Scalaz(34)- Free :算法-Interpretation

来源:互联网 时间:2016-03-23

  我们说过自由数据结构(free structures)是表达数据类型的最简单结构。List[A]是个数据结构,它是生成A类型Monoid的最简单结构,因为我们可以用List的状态cons和Nil来分别代表Monoid的append和zero。Free[S,A]是个代表Monad的最简单数据结构,它可以把任何Functor S升格成Monad。Free的两个结构Suspend,Return分别代表了Monad的基本操作函数flatMap,point,我特别强调结构的意思是希望大家能意识到那就是内存heap上的一块空间。我们同样可以简单的把Functor视为一种算法,通过它的map函数实现运算。我们现在可以把Monad的算法flatMap用Suspend[S[Free[S,A]]来表示,那么一段由Functor S(ADT)形成的程序(AST)可以用一串递归结构表达:Suspend(S(Suspend(S(Suspend(S(....(Return)))))))。我们可以把这样的AST看成是一串链接的内存格,每个格内存放着一个算法ADT,代表下一个运算步骤,每个格子指向下一个形成一串连续的算法,组成了一个完整的程序(AST)。最明显的分别是Free把Monad flatMap这种递归算法化解成内存数据结构,用内存地址指向代替了递归算法必须的内存堆栈(stack)。Free的Interpretation就是对存放在数据结构Suspend内的算法(ADT)进行实际运算。不同方式的Interpreter决定了这段由一连串ADT形成的AST的具体效果。

Free Interpreter的具体功能就是按存放在数据结构Suspend内的算法(ADT)进行运算后最终获取A值。这些算法的实际运算可能会产生副作用,比如IO算法的具体操作。scalaz是通过几个运算函数来提供Free Interpreter,包括:fold,foldMap,foldRun,runFC,runM。我们先看看这几个函数的源代码:

 /** Catamorphism. Run the first given function if Return, otherwise, the second given function. */

final def fold[B](r: A => B, s: S[Free[S, A]] => B)(implicit S: Functor[S]): B =

resume.fold(s, r)

/**

* Catamorphism for `Free`.

* Runs to completion, mapping the suspension with the given transformation at each step and

* accumulating into the monad `M`.

*/

final def foldMap[M[_]](f: S ~> M)(implicit S: Functor[S], M: Monad[M]): M[A] =

this.resume match {

case -\/(s) => Monad[M].bind(f(s))(_.foldMap(f))

case \/-(r) => Monad[M].pure(r)

}

/** Runs to completion, allowing the resumption function to thread an arbitrary state of type `B`. */

final def foldRun[B](b: B)(f: (B, S[Free[S, A]]) => (B, Free[S, A]))(implicit S: Functor[S]): (B, A) = {

@tailrec def foldRun2(t: Free[S, A], z: B): (B, A) = t.resume match {

case -\/(s) =>

val (b1, s1) = f(z, s)

foldRun2(s1, b1)

case \/-(r) => (z, r)

}

foldRun2(this, b)

}

/**

* Runs to completion, using a function that maps the resumption from `S` to a monad `M`.

* @since 7.0.1

*/

final def runM[M[_]](f: S[Free[S, A]] => M[Free[S, A]])(implicit S: Functor[S], M: Monad[M]): M[A] = {

def runM2(t: Free[S, A]): M[A] = t.resume match {

case -\/(s) => Monad[M].bind(f(s))(runM2)

case \/-(r) => Monad[M].pure(r)

}

runM2(this)

}

/** Interpret a free monad over a free functor of `S` via natural transformation to monad `M`. */

def runFC[S[_], M[_], A](sa: FreeC[S, A])(interp: S ~> M)(implicit M: Monad[M]): M[A] =

sa.foldMap[M](new (({type λ[α] = Coyoneda[S, α]})#λ ~> M) {

def apply[A](cy: Coyoneda[S, A]): M[A] =

M.map(interp(cy.fi))(cy.k)

})

我们应该可以看出Interpreter的基本原理就是把不可运算的抽象指令ADT转换成可运算的表达式。在这个转换过程中产生运算结果。我们下面用具体例子一个一个介绍这几个函数的用法。还是用上期的例子:

 1 object qz {

2 sealed trait Quiz[+Next]

3 object Quiz {

4 //问题que:String, 等待String 然后转成数字或操作符号

5 case class Question[Next](que: String, n: String => Next) extends Quiz[Next]

6 case class Answer[Next](ans: String, n: Next) extends Quiz[Next]

7 implicit object QFunctor extends Functor[Quiz] {

8 def map[A,B](qa: Quiz[A])(f: A => B): Quiz[B] =

9 qa match {

10 case q: Question[A] => Question(q.que, q.n andThen f)

11 case Answer(a,n) => Answer(a,f(n))

12 }

13 }

14 //操作帮助方法helper methods

15 def askNumber(q: String) = Question(q, (inputString => inputString.toInt)) //_.toInt

16 def askOperator(q: String) = Question(q, (inputString => inputString.head.toUpper.toChar)) //_.head.toUpper.toChar

17 def answer(fnum: Int, snum: Int, opr: Char) = {

18 def result =

19 opr match {

20 case 'A' => fnum + snum

21 case 'M' => fnum * snum

22 case 'D' => fnum / snum

23 case 'S' => fnum - snum

24 }

25 Answer("my answer is: " + result.toString,())

26 }

27 implicit def quizToFree[A](qz: Quiz[A]): Free[Quiz,A] = Free.liftF(qz)

28 }

29 import Quiz._

30 val prg = for {

31 fn <- askNumber("The first number is:")

32 sn <- askNumber("The second number is:")

33 op <- askOperator("The operation is:")

34 _ <- answer(fn,sn,op)

35 } yield()

prg是一段功能描述:在提示后读取一个数字,重复一次,再读取一个字串,把读取的数字和字串用来做个运算。至于怎么提示、如何读取输入、如何运算输入内容,可能会有种种不同的方式,那要看Interpreter具体是怎么做的了。好了,现在我们看看如何用fold来运算prg:fold需要两个入参数:r:A=>B,一个在运算终止Return状态时运行的函数,另一个是s:S[Free[S,A]]=>B,这个函数在Suspend状态时运算入参数ADT:

1 def runQuiz[A](p: Free[Quiz,A]): Unit= p.fold(_ => (), {

2 case Question(q,f) => {

3 println(q)

4 runQuiz(f(readLine))

5 }

6 case Answer(a,n) => println(a)

7 })

注意runQuiz是个递归函数。在Suspend Question状态下,运算f(readLine)产生下一个运算。在这个函数里我们赋予了提示、读取正真的意义,它们都是通过IO操作println,readLine实现的。

1 object main extends App {

2 import freeRun._

3 import qz._

4 runQuiz(prg)

5 }

运行结果:

The first number is:

3

The second number is:

8

The operation is:

mul

my answer is: 24

结果正是我们期待的。但这个fold方法每调用一次只运算一个ADT,所以使用了递归算法连续约化Suspend直到Return。递归算法很容易造成堆栈溢出异常,不安全。下一个试试foldMap。foldMap使用了Monad.bind连续通过高阶类型转换(natural transformation)将ADT转换成运行指令,并在转换过程中实施运算:

 1 object QuizConsole extends (Quiz ~> Id) {

2 import Quiz._

3 def apply[A](qz: Quiz[A]): Id[A] = qz match {

4 case Question(a,f) => {

5 println(a)

6 f(readLine)

7 }

8 case Answer(a,n) => println(a);n

9 }

10 }

11 //运行foldMap

12 prg.foldMap(QuizConsole)

13 //结果一致

上面的natural transformation是把Quiz类型转成Id类型。Id[A]=A,所以高阶类型Quiz可以被转换成基本类型Unit(println返回Unit)。这个例子同样用IO函数来实现AST功能。我们也可以用一个模拟的输入输出方式来测试AST功能,也就是用另一个Interpreter来运算AST,我们可以用Map[String,String]来模拟输入输出环境:

 1 type Tester[A] = Map[String, String] => (List[String], A)

2 object QuizTester extends (Quiz ~> Tester) {

3 def apply[A](qa: Quiz[A]): Tester[A] = qa match {

4 case Question(q,f) => m => (List(),f(m(q)))

5 case Answer(a,n) => m => (List(a),n)

6 }

7 }

8 implicit object testerMonad extends Monad[Tester] {

9 def point[A](a: => A) = _ => (List(),a)

10 def bind[A,B](ta: Tester[A])(f: A => Tester[B]): Tester[B] =

11 m => {

12 val (o1,a) = ta(m)

13 val (o2,b) = f(a)(m)

14 (o1 ++ o2, b)

15 }

16 }

Tester必须是个Monad,所以我们必须提供隐式对象testerMonad。看看运算结果:

1 val m = Map(

2 "The first number is:" -> "8",

3 "The second number is:" -> "3",

4 "The operation is:" -> "Sub"

5 )

6 println(prg.foldMap(QuizTester).apply(m))

7 //(List(my answer is: 5),())

foldRun通过入参数f:(B,S[Free[S,A]])=>(B,Free[S,A])支持状态跟踪,入参数b:B是状态初始值。我们先实现这个f函数:

 1 type FreeQuiz[A] = Free[Quiz,A]

2 def quizst(track: List[String], prg: Quiz[FreeQuiz[Unit]]): (List[String], FreeQuiz[Unit]) =

3 prg match {

4 case Question(q,f) => {

5 println(q)

6 val input = readLine

7 (q+input :: track, f(input))

8 }

9 case Answer(a,n) => println(a); (a :: track, n)

10 }

运行foldRun的结果如下:

println(prg.foldRun(List[String]())(quizst)._1)

The first number is:

2

The second number is:

4

The operation is:

Mul

my answer is: 8

List(my answer is: 8, The operation is:Mul, The second number is:4, The first number is:2)

下一个是runM了,它的入参数就是一个S[_]到M[_]的转换函数:f: S[Free[S,A]]=>M[Free[S,A]]。我们先实现了这个f函数:

1 type FreeQuiz[A] = Free[Quiz,A]

2 def runquiz[A](prg: Quiz[FreeQuiz[A]]): Id[FreeQuiz[A]] =

3 prg match {

4 case Question(q,f) => {

5 println(q)

6 f(readLine)

7 }

8 case Answer(a,n) => println(a); n

9 }

测试运行runM:

prg.runM(run quiz)

The first number is:

4

The second number is:

2

The operation is:

Mul

my answer is: 8

我们曾经介绍过有些F[_]是无法实现map函数的,因此无法成为Functor,如以下ADT:

 1 sealed trait Calc[+A]

2 object Calc {

3 case class Push(value: Int) extends Calc[Unit]

4 case class Add() extends Calc[Unit]

5 case class Mul() extends Calc[Unit]

6 case class Div() extends Calc[Unit]

7 case class Sub() extends Calc[Unit]

8 implicit def calcToFree[A](ca: Calc[A]) = Free.liftFC(ca)

9 }

10 import Calc._

11 val ast = for {

12 _ <- Push(23)

13 _ <- Push(3)

14 _ <- Add()

15 _ <- Push(5)

16 _ <- Mul()

17 } yield () //> ast : scalaz.Free[[x]scalaz.Coyoneda[Exercises.interact.Calc,x],Unit] = Gosub()

从Calc无法获取B类型值,所以无法实现Calc.map,因而Calc无法成为Functor。runFC就是专门为运算Calc这样的非Functor高阶类型值的。runFC需要一个FreeC[S,A]类型入参数:

/** A free monad over the free functor generated by `S` */

type FreeC[S[_], A] = Free[({type f[x] = Coyoneda[S, x]})#f, A]

}

可以得出runFC是专门为Coyoneda设计的。Coyoneda可以替代Calc[A],又是一个Functor,所以可以用Free产生Calc类型的Monad。我们先把Interpreter实现了:

 1 type Stack = List[Int]

2 type StackState[A] = State[Stack,A]

3 object CalcStack extends (Calc ~> StackState) {

4 def apply[A](ca: Calc[A]): StackState[A] = ca match {

5 case Push(v) => State((s: Stack) => (v :: s, ()))

6 case Add() => State((s: Stack) => {

7 val a :: b :: t = s

8 ((a+b) :: t,())

9 })

10 case Mul() => State((s: Stack) => {

11 val a :: b :: t = s

12 ((a * b) :: t, ())

13 })

14 case Div() => State((s: Stack) => {

15 val a :: b :: t = s

16 ((a / b) :: t,())

17 })

18 case Sub() => State((s: Stack) => {

19 val a :: b :: t = s

20 ((a - b) :: s, ())

21 })

22 }

23 }

这个Interpreter用的是Stack内元素操作的运算方式。用runFC对ast运算的结果:

println(Free.runFC(ast)(CalcStack).apply(List[Int]()))

//(List(130),())

以上示范了针对任何抽象的Monadic Programm,我们如何通过各种Interpreter的具体实现方式来确定程序功能的。

 

 

 

 

 

 

 

 

 

 

 

 

 

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