将无形 HList 转换为元组

Question

我在 Scala 中有这个版本的 try-with-resources。我想知道是否可以使用 Shapeless 和 HList 制作一个通用版本？

import scala.util.{Failure, Success, Try}

class Loan1[A <: AutoCloseable](resource: A) {
  def to[B](block: A => B): B = {
    Try(block(resource)) match {
      case Success(result) =>
        resource.close()
        result
      case Failure(e) =>
        resource.close()
        throw e
    }
  }
}

class Loan2[A <: AutoCloseable, B <: AutoCloseable](r1: A, r2: B){
  def to[R](block: (A,B) => R): R = {
    Try(block(r1,r2)) match {
      case Success(result) =>
        r1.close(); r2.close()
        result
      case Failure(e) =>
        r1.close(); r2.close()
        throw e
    }
  }
}

object Loan {

  def apply[A <: AutoCloseable](resource: A): Loan1[A] = new Loan1(resource)

  def apply[A <: AutoCloseable, B <: AutoCloseable] (r1: A, r2: B)= new Loan2(r1, r2)

}

我猜是签名相似的东西

  def apply[L <: HList](list: L)(implicit con: LUBConstraint[L, AutoCloseable]) = ???

另一个问题是如何在 block: (A,B) => R 部分中以元组的形式提供元素？

这可以实现吗？

Answer 1

其实并没有那么难。您需要一种从元组 (Generic.Aux[Tup, L]) 获取 HList 的方法，以及一种从 Hlist (ToList[L, AutoCloseable]) 获取 List[AutoClosable] 的方法。

除了 ToList 部分之外，可能还有其他方法可以做到这一点，但它很容易融合 LUBConstraint[L, AutoCloseable] 和能够在每个资源上调用 close() 的要求.

scala> :paste
// Entering paste mode (ctrl-D to finish)

import shapeless._, ops.hlist._
import scala.util.{Failure, Success, Try}

class Loan[Tup, L <: HList](resources: Tup)(
  implicit 
  gen: Generic.Aux[Tup, L],
  con: ToList[L, AutoCloseable]
) {
  def to[B](block: Tup => B): B = {
    Try(block(resources)) match {
      case Success(result) =>
        gen.to(resources).toList.foreach { _.close() }
        result
      case Failure(e) =>
        gen.to(resources).toList.foreach { _.close() }
        throw e
    }
  }
}

object Loan {
    def apply[Tup, L <: HList](resources: Tup)(
      implicit 
      gen: Generic.Aux[Tup, L],
      con: ToList[L, AutoCloseable]
    ) = new Loan(resources)
}

// Exiting paste mode, now interpreting.


scala> class Bar() extends AutoCloseable { def close = println("close Bar"); def IAmBar = println("doing bar stuff") }
defined class Bar

scala> class Foo() extends AutoCloseable { def close = println("close Foo"); def IAmFoo = println("doing foo stuff") }
defined class Foo

scala> Loan(new Foo, new Bar).to{ case (f, b) => f.IAmFoo; b.IAmBar }
doing foo stuff
doing bar stuff
close Foo
close Bar

唯一的问题是，对于恰好 1 个资源的情况，您需要编写 Tuple1(new Foo) 和模式匹配，例如 case Tuple1(f)。最简单的解决方案是保留 Loan1 部分并将 Loan2 部分替换为 LoanN 部分，该部分使用 shapeless 实现并且适用于每个 >1 的元数。所以这几乎等于将我的解决方案复制粘贴到您的解决方案并将我的 Loan class 重命名为 LoanN:

import shapeless._, ops.hlist._, ops.nat._
import scala.util.{Failure, Success, Try}

class LoanN[Tup, L <: HList](resources: Tup)(
  implicit 
  gen: Generic.Aux[Tup, L],
  con: ToList[L, AutoCloseable]
) {
  def to[B](block: Tup => B): B = {
    Try(block(resources)) match {
      case Success(result) =>
        gen.to(resources).toList.foreach { _.close() }
        result
      case Failure(e) =>
        gen.to(resources).toList.foreach { _.close() }
        throw e
    }
  }
}

