嵌套方法的成本

Question

嵌套方法的成本

在Scala中,可以在其他方法中定义方法.这将它们的使用范围限制在定义块内.我使用它们来提高使用几个高阶函数的代码的可读性.与匿名函数文字相比,这允许我在传递它们之前给它们有意义的名称.

例如:

class AggregatedPerson extends HashSet[PersonRecord] {
  def mostFrequentName: String = {
    type NameCount = (String, Int)
    def moreFirst(a: NameCount, b: NameCount) = a._2 > b._2
    def countOccurrences(nameGroup: (String, List[PersonRecord])) =
      (nameGroup._1, nameGroup._2.size) 

    iterator.toList.groupBy(_.fullName).
      map(countOccurrences).iterator.toList.
      sortWith(moreFirst).head._1
  }
}

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是否有任何运行时成本,因为我应该知道嵌套方法定义？

关闭的答案有何不同？

Answer 1

ret*_*nym 57

期间compilaton,嵌套功能moveFirst和countOccurences移出到相同的水平mostFrequentName.他们得到编译器合成的名称:moveFirst$1和countOccurences$1.

此外,当您在没有参数列表的情况下引用其中一种方法时,它将被提升为一个函数.所以map(countOccurences)和写作一样map((a: (String, List[PersonRecord])) => countOccurences(a)).这个匿名函数被编译成一个单独的类AggregatedPerson$$anonfun$mostFrequentName$2,除了转发之外什么也没做countOccurences$.

作为旁注,将方法提升为函数的过程称为Eta Expansion.如果在期望函数类型的上下文中省略参数列表(如在您的示例中),或者如果您使用_代替整个参数列表,或者代替每个参数(val f1 = countOccurences _ ; val f2 = countOccurences(_).

如果代码直接在闭包中,那么堆栈中的方法调用将减少一个,并且生成的合成方法会少一个.在大多数情况下,这对性能的影响可能为零.

我发现它是构建代码的一个非常有用的工具,并认为你的例子非常惯用Scala.

另一个有用的工具是使用小块来初始化val:

val a = {
   val temp1, temp2 = ...
   f(temp1, temp2)
}

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您可以使用scalac -print以确切地了解Scala代码如何转换为为JVM准备好的表单.下面是程序的输出:

[[syntax trees at end of cleanup]]// Scala source: nested-method.scala
package <empty> {

  class AggregatedPerson extends scala.collection.mutable.HashSet with ScalaObject {
    def mostFrequentName(): java.lang.String = AggregatedPerson.this.iterator().toList().groupBy({
      (new AggregatedPerson$$anonfun$mostFrequentName$1(AggregatedPerson.this): Function1)
    }).map({
      {
        (new AggregatedPerson$$anonfun$mostFrequentName$2(AggregatedPerson.this): Function1)
      }
    }, collection.this.Map.canBuildFrom()).$asInstanceOf[scala.collection.MapLike]().iterator().toList().sortWith({
      {
        (new AggregatedPerson$$anonfun$mostFrequentName$3(AggregatedPerson.this): Function2)
      }
    }).$asInstanceOf[scala.collection.IterableLike]().head().$asInstanceOf[Tuple2]()._1().$asInstanceOf[java.lang.String]();
    final def moreFirst$1(a: Tuple2, b: Tuple2): Boolean = scala.Int.unbox(a._2()).>(scala.Int.unbox(b._2()));
    final def countOccurrences$1(nameGroup: Tuple2): Tuple2 = new Tuple2(nameGroup._1(), scala.Int.box(nameGroup._2().$asInstanceOf[scala.collection.SeqLike]().size()));
    def this(): AggregatedPerson = {
      AggregatedPerson.super.this();
      ()
    }
  };

  @SerialVersionUID(0) @serializable final <synthetic> class AggregatedPerson$$anonfun$mostFrequentName$1 extends scala.runtime.AbstractFunction1 {
    final def apply(x$1: PersonRecord): java.lang.String = x$1.fullName();
    final <bridge> def apply(v1: java.lang.Object): java.lang.Object = AggregatedPerson$$anonfun$mostFrequentName$1.this.apply(v1.$asInstanceOf[PersonRecord]());
    def this($outer: AggregatedPerson): AggregatedPerson$$anonfun$mostFrequentName$1 = {
      AggregatedPerson$$anonfun$mostFrequentName$1.super.this();
      ()
    }
  };

  @SerialVersionUID(0) @serializable final <synthetic> class AggregatedPerson$$anonfun$mostFrequentName$2 extends scala.runtime.AbstractFunction1 {
    final def apply(nameGroup: Tuple2): Tuple2 = AggregatedPerson$$anonfun$mostFrequentName$2.this.$outer.countOccurrences$1(nameGroup);
    <synthetic> <paramaccessor> private[this] val $outer: AggregatedPerson = _;
    final <bridge> def apply(v1: java.lang.Object): java.lang.Object = AggregatedPerson$$anonfun$mostFrequentName$2.this.apply(v1.$asInstanceOf[Tuple2]());
    def this($outer: AggregatedPerson): AggregatedPerson$$anonfun$mostFrequentName$2 = {
      if ($outer.eq(null))
        throw new java.lang.NullPointerException()
      else
        AggregatedPerson$$anonfun$mostFrequentName$2.this.$outer = $outer;
      AggregatedPerson$$anonfun$mostFrequentName$2.super.this();
      ()
    }
  };
  @SerialVersionUID(0) @serializable final <synthetic> class AggregatedPerson$$anonfun$mostFrequentName$3 extends scala.runtime.AbstractFunction2 {
    final def apply(a: Tuple2, b: Tuple2): Boolean = AggregatedPerson$$anonfun$mostFrequentName$3.this.$outer.moreFirst$1(a, b);
    <synthetic> <paramaccessor> private[this] val $outer: AggregatedPerson = _;
    final <bridge> def apply(v1: java.lang.Object, v2: java.lang.Object): java.lang.Object = scala.Boolean.box(AggregatedPerson$$anonfun$mostFrequentName$3.this.apply(v1.$asInstanceOf[Tuple2](), v2.$asInstanceOf[Tuple2]()));
    def this($outer: AggregatedPerson): AggregatedPerson$$anonfun$mostFrequentName$3 = {
      if ($outer.eq(null))
        throw new java.lang.NullPointerException()
      else
        AggregatedPerson$$anonfun$mostFrequentName$3.this.$outer = $outer;
      AggregatedPerson$$anonfun$mostFrequentName$3.super.this();
      ()
    }
  }
}

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Answer 2

Rex*_*err 14

运行时成本很低.你可以在这里观察它(为长代码道歉):

object NestBench {
  def countRaw() = {
    var sum = 0
    var i = 0
    while (i<1000) {
      sum += i
      i += 1
      var j = 0
      while (j<1000) {
        sum += j
        j += 1
        var k = 0
        while (k<1000) {
          sum += k
          k += 1
          sum += 1
        }
      }
    }
    sum
  }
  def countClosure() = {
    var sum = 0
    var i = 0
    def sumI {
      sum += i
      i += 1
      var j = 0
      def sumJ {
        sum += j
        j += 1
        var k = 0
        def sumK {
          def sumL { sum += 1 }
          sum += k
          k += 1
          sumL
        }
        while (k<1000) sumK
      }
      while (j<1000) sumJ
    }
    while (i<1000) sumI
    sum
  }
  def countInner() = {
    var sum = 0
    def whileI = {
      def whileJ = {
        def whileK = {
          def whileL() = 1
          var ksum = 0
          var k = 0
          while (k<1000) { ksum += k; k += 1; ksum += whileL }
          ksum
        }
        var jsum = 0
        var j = 0
        while (j<1000) {
          jsum += j; j += 1
          jsum += whileK
        }
        jsum
      }
      var isum = 0
      var i = 0
      while (i<1000) {
        isum += i; i += 1
        isum += whileJ
      }
      isum
    }
    whileI
  }
  def countFunc() = {
    def summer(f: => Int)() = {
      var sum = 0
      var i = 0
      while (i<1000) {
        sum += i; i += 1
        sum += f
      }
      sum
    }
    summer( summer( summer(1) ) )()
  }
  def nsPerIteration(f:() => Int): (Int,Double) = {
    val t0 = System.nanoTime
    val result = f()
    val t1 = System.nanoTime
    (result , (t1-t0)*1e-9)
  }
  def main(args: Array[String]) {
    for (i <- 1 to 5) {
      val fns = List(countRaw _, countClosure _, countInner _, countFunc _)
      val labels = List("raw","closure","inner","func")
      val results = (fns zip labels) foreach (fl => {
        val x = nsPerIteration( fl._1 )
        printf("Method %8s produced %d; time/it = %.3f ns\n",fl._2,x._1,x._2)
      })
    }
  }
}

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整数求和有四种不同的方法:

