Question

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

到目前为止，我已经提出了这个问题。

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)
}

这会产生以下编译器错误：

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)

此错误消息令我感到困惑，因为我认为com.github.wpm.cancan.Derived1在此上下文中应与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 polymorphism：

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

只是添加一个小精度（非常好）特拉维斯·布朗回答：C中的trait Base[C <: Base[C]]不是让你引用实现类;它只是坚持写subclass extends Base[subclass]的惯例，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，如果您要将Base的实现作为案例类，那么您可以通过自我类型绑定实现此目的：

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)
}

多态Scala返回类型

2 个答案: