Scala:类型注释使尾递归检查失败 [英] Scala: type annotations make tail recursion check fail

问题描述

我在此模式匹配中添加类型注释只是为了我自己的理解.

I'm adding type annotations to this pattern match just for my own understanding.

@annotation.tailrec def run[A](io: IO[A]): A = {
  io match {
    case Return(a) => a
    case Suspend(r) => r()
    case FlatMap(x, f) => x match {
      case Return(a) => run(f(a))
      case Suspend(r) => run(f(r()))
      case FlatMap(y, g) => 
        run(y flatMap (a => g(a) flatMap f))
    }
  }
}

为什么这些类型注解会破坏尾递归检查?添加新的类型定义和类型注释后，我没有清楚地看到代价高昂的新递归.

Why do these type annotations break the tail recursion check? With the new type definitions and type annotations added, I don't clearly see a costly new recursion.

could not optimize @tailrec annotated method run: it contains a recursive call not in tail position
io match {
^

@annotation.tailrec def run[A](io: IO[A]): A = {
  type rType = Unit => A
  type fType = A => IO[A]
  type gType = A => IO[A]
  io match {
    case Return(a: A) => a
    case Suspend(r: rType) => r()
    case FlatMap(x: IO[A], f: fType) => x match {
      case Return(a: A) => run(f(a))
      case Suspend(r: rType) => run(f(r()))
      case FlatMap(y: IO[A], g: gType) => 
        run(y flatMap (a => g(a) flatMap f))
    }
  }
}

匹配的案例类:

case class Return[A](a: A) extends IO[A]
case class Suspend[A](resume: () => A) extends IO[A]
case class FlatMap[A,B](sub: IO[A], k: A => IO[B]) extends IO[B]

只要省略类型注解，类型'a'就行

As long as the type annotations are omitted, type of 'a' in line

 F.flatMap(r)((a: A) => run(f(a)))

必须是任何":

 [error]  found   : A => F[A]
 [error]  required: Any => F[A]
 [error]         F.flatMap(r)((a: A) => run(f(a)))

编译:

 F.flatMap(r)(a => run(f(a)))

<小时>

奖励问题.

---

Bonus question.

似乎不允许对 case 类中的函数进行模式匹配，如下所示:

It seems that pattern matching against a function inside a case class, like this, is not allowed:

io match {
  ...
  case Suspend(r: Unit => A) => r()
  /* or */
  case Suspend(r: () => A) => r()
  ...
}

编译:

io match {
  ...
  case Suspend(r: Function0[A]) => r()
  ...
}

这是为什么?

由于类型擦除，这些类型注释最终不会有太大用处.注释这些类型后，我可以期待看到这样的编译器警告:

These type annotations won't have much use in the end because of type erasure. After annotating these types, I can expect to see a compiler warning like this:

abstract type pattern ... is unchecked since it is eliminated by erasure

<小时>

此代码来自Scala 中的函数式编程"的第 13 章或包 fpinscala.iomonad.https://github.com/fpinscala/fpinscala

谢谢

推荐答案

奖励答案:关于类型擦除的问题比较多，看看如何解决 Scala 上的类型擦除问题?或者，为什么我不能得到我的集合的类型参数?你可以写一些像

Bonus answer: There are a lot of questions about type erasure, take a look at How do I get around type erasure on Scala? Or, why can't I get the type parameter of my collections? You can write something like

    case FlatMap(y: IO[A], g: gType@unchecked) if g.isInstanceOf[gType] =>

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