泛函编程(29)-泛函实用结构:Trampoline-不再怕StackOverflow

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   泛函编程方式其中一个特点就是普遍地使用递归算法,而且有些地方还无法避免使用递归算法。比如说flatMap就是一种推进式的递归算法,没了它就无法使用for-comprehension,那么泛函编程也就无法被称为Monadic Programming了。虽然递归算法能使代码更简洁易明,但同时又以占用堆栈(stack)方式运作。堆栈是软件程序有限资源,所以在使用递归算法对大型数据源进行运算时系统往往会出现StackOverflow错误。如果不想办法解决递归算法带来的StackOverflow问题,泛函编程模式也就失去了实际应用的意义了。

针对StackOverflow问题,Scala compiler能够对某些特别的递归算法模式进行优化:把递归算法转换成while语句运算,但只限于尾递归模式(TCE, Tail Call Elimination),我们先用例子来了解一下TCE吧:

以下是一个右折叠算法例子:

1 def foldR[A,B](as: List[A], b: B, f: (A,B) => B): B = as match {

2 case Nil => b

3 case h :: t => f(h,foldR(t,b,f))

4 } //> foldR: [A, B](as: List[A], b: B, f: (A, B) => B)B

5 def add(a: Int, b: Int) = a + b //> add: (a: Int, b: Int)Int

6

7 foldR((1 to 100).toList, 0, add) //> res0: Int = 5050

8 foldR((1 to 10000).toList, 0, add) //> java.lang.StackOverflowError

以上的右折叠算法中自引用部分不在最尾部,Scala compiler无法进行TCE,所以处理一个10000元素的List就发生了StackOverflow。

再看看左折叠:

1 def foldL[A,B](as: List[A], b: B, f: (B,A) => B): B = as match {

2 case Nil => b

3 case h :: t => foldL(t,f(b,h),f)

4 } //> foldL: [A, B](as: List[A], b: B, f: (B, A) => B)B

5 foldL((1 to 100000).toList, 0, add) //> res1: Int = 705082704

在这个左折叠例子里自引用foldL出现在尾部位置,Scala compiler可以用TCE来进行while转换:

 1  def foldl2[A,B](as: List[A], b: B,

2 f: (B,A) => B): B = {

3 var z = b

4 var az = as

5 while (true) {

6 az match {

7 case Nil => return z

8 case x :: xs => {

9 z = f(z, x)

10 az = xs

11 }

12 }

13 }

14 z

15 }

经过转换后递归变成Jump,程序不再使用堆栈,所以不会出现StackOverflow。

但在实际编程中,统统把递归算法编写成尾递归是不现实的。有些复杂些的算法是无法用尾递归方式来实现的,加上JVM实现TCE的能力有局限性,只能对本地(Local)尾递归进行优化。

我们先看个稍微复杂点的例子:

 

1 def even[A](as: List[A]): Boolean = as match {

2 case Nil => true

3 case h :: t => odd(t)

4 } //> even: [A](as: List[A])Boolean

5 def odd[A](as: List[A]): Boolean = as match {

6 case Nil => false

7 case h :: t => even(t)

8 } //> odd: [A](as: List[A])Boolean

 

在上面的例子里even和odd分别为跨函数的各自的尾递归,但Scala compiler无法进行TCE处理,因为JVM不支持跨函数Jump:

 

1 even((1 to 100).toList) //> res2: Boolean = true

2 even((1 to 101).toList) //> res3: Boolean = false

3 odd((1 to 100).toList) //> res4: Boolean = false

4 odd((1 to 101).toList) //> res5: Boolean = true

5 even((1 to 10000).toList) //> java.lang.StackOverflowError

 

处理10000个元素的List还是出现了StackOverflowError

我们可以通过设计一种数据结构实现以heap交换stack。Trampoline正是专门为解决StackOverflow问题而设计的数据结构:

 

1 trait Trampoline[+A] {

2 final def runT: A = this match {

3 case Done(a) => a

4 case More(k) => k().runT

5 }

6 }

7 case class Done[+A](a: A) extends Trampoline[A]

8 case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]

 

Trampoline代表一个可以一步步进行的运算。每步运算都有两种可能:Done(a),直接完成运算并返回结果a,或者More(k)运算k后进入下一步运算;下一步又有可能存在Done和More两种情况。注意Trampoline的runT方法是明显的尾递归,而且runT有final标示,表示Scala可以进行TCE。

有了Trampoline我们可以把even,odd的函数类型换成Trampoline:

1 def even[A](as: List[A]): Trampoline[Boolean] = as match {

2 case Nil => Done(true)

3 case h :: t => More(() => odd(t))

4 } //> even: [A](as: List[A])ch13.ex1.Trampoline[Boolean]

5 def odd[A](as: List[A]): Trampoline[Boolean] = as match {

6 case Nil => Done(false)

7 case h :: t => More(() => even(t))

8 } //> odd: [A](as: List[A])ch13.ex1.Trampoline[Boolean]

我们可以用Trampoline的runT来运算结果:

 

1 even((1 to 10000).toList).runT //> res6: Boolean = true

2 even((1 to 10001).toList).runT //> res7: Boolean = false

3 odd((1 to 10000).toList).runT //> res8: Boolean = false

4 odd((1 to 10001).toList).runT //> res9: Boolean = true

 

这次我们不但得到了正确结果而且也没有发生StackOverflow错误。就这么简单?

我们再从一个比较实际复杂一点的例子分析。在这个例子中我们遍历一个List并维持一个状态。我们首先需要State类型:

 

 1 case class State[S,+A](runS: S => (A,S)) {

2 import State._

3 def flatMap[B](f: A => State[S,B]): State[S,B] = State[S,B] {

4 s => {

5 val (a1,s1) = runS(s)

6 f(a1) runS s1

7 }

8 }

9 def map[B](f: A => B): State[S,B] = flatMap( a => unit(f(a)))

10 }

11 object State {

12 def unit[S,A](a: A) = State[S,A] { s => (a,s) }

13 def getState[S]: State[S,S] = State[S,S] { s => (s,s) }

14 def setState[S](s: S): State[S,Unit] = State[S,Unit] { _ => ((),s)}

15 }

 

再用State类型来写一个对List元素进行序号标注的函数:

 1 def zip[A](as: List[A]): List[(A,Int)] = {

2 as.foldLeft(

3 unit[Int,List[(A,Int)]](List()))(

4 (acc,a) => for {

5 xs <- acc

6 n <- getState[Int]

7 _ <- setState[Int](n + 1)

8 } yield (a,n) :: xs

9 ).runS(0)._1.reverse

10 } //> zip: [A](as: List[A])List[(A, Int)]

运行一下这个zip函数:

 

1 zip((1 to 10).toList) //> res0: List[(Int, Int)] = List((1,0), (2,1), (3,2), (4,3), (5,4), (6,5), (7,6

2 //| ), (8,7), (9,8), (10,9))

 

结果正确。如果针对大型的List呢?

 

1 zip((1 to 10000).toList) //> java.lang.StackOverflowError

 

按理来说foldLeft是尾递归的,怎么StackOverflow出现了。这是因为State组件flatMap是一种递归算法,也会导致StackOverflow。那么我们该如何改善呢?我们是不是像上面那样把State转换动作的结果类型改成Trampoline就行了呢?

 

 1 case class State[S,A](runS: S => Trampoline[(A,S)]) {

2 def flatMap[B](f: A => State[S,B]): State[S,B] = State[S,B] {

3 s => More(() => {

4 val (a1,s1) = runS(s).runT

5 More(() => f(a1) runS s1)

6 })

7 }

8 def map[B](f: A => B): State[S,B] = flatMap( a => unit(f(a)))

9 }

10 object State {

11 def unit[S,A](a: A) = State[S,A] { s => Done((a,s)) }

12 def getState[S]: State[S,S] = State[S,S] { s => Done((s,s)) }

13 def setState[S](s: S): State[S,Unit] = State[S,Unit] { _ => Done(((),s))}

14 }

15 trait Trampoline[+A] {

16 final def runT: A = this match {

17 case Done(a) => a

18 case More(k) => k().runT

19 }

20 }

21 case class Done[+A](a: A) extends Trampoline[A]

22 case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]

23

24 def zip[A](as: List[A]): List[(A,Int)] = {

25 as.foldLeft(

26 unit[Int,List[(A,Int)]](List()))(

27 (acc,a) => for {

28 xs <- acc

29 n <- getState[Int]

30 _ <- setState[Int](n + 1)

31 } yield (a,n) :: xs

32 ).runS(0).runT._1.reverse

33 } //> zip: [A](as: List[A])List[(A, Int)]

34 zip((1 to 10).toList) //> res0: List[(Int, Int)] = List((1,0), (2,1), (3,2), (4,3), (5,4), (6,5), (7,

35 //| 6), (8,7), (9,8), (10,9))

 

在这个例子里我们把状态转换函数 S => (A,S) 变成 S => Trampoline[(A,S)]。然后把其它相关函数类型做了相应调整。运行zip再检查结果:结果正确。那么再试试大型List:

1 zip((1 to 10000).toList) //> java.lang.StackOverflowError

还是会出现StackOverflow。这次是因为flatMap中的runT不在尾递归位置。那我们把Trampoline变成Monad看看如何?那我们就得为Trampoline增加一个flatMap函数:

 

 1 trait Trampoline[+A] {

2 final def runT: A = this match {

3 case Done(a) => a

4 case More(k) => k().runT

5 }

6 def flatMap[B](f: A => Trampoline[B]): Trampoline[B] = {

7 this match {

8 case Done(a) => f(a)

9 case More(k) => f(runT)

10 }

11 }

12 }

13 case class Done[+A](a: A) extends Trampoline[A]

14 case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]

这样我们可以把State.flatMap调整成以下这样:

 1 case class State[S,A](runS: S => Trampoline[(A,S)]) {

2 def flatMap[B](f: A => State[S,B]): State[S,B] = State[S,B] {

3 s => More(() => {

4 // val (a1,s1) = runS(s).runT

5 // More(() => f(a1) runS s1)

6 runS(s) flatMap { // runS(s) >>> Trampoline

7 case (a1,s1) => More(() => f(a1) runS s1)

8 }

9 })

10 }

11 def map[B](f: A => B): State[S,B] = flatMap( a => unit(f(a)))

12 }

现在我们把递归算法都推到了Trampoline.flatMap这儿了。不过Trampoline.flatMap的runT引用f(runT)不在尾递归位置,所以这样调整还不足够。看来核心还是要解决flatMap尾递归问题。我们可以再为Trampoline增加一个状态结构FlatMap然后把flatMap函数引用变成类型实例构建(type construction):

1 case class Done[+A](a: A) extends Trampoline[A]

2 case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]

3 case class FlatMap[A,B](sub: Trampoline[A], k: A => Trampoline[B]) extends Trampoline[B]

case class FlatMap这种Trampoline状态意思是先引用sub然后把结果传递到下一步k再运行k:基本上是沿袭flatMap功能。再调整Trampoline.resume, Trampoline.flatMap把FlatMap这种状态考虑进去:

 1 trait Trampoline[+A] {

2 final def runT: A = resume match {

3 case Right(a) => a

4 case Left(k) => k().runT

5 }

6 def flatMap[B](f: A => Trampoline[B]): Trampoline[B] = {

7 this match {

8 // case Done(a) => f(a)

9 // case More(k) => f(runT)

10 case FlatMap(a,g) => FlatMap(a, (x: Any) => g(x) flatMap f)

11 case x => FlatMap(x, f)

12 }

13 }

14 def map[B](f: A => B) = flatMap(a => Done(f(a)))

15 def resume: Either[() => Trampoline[A], A] = this match {

16 case Done(a) => Right(a)

17 case More(k) => Left(k)

18 case FlatMap(a,f) => a match {

19 case Done(v) => f(v).resume

20 case More(k) => Left(() => k() flatMap f)

21 case FlatMap(b,g) => FlatMap(b, (x: Any) => g(x) flatMap f).resume

22 }

23 }

24 }

25 case class Done[+A](a: A) extends Trampoline[A]

26 case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]

27 case class FlatMap[A,B](sub: Trampoline[A], k: A => Trampoline[B]) extends Trampoline[B]

在以上对Trampoline的调整里我们引用了Monad的结合特性(associativity):

FlatMap(FlatMap(b,g),f) == FlatMap(b,x => FlatMap(g(x),f)

重新右结合后我们可以用FlatMap正确表达复数步骤的运算了。

现在再试着运行zip:

