如何确定类型参数是否是特征的子类型?

时间:2012-06-20 19:54:02

标签: scala

假设我有以下类型

class Foo
trait Bar

有没有办法制作一个接受Type参数T的方法,并确定该T是否为Bar?例如,

def isBar[T <: Foo: Manifest] = 
  classOf[Bar].isAssignableFrom(manifest[T].erasure)

可悲的是,isBar[Foo with Bar]false,因为删除似乎会删除mixins。

此外,manifest[Foo with Bar] <:< manifest[Bar]为假

这有可能吗?

我看了这个问题:How to tell if a Scala reified type extends a certain parent class?

但这个答案不适用于混合特征,因为它们似乎已被删除,如上所述。

4 个答案:

答案 0 :(得分:22)

这可以通过TypeTags(至少2.10M7)来实现:

scala> class Foo; trait Bar
defined class Foo
defined trait Bar

scala> import reflect.runtime.universe._
import reflect.runtime.universe._

scala> def isBar[A <: Foo : TypeTag] = typeOf[A].baseClasses.contains(typeOf[Bar].typeSymbol)
isBar: [A <: Foo](implicit evidence$1: reflect.runtime.universe.TypeTag[A])Boolean

scala> isBar[Foo]
res43: Boolean = false

scala> isBar[Foo with Bar]
res44: Boolean = true

TypeTags提供Scala类型的1:1转换,因为它们代表编译器知道的类型。因此,它们比普通的旧Manifest强大得多:

scala> val fooBar = typeTag[Foo with Bar]
fooBar: reflect.runtime.universe.TypeTag[Foo with Bar] = TypeTag[Foo with Bar]

使用方法tpe,我们可以完全访问Scalas new Reflection:

scala> val tpe = fooBar.tpe // equivalent to typeOf[Foo with Bar]
tpe: reflect.runtime.universe.Type = Foo with Bar

scala> val tpe.<tab><tab> // lot of nice methods here
=:=                 asInstanceOf        asSeenFrom          baseClasses         baseType            contains            declaration         
declarations        erasure             exists              find                foreach             isInstanceOf        kind                
map                 member              members             narrow              normalize           substituteSymbols   substituteTypes     
takesTypeArgs       termSymbol          toString            typeConstructor     typeSymbol          widen  

答案 1 :(得分:6)

有可能在2.10之前做到这一点,而不是(据我所知)有清单:

def isBar[T <: Foo](implicit ev: T <:< Bar = null) = ev != null

这有点像黑客,但它可以按照需要运作。

scala> isBar[Foo with Bar]
res0: Boolean = true

scala> isBar[Foo]
res1: Boolean = false

答案 2 :(得分:3)

您可以使用类型类来解决它而无需反射:

trait IsBar[T] {
  def apply():Boolean
}

trait LowerLevelImplicits {
  implicit def defaultIsBar[T] = new IsBar[T]{
    def apply() = false
  }
}

object Implicits extends LowerLevelImplicits {
  implicit def isBarTrue[T <: Bar] = new IsBar[T] {
    def apply() = true
  }
}

def isBar[T<:Foo]( t: T )( implicit ib: IsBar[T] ) = ib.apply()

scala> import Implicits._

scala> isBar( new Foo )
res6: Boolean = false

scala> isBar( new Foo with Bar )
res7: Boolean = true

答案 3 :(得分:0)

另一种类型类用法(更通用):

  trait SubClassGauge[A, B] {
    def A_isSubclassOf_B: Boolean
  }

  implicit class IsSubclassOps[A](a: A) {
    def isSubclassOf[B](implicit ev: SubClassGauge[A, B]): Boolean = ev.A_isSubclassOf_B
  }

  trait LowerLevelImplicits {
    implicit def defaultSubClassGauge[A, B] = new SubClassGauge[A, B] {
      override def A_isSubclassOf_B: Boolean = false
    }
  }

  object Implicits extends LowerLevelImplicits {
    implicit def subClassGauge[A <: B, B]: SubClassGauge[A, B] = new SubClassGauge[A, B] {
      override def A_isSubclassOf_B: Boolean = true
    }
  }

  trait Prime
  class NotSuper
  class Super extends Prime
  class Sub extends Super
  class NotSub

现在,在REPL中:

@ import Implicits._ 
import Implicits._
@ (new Sub).isSubclassOf[NotSuper] 
res29: Boolean = false
@ (new Sub).isSubclassOf[Super] 
res30: Boolean = true
@ (new Sub).isSubclassOf[Prime] 
res31: Boolean = true
@ (new Super).isSubclassOf[Prime] 
res32: Boolean = true
@ (new Super).isSubclassOf[Sub] 
res33: Boolean = false
@ (new NotSub).isSubclassOf[Super] 
res34: Boolean = false

TypeTag现在属于scala reflect package。人们需要添加额外的依赖来使用它。