Question

我真的在scala中与暧昧的类型挣扎......

我想实现一个StackManager，它包含多个不同类型的堆栈......但似乎有些错误。

object StackManager {

  val stacks = mutable.HashMap[String, mutable.Stack[_]]()
  def get[T](id: String): mutable.Stack[T] = {
    stacks.get(id) match {
      case Some(stack: mutable.Stack[T]) => stack
      case None => {
        val newStack = mutable.Stack[T]()
        stacks.put(id, newStack)
        newStack
      }
    }
  }
  def set[T](id: String, stack: mutable.Stack[T]) {
    stacks.put(id, stack)
  }
}

编辑：我想在我的经理中存储不同类型的堆栈：

StackManager.set[ClassA]("stack01", new mutable.Stack[ClassA]())
StackManager.set[ClassB]("stack02", new mutable.Stack[ClassB]())

StackManager.get[ClassA]("stack01") // returns the stack with type mutable.Stack[ClassA]
StackManager.get[ClassB]("stack02") // returns the stack with type mutable.Stack[ClassB]

EDIT2：scala version 2.9.2

Answer 1

因为在JVM上执行scala时不存储类型参数，所以如果需要，您需要自己存储这些类型。例如，

val stacks = mutable.HashMap[String, (Class, mutable.Stack[_])]()

然后检查Class是否等于请求的内容但这可能不是你想要的：堆栈将被不同类型的堆栈覆盖。但你得到了这个想法，并且可以创造出有效的东西。

Answer 2

有一个解决方案，但它有点不稳定。（引用下面Scala 2.10控制台的输出，作为快速演示）

import scala.reflect.runtime.{universe=>ru}
import scala.collection.mutable

scala> :paste
// Entering paste mode (ctrl-D to finish)

object StackManager {

  val stacks = mutable.HashMap[(String, ru.Type), mutable.Stack[_]]()
  def get[T:ru.TypeTag](id: String): mutable.Stack[T] = {
    stacks.get((id,ru.typeOf[T])) match {
      case Some(stack: mutable.Stack[T]) => stack
      case _ => {
        val newStack = mutable.Stack[T]()
        stacks.put((id,ru.typeOf[T]), newStack)
        newStack
      }
    }
  }
  def set[T:ru.TypeTag](id: String, stack: mutable.Stack[T]) {
    stacks.put((id,ru.typeOf[T]), stack)
  }
}


// Exiting paste mode, now interpreting.

defined module StackManager

scala> StackManager.set("example", new mutable.Stack[String])

scala> StackManager.set("example", new mutable.Stack[Int])

scala> StackManager.get[Int]("example")
res15: scala.collection.mutable.Stack[Int] = Stack()

scala> StackManager.get[String]("example")
res16: scala.collection.mutable.Stack[String] = Stack()

scala> StackManager.get[Double]("example")
res17: scala.collection.mutable.Stack[Double] = Stack()

不那么时髦的解决方案，没有类型标签：

object StackManager {

  val stacks = mutable.HashMap[String, mutable.Stack[_ <:Any]]()
  def get[T](id: String): mutable.Stack[T] = {
    stacks.get(id) match {
      case Some(stack: mutable.Stack[_]) => stack.asInstanceOf[mutable.Stack[T]]
      case _ => {
        val newStack = mutable.Stack[T]()
        stacks.put(id,newStack)
        newStack
      }
    }
  }
  def set[T](id: String, stack: mutable.Stack[T]) {
    stacks.put(id,stack)
  }
}

它似乎也在进行类型检查：

scala> StackManager.set[String]("string",new mutable.Stack[String])

scala> StackManager.set[Integer]("integer",new mutable.Stack[String])
<console>:11: error: type mismatch;
 found   : scala.collection.mutable.Stack[String]
 required: scala.collection.mutable.Stack[Integer]
              StackManager.set[Integer]("integer",new mutable.Stack[String])

scala> StackManager.set[Integer]("integer",new mutable.Stack[Integer])

scala> StackManager.get[Integer]("integer")
res11: scala.collection.mutable.Stack[Integer] = Stack()

我没有在scala 2.9中尝试过它。

Answer 3

我找到了合适的解决方案：

object StackManager {

  val stacks = mutable.HashMap[String, S forSome {type S}]()
  def get[S](id: String): S = {
    stacks.get(id) match {
      case Some(stack) => stack.asInstanceOf[S]
      case None => null.asInstanceOf[S]
    }
  }
  def set[S](id: String, stack: S) {
    stacks.put(id, stack)
  }
}


StackManager.set[mutable.Stack[ClassA]]("stack01", new mutable.Stack[ClassA]())
StackManager.set[mutable.Stack[ClassB]]("stack02", new mutable.Stack[ClassB]())

StackManager.get[mutable.Stack[ClassA]]("stack01") // returns the stack with type mutable.Stack[ClassA]
StackManager.get[mutable.Stack[ClassB]]("stack02") // returns the stack with type mutable.Stack[ClassB]

任何改进的想法？某种类型检查mutable.Stack会很棒...

在scala中挣扎着模糊的类型

3 个答案: