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。
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/12914.html