在覆盖方法中使用和扩展类型

Question

我正在尝试创建一个抽象 class，它有一个方法，我想采用与某些基础 class 相同或扩展的参数。我以为会是：

trait InputParams
abstract class A() {
  def input[T <: InputParams](datum: T) // abstract
  ...
}

case class ExtendedInputParams() extends InputParams
class B() extends A() {
  override def input[T <: InputParams](datum: T)...
  // the input(datum) method needs more to be able to treat datum as an
  // ExtendedInputParams, how do I do this?
  ...
}

in class B（我怀疑 A 也需要更改）我如何将 datum 类型定义为 ExtendedInputParamsfor (new B).input(some-datum-of-type-ExtendedInputParams) 以便覆盖 input 可以将 datum 视为 ExtendedInputParams 并仍然强制其扩展 InputParams？我还想强制该方法覆盖 class A 中的抽象方法。

为清楚起见进行了更新

Answer 1

缩小子类中的约束：

scala> class P ; class Q extends P
defined class P
defined class Q

scala> class A { type T <: P ; def f[X <: T](a: X) = a.toString }
defined class A

scala> class B extends A { type T = Q ; override def f[X <: T](a: X) = a.toString * 2 }
defined class B

scala> (new B).f(new Q)
res0: String = Q@1ea930ebQ@1ea930eb

scala> (new B).f(new P)
<console>:14: error: inferred type arguments [P] do not conform to method f's type parameter bounds [A <: Q]
       (new B).f(new P)
               ^
<console>:14: error: type mismatch;
 found   : P
 required: A
       (new B).f(new P)
                 ^

scala> (new A { type T = P }).f(new P)
res3: String = P@4300e240

改写：

scala> class P ; class Q extends P { def q = 42 }
defined class P
defined class Q

scala> class X[T <: P] { def f[A <: T](a: A) = a.toString }
defined class X

scala> class Y[T <: Q] extends X[T] { override def f[A <: T](a: A) = a.q.toString }
defined class Y

scala> (new Y).f(new Q)
res0: String = 42

在覆盖方法中使用和扩展类型

Using and extended type in an overriden method

scala

scala-2.11