如何从 Scala TypeTag 获取通用的简单类名？

Question

How can I get the simple class name including generic using TypeTag ? I think that the method signature should be like:

def getClassName[A: TypeTag](a: A): String

getClassName(Map("a" -> 123)) should return Map[String,Int] .

---

Things I've tried:

def getClassName[A: TypeTag](a: A): String = {
  typeOf[A].typeSymbol.name.toString
}

scala> getClassName(Map("a" -> 123))
res1: String = Map

def getClassName[A: TypeTag](a: A): String = {
  typeOf[A].typeSymbol.toString
}

scala> getClassName(Map("a" -> 123))
res1: String = trait Map

def getClassName[A: TypeTag](a: A): String = {
  typeOf[A].toString
}

scala> getClassName(Map("a" -> 123))
res1: String = scala.collection.immutable.Map[String,Int] // It knows the full type!

def getClassName[A: TypeTag](a: A): String = {
  typeOf[A].typeSymbol.fullName
}

scala> getClassName(Map("a" -> 123))
res1: String = scala.collection.immutable.Map

Answer 1

From here: https://docs.scala-lang.org/overviews/reflection/typetags-manifests.html

import scala.reflect.runtime.universe._

def getClassName[T](x: T)(implicit tag: TypeTag[T]): String = {
    tag.tpe match { case TypeRef(_, t, args) => s"""${t.name} [${args.mkString(",")}]""" }
}

getClassName(Map("a" -> 123))

res5: String = Map [java.lang.String,Int]

UPDATE: Shorter version with full class names

 def getClassName[T: TypeTag](x: T) = typeOf[T].toString

getClassName(Map("a" -> 123))
res1: String = scala.collection.immutable.Map[java.lang.String,Int]

Answer 2

I'll add my answers to this based on Artem Aliev's answer , since it doesn't seems like Scala has this built-in. This method uses context bound syntax instead of an implicit parameter.

def getClassName[A: TypeTag](a: A): String = {
  val typeArgs = typeOf[A].typeArgs
  s"${typeOf[A].typeSymbol.name}${if (typeArgs.nonEmpty) typeArgs.mkString("[",",", "]") else ""}"
}
scala> getClassName(Map("a" -> 123))
res1: String = Map[String,Int]