原始循环("原始")
while循环内部方法是对sum变量的闭包
while循环内部方法返回部分和
一个求和的嵌套函数

我们在内部循环中以纳秒的形式看到我机器上的结果:

scala> NestBench.main(Array[String]())
Method      raw produced -1511174132; time/it = 0.422 ns
Method  closure produced -1511174132; time/it = 2.376 ns
Method    inner produced -1511174132; time/it = 0.402 ns
Method     func produced -1511174132; time/it = 0.836 ns
Method      raw produced -1511174132; time/it = 0.418 ns
Method  closure produced -1511174132; time/it = 2.410 ns
Method    inner produced -1511174132; time/it = 0.399 ns
Method     func produced -1511174132; time/it = 0.813 ns
Method      raw produced -1511174132; time/it = 0.411 ns
Method  closure produced -1511174132; time/it = 2.372 ns
Method    inner produced -1511174132; time/it = 0.399 ns
Method     func produced -1511174132; time/it = 0.813 ns
Method      raw produced -1511174132; time/it = 0.411 ns
Method  closure produced -1511174132; time/it = 2.370 ns
Method    inner produced -1511174132; time/it = 0.399 ns
Method     func produced -1511174132; time/it = 0.815 ns
Method      raw produced -1511174132; time/it = 0.412 ns
Method  closure produced -1511174132; time/it = 2.357 ns
Method    inner produced -1511174132; time/it = 0.400 ns
Method     func produced -1511174132; time/it = 0.817 ns

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所以,底线是:在简单的情况下,嵌套函数根本不会伤害到你--JVM会发现调用可以内联(因此raw并inner给出相同的时间).如果采用更多功能方法,则不能完全忽略函数调用,但所花费的时间非常小(每次调用大约0.4 ns).如果你使用了很多闭包,那么关闭它们会产生每次调用1 ns的开销,至少在写入单个可变变量的情况下.

你可以修改上面的代码来找到其他问题的答案,但最重要的是它总是非常快,介于"无任何惩罚"到"只担心在非常严格的内部循环中,否则只需要很少的工作做".

(PS为了进行比较,在我的机器上创建一个小对象需要大约4 ns.)

Answer 3

use*_*561 7

截至2014年1月的当前

目前的基准测试已经有3年的历史了,Hotspot和编译器已经有了很大的发展.我也在使用Google Caliper来执行基准测试.

码

import com.google.caliper.SimpleBenchmark

class Benchmark extends SimpleBenchmark {
    def timeRaw(reps: Int) = {
        var i = 0
        var result = 0L
        while (i < reps) {
            result += 0xc37e ^ (i * 0xd5f3)
            i = i + 1
        }
        result
    }

    def normal(i: Int): Long = 0xc37e ^ (i * 0xd5f3)
    def timeNormal(reps: Int) = {
        var i = 0
        var result = 0L
        while (i < reps) {
            result += normal(i)
            i = i + 1
        }
        result
    }

    def timeInner(reps: Int) = {
        def inner(i: Int): Long = 0xc37e ^ (i * 0xd5f3)

        var i = 0
        var result = 0L
        while (i < reps) {
            result += inner(i)
            i = i + 1
        }
        result
    }

    def timeClosure(reps: Int) = {
        var i = 0
        var result = 0L
        val closure = () => result += 0xc37e ^ (i * 0xd5f3)
        while (i < reps) {
            closure()
            i = i + 1
        }
        result
    }

    def normal(i: Int, j: Int, k: Int, l: Int): Long = i ^ j ^ k ^ l 
    def timeUnboxed(reps: Int) = {
        var i = 0
        var result = 0L
        while (i < reps) {
            result += normal(i,i,i,i)
            i = i + 1
        }
        result
    }

    val closure = (i: Int, j: Int, k: Int, l: Int) => (i ^ j ^ k ^ l).toLong 
    def timeBoxed(reps: Int) = {
        var i = 0
        var result = 0L
        while (i < reps) {
            closure(i,i,i,i)
            i = i + 1
        }
        result
    }
}

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结果

benchmark     ns linear runtime
   Normal  0.576 =
      Raw  0.576 =
    Inner  0.576 =
  Closure  0.532 =
  Unboxed  0.893 =
    Boxed 15.210 ==============================

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令人惊讶的是,闭合测试比其他测试完成快4ns.这似乎是Hotspot的特质而不是执行环境,多次运行都返回了相同的趋势.

使用执行装箱的闭包是一个巨大的性能打击,执行一次拆箱和重新装箱需要大约3.579ns,足以进行大量的原始数学运算.在这个特定的位置,在新的优化器上完成工作可能会变得更好.在一般情况下,拳击可以通过迷你车减轻.

编辑: 新的优化器在这里没有真正的帮助,它使Closure0.1 ns更慢,Boxed0.1 ns更快:

 benchmark     ns linear runtime
       Raw  0.574 =
    Normal  0.576 =
     Inner  0.575 =
   Closure  0.645 =
   Unboxed  0.889 =
     Boxed 15.107 ==============================

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来自magarciaEPFL/scala的2.11.0-20131209-220002-9587726041演出

版本

JVM

java version "1.7.0_51"
Java(TM) SE Runtime Environment (build 1.7.0_51-b13)
Java HotSpot(TM) 64-Bit Server VM (build 24.51-b03, mixed mode)

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Scalac

Scala compiler version 2.10.3 -- Copyright 2002-2013, LAMP/EPFL

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