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Monad Reader就是一种函数的组合。在scalaz里函数（function）本身就是Monad，自然也就是Functor和applicative。我们可以用Monadic方法进行函数组合：
import scalaz._import Scalaz._object decompose {//两个测试函数val f = (_: Int) + 3 //> f : Int => Int = val g = (_: Int) * 5 //> g : Int => Int = //functorval h = f map g // f andThen g //> h : Int => Int = val h1 = g map f // f compose g //> h1 : Int => Int = h(2) //g(f(2)) //> res0: Int = 25h1(2) //f(g(2)) //> res1: Int = 13//applicativeval k = (f |@| g){_ + _} //> k : Int => Int = k(10) // f(10)+g(10) //> res2: Int = 63//monadval m = g.flatMap{a => f.map(b => a+b)} //> m : Int => Int = val n = for { a <- f b <- g} yield a + b //> n : Int => Int = m(10) //> res3: Int = 63n(10) //> res4: Int = 63} 以上的函数f,g必须满足一定的条件才能实现组合。这个从f(g(2))或g(f(2))可以看出：必需固定有一个输入参数及输入参数类型和函数结果类型必需一致，因为一个函数的输出成为另一个函数的输入。在FP里这样的函数组合就是Monadic Reader。
但是FP里函数运算结果一般都是M[R]这样格式的，所以我们需要对f:A => M[B],g:B => M[C]这样的函数进行组合。这就是scalaz里的Kleisli了。Kleisli就是函数A=>M[B]的类封套，从Kleisli的类定义可以看出：scalaz/Kleisli.scala
1 final case class Kleisli[M[_], A, B](run: A => M[B]) { self =>2 ...3 trait KleisliFunctions {4 /**Construct a Kleisli from a Function1 */5 def kleisli[M[_], A, B](f: A => M[B]): Kleisli[M, A, B] = Kleisli(f)6 ... Kleisli的目的是把Monadic函数组合起来或者更形象说连接起来。Kleisli提供的操作方法如>=>可以这样理解：
(A=>M[B]) >=> (B=>M[C]) >=> (C=>M[D]) 最终运算结果M[D]
可以看出Kleisli函数组合有着固定的模式：
1、函数必需是 A => M[B]这种模式；只有一个输入，结果是一个Monad M[_]
2、上一个函数输出M[B]，他的运算值B就是下一个函数的输入。这就要求下一个函数的输入参数类型必需是B
3、M必须是个Monad；这个可以从Kleisli的操作函数实现中看出：scalaz/Kleisli.scala
/** alias for `andThen` */ def >=>[C](k: Kleisli[M, B, C])(implicit b: Bind[M]): Kleisli[M, A, C] = kleisli((a: A) => b.bind(this(a))(k.run)) def andThen[C](k: Kleisli[M, B, C])(implicit b: Bind[M]): Kleisli[M, A, C] = this >=> k def >==>[C](k: B => M[C])(implicit b: Bind[M]): Kleisli[M, A, C] = this >=> kleisli(k) def andThenK[C](k: B => M[C])(implicit b: Bind[M]): Kleisli[M, A, C] = this >==> k /** alias for `compose` */ def <=<[C](k: Kleisli[M, C, A])(implicit b: Bind[M]): Kleisli[M, C, B] = k >=> this def compose[C](k: Kleisli[M, C, A])(implicit b: Bind[M]): Kleisli[M, C, B] = k >=> this def <==<[C](k: C => M[A])(implicit b: Bind[M]): Kleisli[M, C, B] = kleisli(k) >=> this def composeK[C](k: C => M[A])(implicit b: Bind[M]): Kleisli[M, C, B] = this <==< k
拿操作函数>=>（andThen）举例：implicit b: Bind[M]明确了M必须是个Monad。
kleisli((a: A) => b.bind(this(a))(k.run))的意思是先运算M[A]，接着再运算k，以M[A]运算结果值a作为下一个函数k.run的输入参数。整个实现过程并不复杂。
实际上Reader就是Kleisli的一个特殊案例：在这里kleisli的M[]变成了Id[]，因为Id[A]=A >>> A=>Id[B] = A=>B，就是我们上面提到的Reader，我们看看Reader在scalaz里是如何定义的：scalar/package.scala
type ReaderT[F[_], E, A] = Kleisli[F, E, A] val ReaderT = Kleisli type =?>[E, A] = Kleisli[Option, E, A] type Reader[E, A] = ReaderT[Id, E, A] type Writer[W, A] = WriterT[Id, W, A] type Unwriter[W, A] = UnwriterT[Id, W, A] object Reader { def apply[E, A](f: E => A): Reader[E, A] = Kleisli[Id, E, A](f) } object Writer { def apply[W, A](w: W, a: A): WriterT[Id, W, A] = WriterT[Id, W, A]((w, a)) } object Unwriter { def apply[U, A](u: U, a: A): UnwriterT[Id, U, A] = UnwriterT[Id, U, A]((u, a)) } type ReaderT[F[_], E, A] = Kleisli[F, E, A] >>> type Reader[E,A] = ReaderT[Id,E,A]
