380

我所知道的关于 TypeTags 的只是它们以某种方式取代了 Manifests。Internet 上的信息很少,无法让我很好地了解该主题。

因此,如果有人在 TypeTags 上分享一些有用材料的链接,包括示例和流行用例,我会很高兴。也欢迎详细的回答和解释。

4

1 回答 1

579

ATypeTag解决了 Scala 的类型在运行时被擦除的问题(类型擦除)。如果我们想做

class Foo
class Bar extends Foo

def meth[A](xs: List[A]) = xs match {
  case _: List[String] => "list of strings"
  case _: List[Foo] => "list of foos"
}

我们会收到警告:

<console>:23: warning: non-variable type argument String in type pattern List[String]↩
is unchecked since it is eliminated by erasure
         case _: List[String] => "list of strings"
                 ^
<console>:24: warning: non-variable type argument Foo in type pattern List[Foo]↩
is unchecked since it is eliminated by erasure
         case _: List[Foo] => "list of foos"
                 ^

为了解决这个问题,Scala 引入了Manifest 。但是他们的问题是不能表示很多有用的类型,比如依赖路径的类型:

scala> class Foo{class Bar}
defined class Foo

scala> def m(f: Foo)(b: f.Bar)(implicit ev: Manifest[f.Bar]) = ev
warning: there were 2 deprecation warnings; re-run with -deprecation for details
m: (f: Foo)(b: f.Bar)(implicit ev: Manifest[f.Bar])Manifest[f.Bar]

scala> val f1 = new Foo;val b1 = new f1.Bar
f1: Foo = Foo@681e731c
b1: f1.Bar = Foo$Bar@271768ab

scala> val f2 = new Foo;val b2 = new f2.Bar
f2: Foo = Foo@3e50039c
b2: f2.Bar = Foo$Bar@771d16b9

scala> val ev1 = m(f1)(b1)
warning: there were 2 deprecation warnings; re-run with -deprecation for details
ev1: Manifest[f1.Bar] = Foo@681e731c.type#Foo$Bar

scala> val ev2 = m(f2)(b2)
warning: there were 2 deprecation warnings; re-run with -deprecation for details
ev2: Manifest[f2.Bar] = Foo@3e50039c.type#Foo$Bar

scala> ev1 == ev2 // they should be different, thus the result is wrong
res28: Boolean = true

因此,它们被TypeTags取代,它们使用起来更简单,并且很好地集成到了新的 Reflection API 中。有了它们,我们可以优雅地解决上面关于路径依赖类型的问题:

scala> def m(f: Foo)(b: f.Bar)(implicit ev: TypeTag[f.Bar]) = ev
m: (f: Foo)(b: f.Bar)(implicit ev: reflect.runtime.universe.TypeTag[f.Bar])↩
reflect.runtime.universe.TypeTag[f.Bar]

scala> val ev1 = m(f1)(b1)
ev1: reflect.runtime.universe.TypeTag[f1.Bar] = TypeTag[f1.Bar]

scala> val ev2 = m(f2)(b2)
ev2: reflect.runtime.universe.TypeTag[f2.Bar] = TypeTag[f2.Bar]

scala> ev1 == ev2 // the result is correct, the type tags are different
res30: Boolean = false

scala> ev1.tpe =:= ev2.tpe // this result is correct, too
res31: Boolean = false

它们也很容易用于检查类型参数:

import scala.reflect.runtime.universe._

def meth[A : TypeTag](xs: List[A]) = typeOf[A] match {
  case t if t =:= typeOf[String] => "list of strings"
  case t if t <:< typeOf[Foo] => "list of foos"
}

scala> meth(List("string"))
res67: String = list of strings

scala> meth(List(new Bar))
res68: String = list of foos

此时,了解使用=:=(类型相等)和<:<(子类型关系)进行相等检查非常重要。切勿使用==or !=,除非您绝对知道自己在做什么:

scala> typeOf[List[java.lang.String]] =:= typeOf[List[Predef.String]]
res71: Boolean = true

scala> typeOf[List[java.lang.String]] == typeOf[List[Predef.String]]
res72: Boolean = false

后者检查结构相等性,这通常不是应该做的,因为它不关心诸如前缀之类的事情(如示例中所示)。

ATypeTag完全由编译器生成,这意味着编译器在TypeTag调用期望这样的方法时创建并填充 a TypeTag。存在三种不同形式的标签:

ClassTag替代ClassManifestwhileTypeTag或多或少替代Manifest.

