Scala check type of generics

本秂侑毒 提交于 2019-12-22 10:08:02

问题


How do I do something like this in Scala?

case class Foo[A](x: A) {

  def get[T]: Option[T] = x match {
    case x: T => Some(x)    // if x is of type T i.e. T =:= A
    case _ => None
  }
}

val test = Foo("hi")
assert(test.get[Int] == None)
assert(test.get[String] == Some("hi"))

I tried this and ran into some weird time inference failure:

import scala.util.{Try, Success}
import reflect._

case class Foo[A](x: A) extends Dynamic {

  def get[T: ClassTag]: Option[T] = Try(x.asInstanceOf[T]) match {
    case Success(r) => Some(r) 
    case _ => None
  }
}

object Foo extends App {
  val test = Foo("hi")
  val wtf: Option[Int] = test.get[Int]
  assert(wtf.isInstanceOf[Option[String]])
  assert(wtf == Some("hi"))     // how????
  // val wtf2: Option[String] = wtf  // does not compile even if above assert passes!!
}

回答1:


Surely a dupe, but hastily:

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

import reflect._
case class Foo[A](x: A) {

  def get[T: ClassTag]: Option[T] = x match {
    case x: T => Some(x)    // if x is of type T i.e. T =:= A
    case _ => None
  }
}

// Exiting paste mode, now interpreting.

import reflect._
defined class Foo

scala> val test = Foo("hi")
test: Foo[String] = Foo(hi)

scala> test.get[Int]
res0: Option[Int] = None

scala> test.get[String]
res1: Option[String] = Some(hi)



回答2:


If you can throw out the get:

case class Foo[A](x: A)
Seq(Foo("hi"), Foo(1), Foo(2d)).collect { case f @ Foo(x: String) => f }.foreach(println)

results in:

Foo(hi)

And the others are similarly trivial.



来源:https://stackoverflow.com/questions/22341339/scala-check-type-of-generics

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!