我确实对以下代码有问题。它不编译。
有没有人知道如何在不使用 asInstanceOf[SomeImpl] 或模式匹配的情况下编译它。
我正在考虑使用上限或下限的一些类型参数。
object InherenceTryout extends App {
import someimpl._
val list = List(SomeImpl("A"), SomeImpl("B"))
val result = new SomeHandler().handleBaseList(list)
println("%s->%s" format (list, result))
}
package base {
// Defines that a 'Base' object can create a new
// 'Base' object where another 'Base' object is mixed in
// Does not define how the mixing will take place
trait Base {
def mix(other: Base): Base
}
// Defines how a default 'Base' object gets mixed into a list of 'Base' objects
trait BaseHandler {
def defaultBase: Base
def handleBaseList(list: List[Base]): List[Base] = list.map(b => b.mix(defaultBase))
}
}
package someimpl {
import base._
// Some implementation of 'Base'
// Defines how a new 'SomeImpl' object is created by mixing in another
// 'SomeImpl' object
// ERROR:
// class SomeImpl needs to be abstract, since method mix in trait Base of type (other: base.Base)base.Base is not defined
// (Note that base.Base does not match someimpl.SomeImpl: class SomeImpl in
// package someimpl is a subclass of trait Base in package base, but method parameter types must match exactly.)
case class SomeImpl(id: String) extends Base {
def mix(other: SomeImpl): SomeImpl = SomeImpl("[%s|%s]" format (id, other.id))
}
// Defines the default 'SomeImpl' object
class SomeHandler extends BaseHandler {
def defaultBase = SomeImpl("D")
}
}
最佳答案
使用类型参数:
package base {
trait Base[T <: Base[T]] {
def mix(other: T): T
}
trait BaseHandler[T <: Base[T]] {
def defaultBase: T
def handleBaseList(list: List[T]): List[T] =
list.map(b => b.mix(defaultBase))
}
}
package someimpl {
import base._
case class SomeImpl(id: String) extends Base[SomeImpl] {
def mix(other: SomeImpl): SomeImpl = SomeImpl("[%s|%s]" format (id, other.id))
}
class SomeHandler extends BaseHandler[SomeImpl] {
def defaultBase = SomeImpl("D")
}
}
关于scala - 子类中的特定参数类型不可能，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13251347/