前者允许完全使用泛型数组:

scala> import scala.reflect._
import scala.reflect._

scala> def createArr[A](seq: A*) = Array[A](seq: _*)
<console>:22: error: No ClassTag available for A
       def createArr[A](seq: A*) = Array[A](seq: _*)
                                           ^

scala> def createArr[A : ClassTag](seq: A*) = Array[A](seq: _*)
createArr: [A](seq: A*)(implicit evidence$1: scala.reflect.ClassTag[A])Array[A]

scala> createArr(1,2,3)
res78: Array[Int] = Array(1, 2, 3)

scala> createArr("a","b","c")
res79: Array[String] = Array(a, b, c)

ClassTag仅提供在运行时创建类型所需的信息(类型已擦除):

scala> classTag[Int]
res99: scala.reflect.ClassTag[Int] = ClassTag[int]

scala> classTag[Int].runtimeClass
res100: Class[_] = int

scala> classTag[Int].newArray(3)
res101: Array[Int] = Array(0, 0, 0)

scala> classTag[List[Int]]
res104: scala.reflect.ClassTag[List[Int]] =↩
        ClassTag[class scala.collection.immutable.List]

正如上面所见,他们不关心类型擦除,因此如果想要“完整”类型TypeTag应该使用:

scala> typeTag[List[Int]]
res105: reflect.runtime.universe.TypeTag[List[Int]] = TypeTag[scala.List[Int]]

scala> typeTag[List[Int]].tpe
res107: reflect.runtime.universe.Type = scala.List[Int]

scala> typeOf[List[Int]]
res108: reflect.runtime.universe.Type = scala.List[Int]

scala> res107 =:= res108
res109: Boolean = true

如您所见,方法tpeTypeTag结果是完整的Type,这与我们在typeOf调用时得到的结果相同。当然,可以同时使用ClassTagTypeTag

scala> def m[A : ClassTag : TypeTag] = (classTag[A], typeTag[A])
m: [A](implicit evidence$1: scala.reflect.ClassTag[A],↩
       implicit evidence$2: reflect.runtime.universe.TypeTag[A])↩
      (scala.reflect.ClassTag[A], reflect.runtime.universe.TypeTag[A])

scala> m[List[Int]]
res36: (scala.reflect.ClassTag[List[Int]],↩
        reflect.runtime.universe.TypeTag[List[Int]]) =↩
       (scala.collection.immutable.List,TypeTag[scala.List[Int]])

现在剩下的问题是什么是感觉WeakTypeTag?简而言之,TypeTag表示一个具体类型(这意味着它只允许完全实例化的类型),而WeakTypeTag只允许任何类型。大多数时候人们并不关心哪个是什么(这意味着TypeTag应该使用),但是例如,当使用应该与泛型类型一起使用的宏时,就需要它们:

object Macro {
  import language.experimental.macros
  import scala.reflect.macros.Context

  def anymacro[A](expr: A): String = macro __anymacro[A]

  def __anymacro[A : c.WeakTypeTag](c: Context)(expr: c.Expr[A]): c.Expr[A] = {
    // to get a Type for A the c.WeakTypeTag context bound must be added
    val aType = implicitly[c.WeakTypeTag[A]].tpe
    ???
  }
}

如果一个替换WeakTypeTagTypeTag错误,则抛出:

<console>:17: error: macro implementation has wrong shape:
 required: (c: scala.reflect.macros.Context)(expr: c.Expr[A]): c.Expr[String]
 found   : (c: scala.reflect.macros.Context)(expr: c.Expr[A])(implicit evidence$1: c.TypeTag[A]): c.Expr[A]
macro implementations cannot have implicit parameters other than WeakTypeTag evidences
             def anymacro[A](expr: A): String = macro __anymacro[A]
                                                      ^

有关两者之间差异的更详细说明TypeTagWeakTypeTag请参阅此问题:Scala Macros: “cannot create TypeTag from a type T has unresolved type parameters”</a>

Scala 的官方文档站点也包含反射指南

于 2012-09-01T23:16:37.980 回答