以字符串作为键和通用类作为值的字典 [英] Dictionary with String as Key and Generic Class as value

问题描述

我有插座和插头，插头是任何通用类型。现在我想将任何类型的插件存储到字典中。我试图用我能想到的所有方式在字典中声明插件的可能类型。没有工作。我只找到传递任何Value T的解决方案，而不是任何泛型类的类。另外我担心在任何情况下我的用词都不对。我错了什么名称？

示例将清除所有内容。
那么我如何将这个Plugs存储到套接字的字典插件中：

  import Foundation 
 
 class AClass {
 
} 
 
 class BClass {
 
} 
 
 class Plug< T> {
 init（）{
 
} 
} 
 
类Socket {
 var plugs = [String：Plug]（）//应该接受任何Plug 
 init（）{
 
} 
 func addPlug（plug：Plug）{//接受任何Plug 
 self.plugs [A ] = plug //应该接受任何Plug 
 
 
 $ b $ var plugDouble = Plug< Double>（）
 var plugString = Plug< String>（）
 var plugAClass = Plug< AClass>（）
 var plugBClass = Plug< BClass>（）
 
 var socket = Socket（）
 socket.addPlug（plug：plugDouble ）; //接受任何插件
 socket.plugs [A] = plugDouble //应该接受任何插件
  

BasePlug ），所有 Plug c $ c>类将从中派生。让您的字典存储类 BasePlug ：
$ b

  class BasePlug {
 
} 
 
 class AClass {
 
} 
 
 class BClass {
 
} 
 
 class Plug< T> ：BasePlug {
 
} 
 
 class Socket {
 var plugs = [String：BasePlug]（）//应该接受任何Plug 
 init（） {
 
} 
 func addPlug（plug：BasePlug）{//接受任何Plug 
 self.plugs [A] = plug //应该接受任何Plug 
 
 
 $ var plugDouble = Plug< Double>（）
 var plugString = Plug< String>（）
 var plugAClass = Plug< AClass>（ ）
 var plugBClass = Plug< BClass>（）
 
 var socket = Socket（）
 socket.addPlug（plugDouble）; //接受任何Plug 
 socket.plugs [A] = plugDouble //应该接受任何Plug 
  

I have sockets and plugs, plugs are of any generic type. Now i want to store a Plug of any type into a dictionary. I tried to declare the possible type of the Plug in the dictionary in all ways i can think of. None worked. I do only find solutions for passing any Value T but not a class of any generic type. Also i fear my usage of words is not right in all cases. What did i name wrong?

Example will clear things up. So how could i store this Plugs into the dictionary "plugs" of the socket:

import Foundation

class AClass {

}

class BClass {

}

class Plug<T> {
    init() {

    }
}

class Socket {
    var plugs = [ String: Plug ]() // should accept any Plug
    init() {

    }
    func addPlug( plug : Plug ) { // should accept any Plug
        self.plugs["A"] = plug // should accept any Plug
    }
}

var plugDouble = Plug<Double>()
var plugString = Plug<String>()
var plugAClass = Plug<AClass>()
var plugBClass = Plug<BClass>()

var socket = Socket()
socket.addPlug(plug: plugDouble ); // should accept any Plug
socket.plugs["A"] = plugDouble // should accept any Plug

解决方案

Create a base class (let's call it BasePlug) that all Plug classes will derive from. Have your dictionary store items of class BasePlug:

class BasePlug {

}

class AClass {

}

class BClass {

}

class Plug<T> : BasePlug {

}

class Socket {
    var plugs = [ String: BasePlug ]() // should accept any Plug
    init() {

    }
    func addPlug( plug : BasePlug ) { // should accept any Plug
        self.plugs["A"] = plug // should accept any Plug
    }
}

var plugDouble = Plug<Double>()
var plugString = Plug<String>()
var plugAClass = Plug<AClass>()
var plugBClass = Plug<BClass>()

var socket = Socket()
socket.addPlug(plugDouble ); // should accept any Plug
socket.plugs["A"] = plugDouble // should accept any Plug

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