多态 Scala 返回类型 [英] Polymorphic Scala return type

问题描述

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()，它返回一个与其实现类相同类型的对象.所以 Derived1.f() 返回 Derived1 并且 Derived2.f() 返回 Derived2.如何在 Scala 中编写此代码?

I have an abstract Scala class Base which has subclasses Derived1 and Derived2. Base defines a function f() which returns an object of the same type as its implementing class. So Derived1.f() returns Derived1 and Derived2.f() returns Derived2. How do I write this in Scala?

这是我到目前为止的想法.

Here is what I have come up with so far.

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:

This gives the following compiler errors:

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)

这个错误信息让我很困惑，因为我认为 com.github.wpm.cancan.Derived1 在这种情况下应该与 Derived1 相同.

This error message is confusing to me because I think com.github.wpm.cancan.Derived1 should be the same as Derived1 in this context.

推荐答案

Randall Schulz 指出了您当前代码不起作用的原因之一.不过，使用 F-有界多态性可以得到你想要的:

Randall Schulz pointed out one of the reasons your current code doesn't work. It is possible to get what you want, though, with F-bounded polymorphism:

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)
}

基础 trait 上的类型参数允许你在那里谈论实现类——例如在 f 的返回类型中.

The type parameter on the base trait allows you to talk about the implementing class there—e.g. in the return type for f.

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