Question

abstract class A {
  type ThisType <: A
  def assign(other: ThisType): ThisType
}

class B extends A {
  override type ThisType = B
  override def assign(other: B): B = ???
}

可能声明p1和p2的地方。从本质上讲，他们可以在任何地方宣布。我想这将涵盖大多数使用场景：

1a）的

implicit val p1: /* ??? */
implicit val p2: /* ??? */
implicit val f = { p1, p2 => p1.assign(p2) }
val c: => Unit = { p1, p2 => p1.assign(p2) }

object R1a {
  apply(implicit p1: B, p2: B, f: (B, B) => B) = new R1a(p1, p2, f)
}

val f: (B, B) => B = { p1, p2 => p1.assign(p2) }
class R1a(val p1: B, val p2: B, val f: (B, B) => B) extends Runnable {
  override def run: Unit = p1.assign(p2)
  // and val f: override def run: Unit = f(p1, p2)
}

1b）的

implicit val p1: /* ??? */
implicit val p2: /* ??? */
implicit val f = { p1, p2 => p1.assign(p2) }
val c: => Unit = { p1, p2 => p1.assign(p2) }

object R1b {
  apply[T <: A](implicit p1: T, p2: T, f: (T, T) => T) = new R1a(p1, p2, f)
}

class R1b[T <: A](val p1: /* ??? */, val p2: /* ??? */, val f: (/* ??? */, /* ??? */T) => /* ??? */) extends Runnable {
  override def run: Unit = p1.assign(p2)
  // and val f: override def run: Unit = f(p1, p2)
}

2a）的

class R2a extends Runnable {
  type T /* ??? */
  val p1: T = /* ??? */
  val p2: T = /* ??? */
  val f: (T, T) => T = { p1, p2 => p1.assign(p2) }
  override def run: Unit = p1.assign(p2)
  // and val f: override def run: Unit = f(p1, p2)
}

2b）的

class R2b[T <: A] extends Runnable {
  val p1: /* ??? */
  val p2: /* ??? */
  val f: (T, T) => T = { p1, p2 => p1.assign(p2) }
  override def run: Unit = p1.assign(p2)
  // and val f: override def run: Unit = f(p1, p2)
}

3a）的

class R3a extends Runnable {
  override def run: Unit = {
    type T /* ??? */
    val p1: T = /* ??? */
    val p2: T = /* ??? */
    val f: (T, T) => T = { p1, p2 => p1.assign(p2) }
    p1.assign(p2)
    // and f(p1, p2)
  }
}

3b）的

class R3b[T <: A] extends Runnable {
  override def run: Unit = {
    val p1: /* ??? */ = /* ??? */
    val p2: /* ??? */ = /* ??? */
    val f: (T, T) => T = { p1, p2 => p1.assign(p2) }
    p1.assign(p2)
    // and f(p1, p2)
  }
}

/* ??? */表示占位符。无需涵盖所有场景，只要有解释即可。我试图在DSL中使用它，因此有时我不想引入新的类型参数，并且当它们适合DSL设计时也使用它们。

这是一个跟进问题：How do I make sure a function receives the same parameter type as the current object?

如果我必须在run函数中执行函数赋值，我该如何实现。

你如何使用类型投影p1和p2是p1.assign（p2）中的相同类型

0 个答案: