多态Scala返回类型

Question

我有一个抽象的Scala类Base ，它有Derived1和Derived2子类。 Base定义了一个函数f（），它返回与其实现类相同类型的对象。 所以Derived1.f()返回Derived1 ， Derived2.f()返回Derived2 。 我如何在Scala中写这个？

这是我到目前为止所提出的。

package com.github.wpm.cancan

abstract class Base {
  def f[C <: Base]: C
}

case class Derived1(x: Int) extends Base {
  def f[Derived1] = Derived1(x + 1)
}

case class Derived2(x: Int) extends Base {
  def f[Derived2] = Derived2(x + 2)
}

这给出了以下编译器错误：

type mismatch;
[error]  found   : com.github.wpm.cancan.Derived1
[error]  required: Derived1
[error]   def f[Derived1] = Derived1(x + 1)

type mismatch;
[error]  found   : com.github.wpm.cancan.Derived2
[error]  required: Derived2
[error]   def f[Derived2] = Derived2(x + 2)

此错误消息让我感到困惑，因为我认为com.github.wpm.cancan.Derived1应该与此上下文中的Derived1相同。

Answer 1

Randall Schulz指出了当前代码不起作用的原因之一。 但是，使用F -bounded多态性可以得到你想要的东西 ：

trait Base[C <: Base[C]] { def f: C }

case class Derived1(x: Int) extends Base[Derived1] {
  def f: Derived1 = Derived1(x + 1)
}

case class Derived2(x: Int) extends Base[Derived2] {
  // Note that you don't have to provide the return type here.
  def f = Derived2(x + 2)
}

基本特征上的类型参数允许您在那里讨论实现类 - 例如，在f的返回类型中。

Answer 2

只是为了增加一个小的精度（非常好）特拉维斯·布朗回答：这不是特征中的C trait Base[C <: Base[C]]让你引用实现类; 它只是坚持编写subclass extends Base[subclass]的约定，让你这样做。 我不知道要提到这种类型。 为了澄清我的意思，这个编译

trait Base[C <: Base[C]] { def f: C }

case class Derived1(x: Int) extends Base[Derived1] {
  def f: Derived1 = Derived1(x + 1)
}
// a Derived2 where f returns Derived1!!
case class Derived2(x: Int) extends Base[Derived1] {
  def f = Derived1(x + 2)
}

现在，如果所有你将要作为Base实现都是case类，你可以通过自我类型绑定得到这个：

trait Base[C <: Base[C]] { self: C => 
  def f: C 
}

case class Derived1(x: Int) extends Base[Derived1] {
  def f: Derived1 = Derived1(x + 1)
}
// a Derived2 where f returns Derived1!!
// this won't compile now
case class Derived2(x: Int) extends Base[Derived1] {
  def f = Derived1(x + 2)
}