如何获取在指定类中混合的特征列表?

时间:2010-05-11 09:28:42

标签: scala traits

更具体的例子:

abstract trait A
trait B extends A
trait C extends A

如何检查延伸特征A(可以从0到多数)的特征是否在指定的类中混合?

3 个答案:

答案 0 :(得分:3)

其他答案的混合怎么样

abstract trait A //interested in this one
trait B extends A //and this one
trait C extends A //this one too
trait D //don't care about this one though

val x = new A with B with D
x.getClass.getInterfaces.filter(classOf[A].isAssignableFrom(_))

返回

Array[java.lang.Class[_]] = Array(interface A, interface B)

答案 1 :(得分:1)

scala> val x = new A with B with C
x: java.lang.Object with A with B with C = $anon$1@8ea25fa

scala> x.getClass.getInterfaces
res11: Array[java.lang.Class[_]] = Array(interface A, interface B, interface C)

答案 2 :(得分:1)

这样的事情怎么样:

def getTraitsExtending(clazz:Class[_], baseTrait:Class[_]): Seq[Class[_]] = {
  clazz.getInterfaces().filter { baseTrait isAssignableFrom _ }
}

这会找到clazz实现的所有特征,这些特征本身就是baseTrait的子特征。具有以下特征:

trait A
trait B extends A
trait C extends A
trait D

使用如下:

scala> val x1 = new C with B
x1: java.lang.Object with C with B = $anon$1@51d92803

scala> getTraitsExtending(x1.getClass, classOf[A])
res0: Seq[Class[_]] = WrappedArray(interface C, interface B)

scala> val x2 = new C with A            
x2: java.lang.Object with C with A = $anon$1@f8db08

scala> getTraitsExtending(x2.getClass, classOf[A])
res1: Seq[Class[_]] = WrappedArray(interface C, interface A)

scala> val x3 = new C with D             
x3: java.lang.Object with C with D = $anon$1@2bbd83d

scala> getTraitsExtending(x3.getClass, classOf[A])
res3: Seq[Class[_]] = WrappedArray(interface C)

这仅查看由传入的实例的类直接实现的接口,但可以扩展为递归查找继承层次结构。