5

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)

}

有类似签名的东西，我猜

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

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

这有可能实现吗？

4

1 回答 1

3

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

除了部件之外，可能还有其他方法可以做到这一点ToList，但它很容易融合LUBConstraint[L, AutoCloseable]和能够调用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，例如case Tuple1(f). 最简单的解决方案是保留零件并用 aLoan1替换零件，该零件以无形实现并适用于每个元数 >1。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中并将我的课程重命名为：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)
}

我还添加了输入的长度必须大于 1 的约束。否则会出现一个漏洞，您传入的 acase class Baz()可以转换为 aList[Nothing]的子类型List[AutoClosable]。

毫无疑问，仍然可以通过自己编写一个更复杂的类型类来消除额外的样板文件，该Loan1类型类能够区分单个参数和参数元组。

您建议接受一个HListas 参数并将其转换为一个元组。这也是可能的，使用shapeless.ops.hlist.Tupler. 然后，该 API 的用户当然必须自己构建HList，而且您仍然会遇到 scala 没有漂亮的语法来解开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 解包Tuple1[A]到 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
  }
}

结合它，Tupler你有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
于 2017-02-14T12:51:31.497 回答