class Loan1[A <: AutoCloseable](resource: A) {
  def to[B](block: A => B): B = {
    Try(block(resource)) match {
      case Success(result) =>
        resource.close()
        result
      case Failure(e) =>
        resource.close()
        throw e
    }
  }
}


object Loan {
    def apply[A <: AutoCloseable](resource: A): Loan1[A] = new Loan1(resource)
    def apply[Tup, L <: HList, Len <: Nat](resources: Tup)(
      implicit 
      gen: Generic.Aux[Tup, L],
      con: ToList[L, AutoCloseable],
      length: Length.Aux[L, Len],
      gt: GT[Len, nat._1]
    ) = new LoanN(resources)
}

我还添加了输入长度必须大于 1 的约束。否则在你传入 case class Baz() 可以转换为 List[Nothing] 的地方存在漏洞是 List[AutoClosable].

的子类型

毫无疑问，带有 Loan1 东西的额外样板仍然可以通过自己编写更复杂的类型class 来消除，它能够区分单个参数和参数元组。

您建议接受一个 HList 作为参数并将其转换为一个元组。这也是可能的，使用 shapeless.ops.hlist.Tupler。那么当然 API 的用户将不得不自己构造 HList，并且你仍然有 scala 没有用于展开 Tuple1 的漂亮语法的问题。第二个问题可以用一个非常简单的自定义类型 class 来解决，它将 Tuple1[A] 解包为 A 并保持其他一切不变：

sealed trait Unwrap[In] { 
  type Out
  def apply(in: In): Out 
}

object Unwrap extends DefaultUnwrap {
  type Aux[In, Out0] = Unwrap[In] { type Out = Out0 }
  def apply[T](implicit unwrap: Unwrap[T]): Unwrap.Aux[T, unwrap.Out] = unwrap

  implicit def unwrapTuple1[A]: Unwrap.Aux[Tuple1[A], A] = new Unwrap[Tuple1[A]] {
    type Out = A
    def apply(in: Tuple1[A]) = in._1
  }
}
trait DefaultUnwrap {
  implicit def dontUnwrapOthers[A]: Unwrap.Aux[A, A] = new Unwrap[A] {
    type Out = A
    def apply(in: A) = in
  }
}

将其与 Tupler 相结合，您将得到一个相对简单的解决方案：

scala> :paste
// Entering paste mode (ctrl-D to finish)

import shapeless._, ops.hlist._
import scala.util.{Failure, Success, Try}

class LoanN[Tup, L <: HList, Res](resources: L)(
  implicit 
  tupler: Tupler.Aux[L, Tup],
  con: ToList[L, AutoCloseable],
  unwrap: Unwrap.Aux[Tup, Res]
) {
  def to[B](block: Res => B): B = {
    Try(block(unwrap(tupler(resources)))) match {
      case Success(result) =>
        resources.toList.foreach { _.close() }
        result
      case Failure(e) =>
        resources.toList.foreach { _.close() }
        throw e
    }
  }
}


object Loan {
    def apply[Tup, L <: HList, Res](resources: L)(
      implicit 
      tupler: Tupler.Aux[L, Tup],
      con: ToList[L, AutoCloseable],
      unwrap: Unwrap.Aux[Tup, Res]
    ) = new LoanN(resources)
}

// Exiting paste mode, now interpreting.


scala> Loan(new Foo :: new Bar ::  HNil).to{ case (f,b) => f.IAmFoo; b.IAmBar }
doing foo stuff
doing bar stuff
close Foo
close Bar

scala> Loan(new Foo :: HNil).to{ case (f) => f.IAmFoo }
doing foo stuff
close Foo

将无形 HList 转换为元组

Convert Shapeless HList to a Tuple

scala

shapeless

hlist