[英]In Scala, why it is impossible to infer TypeTag from type alias or dependent type?
I have a simple scala program to test the Capability of Scala to infer type classes:我有一个简单的 Scala 程序来测试 Scala 推断类型类的能力:
import scala.reflect.ClassTag
object InferTypeTag {
import org.apache.spark.sql.catalyst.ScalaReflection.universe._
def infer(): Unit = {
type U = (Int, String)
val ttg1 = implicitly[TypeTag[(Int, String)]]
val ttg2 = implicitly[TypeTag[U]]
val ctg = implicitly[ClassTag[U]]
}
}
ttg1
got inferred without problem. ttg1
被推断没有问题。
ttg2
triggered the following compilation error: ttg2
触发了以下编译错误:
Error:(14, 26) No TypeTag available for U
val ttg2 = implicitly[TypeTag[U]]
Question 1 : why it doesn't work?问题1 :为什么它不起作用? type U is already final and is impossible to override类型 U 已经是最终的并且不可能被覆盖
Question 2 : if type U is not final and path-dependent, why ctg can be inferred successfully?问题2 :如果类型U不是final且是path-dependent,为什么ctg可以推断成功?
Maybe someone will explain why a local type looks abstract.也许有人会解释为什么局部类型看起来很抽象。 Maybe just a bug.也许只是一个错误。
scala> import scala.reflect.runtime.universe._
import scala.reflect.runtime.universe._
scala> { type X = Tuple2[Int, String] ; weakTypeTag[X] }
res0: reflect.runtime.universe.WeakTypeTag[X] = WeakTypeTag[X]
scala> { type X = Tuple2[Int, String] ; typeTag[X] }
^
error: No TypeTag available for X
scala> { type X = Tuple2[Int, String] ; weakTypeTag[X].tpe }
res2: reflect.runtime.universe.Type = X
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