Lac*_*lan 8 scala type-inference scalaz
我正在尝试traverse_在下面的代码中改进函数的类型推断:
import scala.language.higherKinds
trait Applicative[AF[_]] {
def ap[A, B](a: AF[A])(f: AF[A => B]): AF[B]
def pure[A](a: A): AF[A]
def fmap[A, B](a: AF[A])(f: A => B): AF[B]
}
def traverse_[AP[_]: Applicative, A](xs: Iterable[A])(f: A => AP[Unit]): AP[Unit] = {
val ap = implicitly[Applicative[AP]]
(xs :\ ap.pure(())) { (x, acc) =>
val apFunc = ap.fmap(f(x))(a => identity[Unit] _)
ap.ap(acc)(apFunc)
}
}
implicit def optionAp = new Applicative[Option] {
def ap[A, B](a: Option[A])(f: Option[A => B]): Option[B] = f flatMap (a map _)
def pure[A](a: A) = Some(a)
def fmap[A, B](a: Option[A])(f: A => B) = a map f
}
implicit def eitherAp[L] = new Applicative[({type l[x]=Either[L, x]})#l] {
def ap[A, B](a: Either[L, A])(f: Either[L, A => B]): Either[L, B] = f.right flatMap (a.right map _)
def pure[A](a: A) = Right(a)
def fmap[A, B](a: Either[L, A])(f: A => B) = a.right map f
}
// silly, but compiles
val x = traverse_(1 to 10) {
case 5 => None
case _ => Some(())
}
println(x)
// also silly, but does not compile
val y = traverse_(1 to 10) {
case 5 => Left("x")
case _ => Right(())
}
println(y)
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运行上面给出:
/Users/lodea/tmp/traverse.scala:49: error: no type parameters for method traverse_: (f: Int => AP[Unit])(implicit evidence$1: this.Applicative[AP])AP[Unit] exist so that it can be applied to arguments (Int => Product with Serializable with scala.util.Either[String,Unit])
--- because ---
argument expression's type is not compatible with formal parameter type;
found : Int => Product with Serializable with scala.util.Either[String,Unit]
required: Int => ?AP
val y = traverse_(1 to 10) {
^
/Users/lodea/tmp/traverse.scala:49: error: type mismatch;
found : Int => Product with Serializable with scala.util.Either[String,Unit]
required: Int => AP[Unit]
val y = traverse_(1 to 10) {
^
two errors found
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为了让它编译,我必须指定类型参数traverse_:
val y = traverse_[({type l[x]=Either[String, x]})#l, Int](1 to 10) {
case 5 => Left("x")
case _ => Right(())
}
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有没有一种方法可以重构traverse_或代码的任何其他部分,以使类型推断工作?当类型开始变得越来越复杂时,这会很快变得烦人.
正如Ben James所指出的那样,你正在寻找Miles Sabin的Unapply伎俩.这是scalaz回购.这是 traverseU在其帮助下实施的.以下是一些示例用法.这是我对你的特定情况的粗略(希望是正确的)实现(注意:我已经重命名Applicative为ApplicativeTest不要干扰Applicative,在scalaz中定义):
scalaz> core/console
[warn] Credentials file /home/folone/.ivy2/.credentials does not exist
[info] Starting scala interpreter...
[info]
Welcome to Scala version 2.9.2 (OpenJDK 64-Bit Server VM, Java 1.7.0_15).
Type in expressions to have them evaluated.
Type :help for more information.
scala> :paste
// Entering paste mode (ctrl-D to finish)
import scalaz._
trait ApplicativeTest[AF[_]] {
def ap[A, B](a: AF[A])(f: AF[A => B]): AF[B]
def pure[A](a: A): AF[A]
def fmap[A, B](a: AF[A])(f: A => B): AF[B]
}
def traverse_[AP, A](xs: Iterable[A])(f: A => AP)(implicit G: Unapply[ApplicativeTest, AP]): G.M[Unit] = {
(xs :\ G.TC.pure(())) { (x, acc) =>
val apFunc = G.TC.fmap(G(f(x)))(a => identity[Unit] _)
G.TC.ap(acc)(apFunc)
}
}
implicit def optionAp = new ApplicativeTest[Option] {
def ap[A, B](a: Option[A])(f: Option[A => B]): Option[B] = f flatMap (a map _)
def pure[A](a: A) = Some(a)
def fmap[A, B](a: Option[A])(f: A => B) = a map f
}
implicit def eitherAp[L]: ApplicativeTest[({type l[x]=Either[L, x]})#l] =
new ApplicativeTest[({type l[x]=Either[L, x]})#l] {
def ap[A, B](a: Either[L, A])(f: Either[L, A => B]): Either[L, B] = f.right flatMap (a.right map _)
def pure[A](a: A) = Right(a)
def fmap[A, B](a: Either[L, A])(f: A => B) = a.right map f
}
implicit def iterAp = new ApplicativeTest[Iterable] {
def ap[A, B](a: Iterable[A])(f: Iterable[A ? B]): Iterable[B] = f flatMap(a map _)
def pure[A](a: A) = Iterable(a)
def fmap[A, B](a: Iterable[A])(f: A ? B) = a map f
}
// Exiting paste mode, now interpreting.
import scalaz._
defined trait ApplicativeTest
traverse_: [AP, A](xs: Iterable[A])(f: A => AP)(implicit G: scalaz.Unapply[ApplicativeTest,AP])G.M[Unit]
optionAp: java.lang.Object with ApplicativeTest[Option]{def pure[A](a: A): Some[A]}
eitherAp: [L]=> ApplicativeTest[[x]Either[L,x]]
iterAp: java.lang.Object with ApplicativeTest[Iterable]
scala> val x = traverse_(1 to 10) {
| case 5 => None
| case _ => Some(())
| }
x: Option[Unit] = None
scala> val y = traverse_(1 to 10) {
| case 5 => Left("x"): Either[String, Unit]
| case _ => Right(())
| }
y: Either[String,Unit] = Left(x)
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我仍然不知道如何推断它Either[String, Unit]而Product with Serializable with scala.util.Either[String,Unit]不是严格指定类型,就像我在这一行中所做的那样:case 5 => Left("x"): Either[String, Unit].