将Shapeless HList转换为元组

Question

我try-with-resources在Scala中有这个版本。我想知道是否可以使用Shapeless和HList制作通用版本？

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)

}

我猜有类似签名的东西

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

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

这可能实现吗？

Answer 1

其实没那么难。HList您需要一种从元组 ( )中获取 a 的方法以及一种从( )Generic.Aux[Tup, L]中获取 a 的方法。List[AutoClosable]HlistToList[L, AutoCloseable]

除了该部分之外，可能还有其他方法可以做到这ToList一点，但它很容易融合LUBConstraint[L, AutoCloseable]以及能够调用close()每个资源的要求。

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，如case Tuple1(f). 最简单的解决方案是保留该部件，并将该部件Loan1替换为使用 Shapeless 实现且适用于每个 >1 的部件。因此，这几乎等于将我的解决方案复制粘贴到您的解决方案中并将我的类重命名为：Loan2LoanNLoanLoanN

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。否则会存在一个漏洞，即您传入的 acase class Baz()可以转换为List[Nothing]的子类型 a List[AutoClosable]。

毫无疑问，Loan1仍然可以通过自己编写一个更复杂的类型类来消除这些额外的样板，该类型类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元组。这也是可能的，通过shapeless.ops.hlist.Tupler. 当然，该 API 的用户必须HList自己构建，并且仍然存在 scala 没有漂亮的语法来解包Tuple1. 第二个问题可以通过一个非常简单的自定义类型类来解决，该类型类将 a 展开Tuple1[A]为 anA并保持其他所有内容不变：

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
  }
}

结合起来Tupler，你就有了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