在Scala reduceLeft实现中0.asInstanceOf [B]发生了什么

下面是Scala的TraversableOnce trait的reduceLeft方法的来源 . 读取 var acc: B = 0.asInstanceOf[B] 的行发生了什么?

对我来说,似乎如果我在一个字符串列表上调用它,例如 List("a", "b", "c") ,这将导致像 0.asInstanceOf[String] 这样的东西 . 但是,如果直接尝试, 0.asInstanceOf[String] 会在运行时抛出 ClassCastException .

该行发生了什么,为什么它与在字符串列表中调用时直接调用 0.asInstanceOf[String] 不同?

def reduceLeft[B >: A](op: (B, A) => B): B = {
  if (isEmpty)
    throw new UnsupportedOperationException("empty.reduceLeft")

  var first = true
  var acc: B = 0.asInstanceOf[B]

  for (x <- self) {
    if (first) {
      acc = x
      first = false
    }
    else acc = op(acc, x)
  }
  acc
}

奖金问题:为什么 acc 甚至被初始化为该值?看起来 for 循环的第一次迭代中的代码将使用 TraversableOnce 对象中的第一个元素覆盖该值 .

回答(2)

3 years ago

好吧,这在运行时没有失败 ClassCastException 的原因是因为:

  • 编译器删除了从未使用过的初始化推理(我对此表示怀疑)

  • B是erasedObject (或 AnyRef ),因此演员阵容真的 0.asInstanceOf[Object]

您可以通过检查字节码来测试第一种可能性,但它似乎是一个毫无意义的代码 . 如果它没有被省略,那么每次调用该方法时都会产生不必要的开销 Int (尽管不是对象创建 - 见下文)!

弄清楚编译器产生的内容:1

Welcome to Scala version 2.9.1.final (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_29).
Type in expressions to have them evaluated.
Type :help for more information.

scala> trait X[A] {
 | def foreach[U](f: A => U): U
 | def isEmpty = {
 | var b = false
 | foreach(_ => b = true)
 | !b
 | }
 | def reduceLeft[B >: A](op: (B, A) => B): B = {
 | if (isEmpty) sys.error("Bad")
 | var first = true
 | var acc: B = 0.asInstanceOf[B]
 | for (x <- this) {
 | if (first) { acc = x; first = false } else acc = op(acc, x)
 | }
 | acc
 | }
 | }
defined trait X

所以现在:

scala> :javap -v X
Compiled from "<console>"
public interface X extends scala.ScalaObject
  SourceFile: "<console>"
  Scala: length = 0x

  Signature: length = 0x2
   00 0D 
  InnerClass: 
   public abstract #19= #16 of #18; //X=class X of class 
   public final #21; //class X$$anonfun$isEmpty$1
   public final #23; //class X$$anonfun$reduceLeft$1
  minor version: 0
  major version: 49
  Constant pool:
const #1 = Asciz    SourceFile;
const #2 = Asciz    <console>;
const #3 = Asciz    foreach;
const #4 = Asciz    (Lscala/Function1;)Ljava/lang/Object;;
const #5 = Asciz    <U:Ljava/lang/Object;>(Lscala/Function1<TA;TU;>;)TU;;
const #6 = Asciz    Signature;
const #7 = Asciz    isEmpty;
const #8 = Asciz    ()Z;
const #9 = Asciz    reduceLeft;
const #10 = Asciz   (Lscala/Function2;)Ljava/lang/Object;;
const #11 = Asciz   <B:Ljava/lang/Object;>(Lscala/Function2<TB;TA;TB;>;)TB;;
const #12 = Asciz   Scala;
const #13 = Asciz   <A:Ljava/lang/Object;>Ljava/lang/Object;Lscala/ScalaObject;;
const #14 = Asciz   InnerClasses;
const #15 = Asciz   X;
const #16 = class   #15;    //  X
const #17 = Asciz   ;
const #18 = class   #17;    //  
const #19 = Asciz   X;
const #20 = Asciz   X$$anonfun$isEmpty$1;
const #21 = class   #20;    //  X$$anonfun$isEmpty$1
const #22 = Asciz   X$$anonfun$reduceLeft$1;
const #23 = class   #22;    //  X$$anonfun$reduceLeft$1
const #24 = Asciz   java/lang/Object;
const #25 = class   #24;    //  java/lang/Object
const #26 = Asciz   scala/ScalaObject;
const #27 = class   #26;    //  scala/ScalaObject

{
public abstract java.lang.Object foreach(scala.Function1);
  Signature: length = 0x2
   00 05 

public abstract boolean isEmpty();

public abstract java.lang.Object reduceLeft(scala.Function2);
  Signature: length = 0x2
   00 0B 

}

做你想做的!

弄清楚编译器产生的内容:2

另一种可能性是将它放在源文件和_764969中:

./scalac -Xprint:icode X.scala

然后你得到......

def reduceLeft($this: X, op$1: Function2): java.lang.Object = {
  if ($this.isEmpty())
    scala.sys.`package`.error("Bad")
  else
    ();
  var first$1: scala.runtime.BooleanRef = new scala.runtime.BooleanRef(true);
  var acc$1: scala.runtime.ObjectRef = new scala.runtime.ObjectRef(scala.Int.box(0));
  $this.foreach({
    (new anonymous class X$$anonfun$reduceLeft$1($this, op$1, first$1, acc$1): Function1)
  });
  acc.elem
};

它看起来非常像在那里不必要的拳击!一点是,这不会涉及对象创建,只是查找,因为缓存了-127到127的装箱值 .

您可以通过将上述打印命令中的编译器阶段更改为“擦除”来检查已擦除的行 . 嘿presto:

var acc: java.lang.Object = scala.Int.box(0).$asInstanceOf[java.lang.Object]();

3 years ago

可以通过尝试提出替代方案来推断奖金问题的答案 . 累加器是B型 . 你会分配给它什么B? (有一个更好的答案,即null.asInstanceOf [B]这是通常做的,但我认为这会留下你的问题 . )