,B.self))//nil

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