 1 def zip[A](as: List[A]): List[(A,Int)] = {

2 as.foldLeft(

3 unit[Int,List[(A,Int)]](List()))(

4 (acc,a) => for {

5 xs <- acc

6 n <- getState[Int]

7 _ <- setState[Int](n + 1)

8 } yield (a,n) :: xs

9 ).runS(0).runT._1.reverse

10 } //> zip: [A](as: List[A])List[(A, Int)]

11 zip((1 to 10000).toList) //> res0: List[(Int, Int)] = List((1,0), (2,1), (3,2), (4,3), (5,4), (6,5), (7,

这次运行正常,再不出现StackOverflowError了。

实际上我们可以考虑把Trampoline当作一种通用的堆栈溢出解决方案。

我们首先可以利用Trampoline的Monad特性来调控函数引用,如下:

1 val x = f()

2 val y = g(x)

3 h(y)

4 //以上这三步函数引用可以写成:

5 for {

6 x <- f()

7 y <- g(x)

8 z <- h(y)

9 } yield z

举个实际例子:

 1 implicit def step[A](a: => A): Trampoline[A] = {

2 More(() => Done(a))

3 } //> step: [A](a: => A)ch13.ex1.Trampoline[A]

4 def getNum: Double = 3 //> getNum: => Double

5 def addOne(x: Double) = x + 1 //> addOne: (x: Double)Double

6 def timesTwo(x: Double) = x * 2 //> timesTwo: (x: Double)Double

7 (for {

8 x <- getNum

9 y <- addOne(x)

10 z <- timesTwo(y)

11 } yield z).runT //> res6: Double = 8.0

又或者:

1 def fib(n: Int): Trampoline[Int] = {

2 if (n <= 1) Done(n) else for {

3 x <- More(() => fib(n-1))

4 y <- More(() => fib(n-2))

5 } yield x + y

6 } //> fib: (n: Int)ch13.ex1.Trampoline[Int]

7 (fib(10)).runT //> res7: Int = 55

从上面得出我们可以用flatMap来对Trampoline运算进行流程控制。另外我们还可以通过把多个Trampoline运算交叉组合来实现并行运算:

 1 trait Trampoline[+A] {

2 final def runT: A = resume match {

3 case Right(a) => a

4 case Left(k) => k().runT

5 }

6 def flatMap[B](f: A => Trampoline[B]): Trampoline[B] = {

7 this match {

8 // case Done(a) => f(a)

9 // case More(k) => f(runT)

10 case FlatMap(a,g) => FlatMap(a, (x: Any) => g(x) flatMap f)

11 case x => FlatMap(x, f)

12 }

13 }

14 def map[B](f: A => B) = flatMap(a => Done(f(a)))

15 def resume: Either[() => Trampoline[A], A] = this match {

16 case Done(a) => Right(a)

17 case More(k) => Left(k)

18 case FlatMap(a,f) => a match {

19 case Done(v) => f(v).resume

20 case More(k) => Left(() => k() flatMap f)

21 case FlatMap(b,g) => FlatMap(b, (x: Any) => g(x) flatMap f).resume

22 }

23 }

24 def zip[B](tb: Trampoline[B]): Trampoline[(A,B)] = {

25 (this.resume, tb.resume) match {

26 case (Right(a),Right(b)) => Done((a,b))

27 case (Left(f),Left(g)) => More(() => f() zip g())

28 case (Right(a),Left(k)) => More(() => Done(a) zip k())

29 case (Left(k),Right(a)) => More(() => k() zip Done(a))

30 }

31 }

32 }

33 case class Done[+A](a: A) extends Trampoline[A]

34 case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]

35 case class FlatMap[A,B](sub: Trampoline[A], k: A => Trampoline[B]) extends Trampoline[B]

我们可以用这个zip函数把几个Trampoline运算交叉组合起来实现并行运算:

1 def hello: Trampoline[Unit] = for {

2 _ <- print("Hello ")

3 _ <- println("World!")

4 } yield () //> hello: => ch13.ex1.Trampoline[Unit]

5

6 (hello zip hello zip hello).runT //> Hello Hello Hello World!

7 //| World!

8 //| World!

9 //| res8: ((Unit, Unit), Unit) = (((),()),())

用Trampoline可以解决StackOverflow这个大问题。现在我们可以放心地进行泛函编程了。

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


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