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:
def foldR[A,B](as: List[A], b: B, f: (A,B) => B): B = as match {case Nil => bcase h :: t => f(h,foldR(t,b,f))} //> foldR: [A, B](as: List[A], b: B, f: (A, B) => B)Bdef add(a: Int, b: Int) = a + b //> add: (a: Int, b: Int)IntfoldR((1 to 100).toList, 0, add) //> res0: Int = 5050foldR((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:
def foldL[A,B](as: List[A], b: B, f: (B,A) => B): B = as match {case Nil => bcase h :: t => foldL(t,f(b,h),f)} //> foldL: [A, B](as: List[A], b: B, f: (B, A) => B)BfoldL((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:
def foldl2[A,B](as: List[A], b: B, f: (B,A) => B): B = { var z = b var az = as while (true) { az match { case Nil => return z case x :: xs => { z = f(z, x) az = xs } } } z }
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:
def even[A](as: List[A]): Boolean = as match {case Nil => truecase h :: t => odd(t)} //> even: [A](as: List[A])Booleandef odd[A](as: List[A]): Boolean = as match {case Nil => falsecase h :: t => even(t)} //> 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:
even((1 to 100).toList) //> res2: Boolean = trueeven((1 to 101).toList) //> res3: Boolean = falseodd((1 to 100).toList) //> res4: Boolean = falseodd((1 to 101).toList) //> res5: Boolean = trueeven((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:
trait Trampoline[+A] { final def runT: A = this match { case Done(a) => a case More(k) => k().runT }}case class Done[+A](a: A) extends Trampoline[A]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:
def even[A](as: List[A]): Trampoline[Boolean] = as match {case Nil => Done(true)case h :: t => More(() => odd(t))} //> even: [A](as: List[A])ch13.ex1.Trampoline[Boolean]def odd[A](as: List[A]): Trampoline[Boolean] = as match {case Nil => Done(false)case h :: t => More(() => even(t))} //> odd: [A](as: List[A])ch13.ex1.Trampoline[Boolean]
我们可以用Trampoline的runT来运算结果:
even((1 to 10000).toList).runT //> res6: Boolean = trueeven((1 to 10001).toList).runT //> res7: Boolean = falseodd((1 to 10000).toList).runT //> res8: Boolean = falseodd((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:
case class State[S,+A](runS: S => (A,S)) {import State._def flatMap[B](f: A => State[S,B]): State[S,B] = State[S,B] {s => {val (a1,s1) = runS(s)f(a1) runS s1}} def map[B](f: A => B): State[S,B] = flatMap( a => unit(f(a)))}object State {def unit[S,A](a: A) = State[S,A] { s => (a,s) }def getState[S]: State[S,S] = State[S,S] { s => (s,s) }def setState[S](s: S): State[S,Unit] = State[S,Unit] { _ => ((),s)}}
再用State类型来写一个对List元素进行序号标注的函数:
def zip[A](as: List[A]): List[(A,Int)] = {as.foldLeft( unit[Int,List[(A,Int)]](List()))( (acc,a) => for { xs <- acc n <- getState[Int] _ <- setState[Int](n + 1) } yield (a,n) :: xs).runS(0)._1.reverse} //> zip: [A](as: List[A])List[(A, Int)]
运行一下这个zip函数:
zip((1 to 10).toList) //> res0: List[(Int, Int)] = List((1,0), (2,1), (3,2), (4,3), (5,4), (6,5), (7,6 //| ), (8,7), (9,8), (10,9))
结果正确。如果针对大型的List呢?
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?