好了，说了半天还是回到如何使用Kleisli进行函数组合的吧：
//Kleisli款式函数kf,kgval kf: Int => Option[String] = (i: Int) => Some((i + 3).shows) //> kf : Int => Option[String] =val kg: String => Option[Boolean] = { case "3" => true.some; case _ => false.some } //> kg : String => Option[Boolean] = //Kleisli函数组合操作import Kleisli._val kfg = kleisli(kf) >=> kleisli(kg) //> kfg : scalaz.Kleisli[Option,Int,Boolean] = Kleisli( )kfg(1) //> res5: Option[Boolean] = Some(false)kfg(0) //> res6: Option[Boolean] = Some(true)
例子虽然很简单，但它说明了很多重点：上一个函数输入的运算值是下一个函数的输入值 Int=>String=>Boolean。输出Monad一致统一，都是Option。
那么，Kleisli到底用来干什么呢？它恰恰显示了FP函数组合的真正意义：把功能尽量细分化，通过各种方式的函数组合实现灵活的函数重复利用。也就是在FP领域里，我们用Kleisli来组合FP函数。
下面我们就用scalaz自带的例子scalaz.example里的KleisliUsage.scala来说明一下Kleisli的具体使用方法吧：
下面是一组地理信息结构：
1 // just some trivial data structure ,2 // Continents contain countries. Countries contain cities.3 case class Continent(name: String, countries: List[Country] = List.empty)4 case class Country(name: String, cities: List[City] = List.empty)5 case class City(name: String, isCapital: Boolean = false, inhabitants: Int = 20)
分别是：洲（Continent）、国家（Country）、城市（City）。它们之间的关系是层级的：Continent(Country(City))
下面是一组模拟数据：
val data: List[Continent] = List( Continent("Europe"), Continent("America", List( Country("USA", List( City("Washington"), City("New York"))))), Continent("Asia", List( Country("India", List(City("New Dehli"), City("Calcutta")))))) 从上面的模拟数据也可以看出Continent,Country,City之间的隶属关系。我们下面设计三个函数分别对Continent,Country,City进行查找：
def continents(name: String): List[Continent] = data.filter(k => k.name.contains(name)) //> continents: (name: String)List[Exercises.kli.Continent] //查找名字包含A的continent continents("A") //> res7: List[Exercises.kli.Continent] = List(Continent(America,List(Country(U //| SA,List(City(Washington,false,20), City(New York,false,20))))), Continent(A //| sia,List(Country(India,List(City(New Dehli,false,20), City(Calcutta,false,2 //| 0)))))) //找到两个：List(America,Asia) def countries(continent: Continent): List[Country] = continent.countries //> countries: (continent: Exercises.kli.Continent)List[Exercises.kli.Country] //查找America下的国家 val america = Continent("America", List( Country("USA", List( City("Washington"), City("New York"))))) //> america : Exercises.kli.Continent = Continent(America,List(Country(USA,Lis //| t(City(Washington,false,20), City(New York,false,20))))) countries(america) //> res8: List[Exercises.kli.Country] = List(Country(USA,List(City(Washington,f //| alse,20), City(New York,false,20)))) def cities(country: Country): List[City] = country.cities //> cities: (country: Exercises.kli.Country)List[Exercises.kli.City] val usa = Country("USA", List( City("Washington"), City("New York"))) //> usa : Exercises.kli.Country = Country(USA,List(City(Washington,false,20), //| City(New York,false,20))) cities(usa) //> res9: List[Exercises.kli.City] = List(City(Washington,false,20), City(New Y //| ork,false,20)) 从continents,countries,cities这三个函数运算结果可以看出它们都可以独立运算。这三个函数的款式如下：
