Question

Scala库是否支持将给定类型的方法提升为函数值？

例如，假设我想举起String.length。我可以写

val f: String => Int = _.length

或

val f = { s: String => s.length }

但是，这种语法并不总是理想的（特别是在较大的表达式中）。我想我正在寻找能够启用像

这样的表达式的东西

Lift[String](_.length)
Lift[Option[Int]].lift(_.filter)

我记得这样的事情：

class Lift[T] {                                                          
   def apply[R](f: T => R): T => R = f

   def lift[A, R](f: (T) => (A) => R): (T, A) => R = 
         f(_)(_) 
   def lift[A1, A2, R](f: (T) => (A1, A2) => R): (T, A1, A2) => R =
         f(_)(_,_)
   // ... etc. ...
}
object Lift {
   def apply[T] = new Lift[T]
}

问题1： 标准库（或任何库）是否提供类似的内容？

问题2：如果没有，是否可以以Option.filter可以解除上述的方式编写它（而不是{{ 1}}）？如果不在Lift[Option[Int]].lift[Int => Boolean, Option[Int]](_.filter)方法上提供类型参数，我会收到以下错误：

error: missing parameter type for expanded function ((x$1) => x$1.filter)
       Lift[Option[Int]].lift(_.filter)
                              ^

更新：

显然，我遇到的问题与重载的lift方法有关。如果我重命名重载，我可以解除lift而没有所有额外的类型参数。

Answer 1

我终于想出了一个我很满意的解决方案。此版本支持简单的语法和API的单一入口点，同时还提供对提升功能的形式的控制（即，未经处理，部分为咖喱或完全咖喱）。

<强>实施例：

我将在下面的示例中使用以下类定义：

class Foo {
   def m1: Int = 1
   def m2(i: Int): Int = i
   def m3(i: Int, j: Int): Int = i + j
}

最简单的提升形式是将方法作为部分应用函数返回，相当于调用((_: Foo).method _)：

scala> lift[Foo](_.m1)                         // NOTE: trailing _ not required
res0: (Foo) => Int = <function1>

scala> lift[Foo](_.m2 _)                       // NOTE: trailing _ required
res1: (Foo) => (Int) => Int = <function1>

scala> lift[Foo](_.m3 _)
res2: (Foo) => (Int, Int) => Int = <function1> // NOTE: the result is partly curried

通过导入一些含义，可以请求curry或uncurried表单：

scala> {                        
     | import CurriedLiftables._
     | lift[Foo](_.m3 _)        
     | }
res3: (Foo) => (Int) => (Int) => Int = <function1>

scala> {                          
     | import UncurriedLiftables._
     | lift[Foo](_.m3 _)          
     | }
res4: (Foo, Int, Int) => Int = <function3>

实现：

class Lift[T] {
   def apply[R,F](f: T => R)(implicit e: (T => R) Liftable F): F = e.lift(f)
}
object lift {
   def apply[T] = new Lift[T]
}

class Liftable[From, To](val lift: From => To)

class DefaultLiftables {
   implicit def lift[F]: F Liftable F = new Liftable(identity)
}
object Liftable extends DefaultLiftables

class UncurriedLiftable1 extends DefaultLiftables {
   implicit def lift1[T, A, R]: (T => A => R) Liftable ((T, A) => R) = 
      new Liftable( f => f(_)(_) )
}
class UncurriedLiftable2 extends UncurriedLiftable1 {
   implicit def lift2[T, A1, A2, R]: (T => (A1, A2) => R) Liftable ((T, A1, A2) => R) = 
      new Liftable ( f => f(_)(_,_) )
}
// UncurriedLiftable3, UncurriedLiftable4, ...
object UncurriedLiftables extends UncurriedLiftable2

class CurriedLiftable2 extends DefaultLiftables {
   implicit def lift2[T, A1, A2, R]: (T => (A1, A2) => R) Liftable (T => A1 => A2 => R) =
      new Liftable( f => (x: T) => (a1: A1) => (a2: A2) => f(x)(a1, a2) )
}
// CurriedLiftable3, CurriedLiftable4, ...
object CurriedLiftables extends CurriedLiftable2

我之前的解决方案需要为每个arity提供单独的提升方法：

import Lift._
val f1 = lift0[String](_.length)
val f2 = lift1[Option[Int]](_.filter)
val f3 = lift2[Either[String, Int]](_.fold)

实现：

class Lift0[T] {
   def apply[R](f: T => R): T => R = f
}
class Lift1[T] {
   def apply[A, R](f: (T) => (A) => R): (T, A) => R = 
      f(_)(_) 
}
class Lift2[T] {
   def apply[A1, A2, R](f: (T) => (A1, A2) => R): (T, A1, A2) => R =
      f(_)(_,_)
}
// ... etc. ...

object Lift {
   def lift0[T] = new Lift0[T]
   def lift1[T] = new Lift1[T]
   def lift2[T] = new Lift2[T]
   // ... etc. ...
}

Answer 2

有什么问题

(_: String).length
(_: Option[Int]).filter _

Answer 3

传入过滤器作为部分应用的方法似乎可以完成这项工作：

scala> class Lift[T] {                                        
     |    def apply[R](f: T => R): T => R = f
     | }
defined class Lift

scala> object Lift {
     |    def apply[T] = new Lift[T]
     | }
defined module Lift

scala> val ls = Lift[String](_.length)
ls: (String) => Int = <function1>

scala> val los = Lift[Option[Int]](_.filter _)     
los: (Option[Int]) => ((Int) => Boolean) => Option[Int] = <function1>

在Scala中提升函数值的方法

3 个答案: