问题描述
我正在尝试制作自定义的异构列表和地图.尽管有使用 Manifest 的示例，但在 Scala 2.10 中它们已被弃用，我应该使用 TypeTags(或 Classtags).在地图的情况下，我似乎可以使用(比如说)元组 String->(TypeTag[ _ <: Any ], Any ) 来保留 Any 与 Type 的绑定.
I am trying to make custom heterogeneous lists and maps. Although there are examples around using Manifest, with Scala 2.10 they are deprecated and I should use TypeTags (or Classtags). In the case of maps it seems I can preserve the binding of an Any to a Type using (say) a tuple String->(TypeTag[ _ <: Any ], Any ).
我的问题是如何从恢复的 TypeTag 和未定义的 T 中获取，以便能够返回 TypeTag.tpe 的实例 - 在代码中我有//** 这里如何使用保存的 typeTag 定义 T?**
My problem is how to get from the recovered TypeTag and an undefined T, to be able to return an instance of TypeTag.tpe - at the point in the code where I have //** How do I use saved typeTag to define T here?**
正如所写，方法 get 中没有编译器错误，但 T 设置为 Nothing 并返回 Some(Nothing).我希望我注释掉的行能够工作:case Some( x ) =>//println( "Get 2*'pi'=" + x*2 ) 其中有一个编译器消息，value * 不是 Nothing 的成员".我意识到我可以写得更紧凑，但完成后，我可以将鼠标悬停在我的 IDE 中并一步一步地进行.有一个相关的问题 - Scala:什么是 TypeTag，我该如何使用它? 但它似乎没有走到最后一英里" - 重新标记 Any.
As written, there are no compiler errors in method get, but T is set to Nothing and returns Some(Nothing). I would like my commented-out line to work: case Some( x ) => // println( "Get 2*'pi'=" + x*2 ) where there is a complier message, "value * is not a member of Nothing".I realise I could write more compactly, but as done, I can mouse-over in my IDE and follow step by step. There is a related question - Scala: What is a TypeTag and how do I use it? but it does not seem to go the 'last mile' - retagging an Any.
如何做到这一点?
这是我到目前为止的代码:
Here is the code I have so far:
import scala.reflect._
import scala.reflect.runtime.universe._
import collection.mutable.Map
object Test extends HMap {
def main( args: Array[ String ] ) {
var hmap = new HMap
hmap( "anInt" ) = 1
hmap( "pi" ) = 3.1416f
hmap( "c" ) = "hello"
// Test
val result = hmap.get( "pi" )
result match {
case Some( x ) =>
println( "Get 'pi'=" + x*2 )
case _ =>
}
}
}
class HMap {
private var coreMap =
Map.empty[ String, ( TypeTag[ _ <: Any ], Any ) ]
// Save the type tag with the value
def update[ T: TypeTag ]( key: String, value: T ) =
coreMap.put( key, ( typeTag[ T ], value ) )
override def toString = coreMap.toString
def get[ T: TypeTag ]( key: String ): Option[ T ] = {
val option = coreMap.get( key )
val result = option match {
case None => None
case Some( x ) => {
val typeTag = x._1; val value = x._2
println( "Matched Type = " +
typeTag.tpe + " Value=" + value )
// **** How do I use saved typeTag to define T here? ****
val v = value.asInstanceOf[ T ]
val s = Some( v )
println( "Returning " + s )
s
}
}
result
}
}
推荐答案
T 是在调用 get 方法时定义的，不能在函数内部将其更改为其他类型.编译器需要信息来获取 T 的类型信息，否则您必须明确提供:
T is defined when you call the method get, you cant change it inside a function to another type. Compiler need information to get type information for T or you have to provide it explicitly:
def get[T](key: String) = m.get(key).map(_.asInstanceOf[T])
get[Int]("anInt")
如果键入一个键，则可以推断出 T:
If a key is typed, then T can be inferred:
class Key[T](name: String)
def get[T](key: Key[T]) = ...
get(Key[Int]("anInt"))
要在从地图获取时检查类型是否正确，您可以执行您最初所做的操作，保存类型和值:
To check the type is correct when getting from map you can do what you did originally, save a type and a value:
val m = Map.empty[String, (Type, Any)]
def put[T: TypeTag](key: String, value: T) = m.put(key, (typeOf[T], value))
def get[T: TypeTag](key: String) = m.get(key) match {
case Some((t, v)) if t =:= typeOf[T] => Some(v.asInstanceOf[T])
case _ => None
}
这篇关于如何保存 TypeTag，然后稍后使用它将类型重新附加到 Any (Scala 2.10)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！