case class State[S,A](runS: S => Trampoline[(A,S)]) {def flatMap[B](f: A => State[S,B]): State[S,B] = State[S,B] {s => More(() => {val (a1,s1) = runS(s).runTMore(() => f(a1) runS s1)})} def map[B](f: A => B): State[S,B] = flatMap( a => unit(f(a)))}object State {def unit[S,A](a: A) = State[S,A] { s => Done((a,s)) }def getState[S]: State[S,S] = State[S,S] { s => Done((s,s)) }def setState[S](s: S): State[S,Unit] = State[S,Unit] { _ => Done(((),s))}}trait Trampoline[+A] { final def runT: A = this match { case Done(a) => a case More(k) => k().runT }}case class Done[+A](a: A) extends Trampoline[A]case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]def zip[A](as: List[A]): List[(A,Int)] = {as.foldLeft( unit[Int,List[(A,Int)]](List()))( (acc,a) => for { xs <- acc n <- getState[Int] _ <- setState[Int](n + 1) } yield (a,n) :: xs).runS(0).runT._1.reverse} //> zip: [A](as: List[A])List[(A, Int)]zip((1 to 10).toList) //> res0: List[(Int, Int)] = List((1,0), (2,1), (3,2), (4,3), (5,4), (6,5), (7, //| 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:
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:
trait Trampoline[+A] { final def runT: A = this match { case Done(a) => a case More(k) => k().runT } def flatMap[B](f: A => Trampoline[B]): Trampoline[B] = { this match { case Done(a) => f(a) case More(k) => f(runT) } }}case class Done[+A](a: A) extends Trampoline[A]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:
case class State[S,A](runS: S => Trampoline[(A,S)]) {def flatMap[B](f: A => State[S,B]): State[S,B] = State[S,B] {s => More(() => {//val (a1,s1) = runS(s).runT//More(() => f(a1) runS s1) runS(s) flatMap { // runS(s) >>> Trampoline case (a1,s1) => More(() => f(a1) runS s1) }})} def map[B](f: A => B): State[S,B] = flatMap( a => unit(f(a)))}
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):
case class Done[+A](a: A) extends Trampoline[A]case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]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:
trait Trampoline[+A] { final def runT: A = resume match { case Right(a) => a case Left(k) => k().runT } def flatMap[B](f: A => Trampoline[B]): Trampoline[B] = { this match {// case Done(a) => f(a)// case More(k) => f(runT) case FlatMap(a,g) => FlatMap(a, (x: Any) => g(x) flatMap f) case x => FlatMap(x, f) } } def map[B](f: A => B) = flatMap(a => Done(f(a))) def resume: Either[() => Trampoline[A], A] = this match { case Done(a) => Right(a) case More(k) => Left(k) case FlatMap(a,f) => a match { case Done(v) => f(v).resume case More(k) => Left(() => k() flatMap f) case FlatMap(b,g) => FlatMap(b, (x: Any) => g(x) flatMap f).resume } }}case class Done[+A](a: A) extends Trampoline[A]case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]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:
def zip[A](as: List[A]): List[(A,Int)] = {as.foldLeft( unit[Int,List[(A,Int)]](List()))( (acc,a) => for { xs <- acc n <- getState[Int] _ <- setState[Int](n + 1) } yield (a,n) :: xs).runS(0).runT._1.reverse} //> zip: [A](as: List[A])List[(A, Int)]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:
val x = f() val y = g(x) h(y)//以上这三步函数引用可以写成:for { x <- f() y <- g(x) z <- h(y) } yield z
举个实际例子:
implicit def step[A](a: => A): Trampoline[A] = {More(() => Done(a))} //> step: [A](a: => A)ch13.ex1.Trampoline[A]def getNum: Double = 3 //> getNum: => Doubledef addOne(x: Double) = x + 1 //> addOne: (x: Double)Doubledef timesTwo(x: Double) = x * 2 //> timesTwo: (x: Double)Double(for {x <- getNumy <- addOne(x)z <- timesTwo(y)} yield z).runT //> res6: Double = 8.0
又或者:
def fib(n: Int): Trampoline[Int] = {if (n <= 1) Done(n) else for {x <- More(() => fib(n-1))y <- More(() => fib(n-2))} yield x + y} //> fib: (n: Int)ch13.ex1.Trampoline[Int](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:
trait Trampoline[+A] { final def runT: A = resume match { case Right(a) => a case Left(k) => k().runT } def flatMap[B](f: A => Trampoline[B]): Trampoline[B] = { this match {// case Done(a) => f(a)// case More(k) => f(runT) case FlatMap(a,g) => FlatMap(a, (x: Any) => g(x) flatMap f) case x => FlatMap(x, f) } } def map[B](f: A => B) = flatMap(a => Done(f(a))) def resume: Either[() => Trampoline[A], A] = this match { case Done(a) => Right(a) case More(k) => Left(k) case FlatMap(a,f) => a match { case Done(v) => f(v).resume case More(k) => Left(() => k() flatMap f) case FlatMap(b,g) => FlatMap(b, (x: Any) => g(x) flatMap f).resume } } def zip[B](tb: Trampoline[B]): Trampoline[(A,B)] = { (this.resume, tb.resume) match { case (Right(a),Right(b)) => Done((a,b)) case (Left(f),Left(g)) => More(() => f() zip g()) case (Right(a),Left(k)) => More(() => Done(a) zip k()) case (Left(k),Right(a)) => More(() => k() zip Done(a)) } }}case class Done[+A](a: A) extends Trampoline[A]case class More[+A](k: () => Trampoline[A]) extends Trampoline[A]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:
def hello: Trampoline[Unit] = for {_ <- print("Hello ")_ <- println("World!")} yield () //> hello: => ch13.ex1.Trampoline[Unit](hello zip hello zip hello).runT //> Hello Hello Hello World! //| World! //| World! //| 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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