如何在 Scala 中获取泛型函数的实际类型？

Question

如何获取调用泛型函数的实际类型？

以下示例应打印给定函数的类型 f returns:

def find[A](f: Int => A): Unit = {
  print("type returned by f:" + ???)
}

如果用 find(x => "abc") 调用 find 我想得到“"type returned by f: String"。如何在 Scala 2.11 中实现 ???？

Answer 1

使用类型标签

import scala.reflect.runtime.universe._
def func[A: TypeTag](a: A): Unit = println(typeOf[A])

scala> func("asd")
String

查看更多：http://docs.scala-lang.org/overviews/reflection/typetags-manifests.html

Answer 2

使用 TypeTag。当您需要类型参数的隐式 TypeTag（或尝试为任何类型寻找一个）时，编译器将自动生成一个并为您填充值。

import scala.reflect.runtime.universe.{typeOf, TypeTag}

def find[A: TypeTag](f: Int => A): Unit = {
    println("type returned by f: " + typeOf[A])
}

scala> find(x => "abc")
type returned by f: String

scala> find(x => List("abc"))
type returned by f: List[String]

scala> find(x => List())
type returned by f: List[Nothing]

scala> find(x => Map(1 -> "a"))
type returned by f: scala.collection.immutable.Map[Int,String]

以上定义等同于：

def find[A](f: Int => A)(implicit tt: TypeTag[A]): Unit = {
     println("type returned by f: " + typeOf[A])
}

如何在 Scala 中获取泛型函数的实际类型？

How to get the actual type of a generic function in Scala?

generics

reflection

scala

type-erasure