Swift类型和泛型：它真的很麻烦吗？

Question

这些天掌握Swift，我已经实现了一个字典＆lt;类型：任何＆gt;在斯威夫特。一切正常，但代码看起来非常难看。我应该把它归咎于语言的年轻人，还是我对它缺乏经验？任何见解都非常赞赏！

步骤1：创建一个符合Hashable的Type包装器（因为Any.Type和AnyClass都没有）

class HashableType : Hashable
{
    let type : Any.Type

    //Piggybacking String's Hashable implementation
    var hashValue : Int
    {
        return toString( type ).hashValue
    }

    init( type : Any.Type )
    {
        self.type = type
    }

    init( instance: Any )
    {
        self.type = instance.dynamicType
    }
}

//Equatable implementation, wish it could be nested in the class declaration...
func ==( lhs: HashableType, rhs: HashableType )->Bool
{
    return lhs.hashValue == rhs.hashValue
}

第2步：真实世界用例 - 注册类型的默认实例

final class DefaultInstances
{
    private static var __defaultInstances = [ HashableType: Any ]()

    class func registerDefaultInstance( instance: Any ) -> Bool
    {
        let hashableType = HashableType( instance: instance )

        if let defaultInstance = __defaultInstances[ hashableType ]
        {
            return false
        }

        __defaultInstances[ hashableType ] = instance

        return true
    }

    // T, T.Type and T.self can all coexist, not a contrived example.
    // T.Type:  the "cannot explicitly specialize a generic function" compile error is quite a shame here -
    // would be much cleaner in C# for example
    class func defaultInstance< T >( type: T.Type ) -> T?
    {
        return __defaultInstances[ HashableType( type: T.self ) ] as? T

        // could also work as
        // return __defaultInstances[ HashableType( type: type ) ] as? T
    }
}

第3步：测试

func TestDefaultInstances()
{
    if DefaultInstances.registerDefaultInstance( Int( 2 ) ) == true
    {
        println( "did register default int." )
    }
    else
    {
        if let defaultInt = DefaultInstances.defaultInstance( Int )
        {
            println( "Already registered Int default instance: \(defaultInt)")
        }
    }
}