String => List[Continent]
Continent => List[Country]
Country => List[City]
无论函数款式或者类封套（List本来就是Monad）都适合Kleisli。我们可以用Kleisli把这三个局部函数用各种方法组合起来实现更广泛功能：
val allCountry = kleisli(continents) >==> countries //> allCountry : scalaz.Kleisli[List,String,Exercises.kli.Country] = Kleisli(< //| function1>) val allCity = kleisli(continents) >==> countries >==> cities //> allCity : scalaz.Kleisli[List,String,Exercises.kli.City] = Kleisli(还有个=<<符号挺有意思：) allCountry("Amer") //> res10: List[Exercises.kli.Country] = List(Country(USA,List(City(Washington, //| false,20), City(New York,false,20)))) allCity("Amer") //> res11: List[Exercises.kli.City] = List(City(Washington,false,20), City(New //| York,false,20))
1 def =<<(a: M[A])(implicit m: Bind[M]): M[B] = m.bind(a)(run)意思是用包嵌的函数flatMap一下输入参数M[A]：
1 allCity =<< List("Amer","Asia") //> res12: List[Exercises.kli.City] = List(City(Washington,false,20), City(New 2 //| York,false,20), City(New Dehli,false,20), City(Calcutta,false,20)) 那么如果我想避免使用List()，用Option［List]作为函数输出可以吗？Option是个Monad，第一步可以通过。下一步是把函数款式对齐了：
List[String] => Option[List[Continent]]
List[Continent] => Option[List[Country]]
List[Country] => Option[List[City]]
下面是这三个函数的升级版：
//查找Continent List[String] => Option[List[Continent]] def maybeContinents(names: List[String]): Option[List[Continent]] = names.flatMap(name => data.filter(k => k.name.contains(name))) match { case h :: t => (h :: t).some case _ => none } //> maybeContinents: (names: List[String])Option[List[Exercises.kli.Continent]] //| //测试运行 maybeContinents(List("Amer","Asia")) //> res13: Option[List[Exercises.kli.Continent]] = Some(List(Continent(America, //| List(Country(USA,List(City(Washington,false,20), City(New York,false,20)))) //| ), Continent(Asia,List(Country(India,List(City(New Dehli,false,20), City(Ca //| lcutta,false,20))))))) //查找Country List[Continent] => Option[List[Country]] def maybeCountries(continents: List[Continent]): Option[List[Country]] = continents.flatMap(continent => continent.countries.map(c => c)) match { case h :: t => (h :: t).some case _ => none } //> maybeCountries: (continents: List[Exercises.kli.Continent])Option[List[Exer //| cises.kli.Country]] //查找City List[Country] => Option[List[Country]] def maybeCities(countries: List[Country]): Option[List[City]] = countries.flatMap(country => country.cities.map(c => c)) match { case h :: t => (h :: t).some case _ => none } //> maybeCities: (countries: List[Exercises.kli.Country])Option[List[Exercises. //| kli.City]] val maybeAllCities = kleisli(maybeContinents) >==> maybeCountries >==> maybeCities //> maybeAllCities : scalaz.Kleisli[Option,List[String],List[Exercises.kli.Cit //| y]] = Kleisli( ) maybeAllCities(List("Amer","Asia")) //> res14: Option[List[Exercises.kli.City]] = Some(List(City(Washington,false,2 //| 0), City(New York,false,20), City(New Dehli,false,20), City(Calcutta,false, //| 20))) 我们看到，只要Monad一致，函数输入输出类型匹配，就能用Kleisli来实现函数组合。 转载地址：http://uokdo.baihongyu.com/