3

我无法更改的函数返回 Scalaz Reader，

type Action[A] = Reader[Session, A]

def findAccount(s: String): Action[Account] = 
  Reader((session: Session) => Account(s))

我想创建一个基于findAccount(...)返回ReaderT[Option, Session, A]的新函数

type ActionT[A] = ReaderT[Option, Session, A]

def findAccountT(s: String): ActionT[Account] = findAccount(s).map(Option(_))

因为最终我想这样做，

def findAccBalT(accountNumber: String) = for {
  acc <- findAccountT(accountNumber)
  bal <- findBalanceT(acc)
}  yield bal

我该如何进行？是否有意义？谢谢

全面披露，

import scalaz._
import Scalaz._        

trait Session {
  def doSomething(): Unit
}

case class Account(number: String) extends AnyVal
case class Amount(value: Int, currency: String) 
case class DBSession() extends Session {
  override def doSomething = println("writing to db")
}

type Action[A] = Reader[Session, A]
type ActionT[A] = ReaderT[Option, Session, A]

def findAccount(s: String): Action[Account] = 
  Reader((session: Session) => Account(s))

def findBalance(account: Account): Action[Amount] = 
  Reader((session: Session) => Amount(333, "S$"))

// failed
def findAccountT(s: String): ActionT[Account] = findAccount(s).map(Option(_))

// failed
def findBalanceT(account: Account): ActionT[Amount] = findBalance(account).map(Option(_))

// failed
def findAccBalT(accountNumber: String) = for {
  acc <- findAccountT(accountNumber)
  bal <- findBalanceT(acc)
}  yield bal
4

1 回答 1

2

简短回答：您可以使用mapK.

Reader[A]是 的类型别名ReaderT[Id, A]并且ReaderT是 的别名Kleisli。Id[A]是一样的A。

在KleisliScalaDoc 中，我们发现mapK：

def mapK[N[_], C](f: (M[B]) => N[C]): Kleisli[N, A, C]

因为我们知道这Reader[Session, A]与Kleisli[Id, Session, A]我们可以使用mapKgo to相同Kleisli[Option, Session, A]：

import scalaz._, Scalaz._

type Session = String
type Action[A] = Reader[Session, A]
type ActionT[A] = ReaderT[Option, Session, A]

val action: Action[String] = Reader(s => s)
val actionT: ActionT[String] = action mapK Option.apply
于 2016-06-05T13:22:50.840 回答