如何在Scala中发现装箱/拆箱

Question

以下建议通过即兴最近有关如何获取斯卡拉告诉我是否有拳击通过查看字节码的时候,我创建了这个类:

class X { def foo(ls : Array[Long]) = ls map (_.toDouble)

看看字节码foo:

public double[] foo(long[]);
  Code:
   Stack=4, Locals=2, Args_size=2
   0:   getstatic       #11; //Field scala/Predef$.MODULE$:Lscala/Predef$;
   3:   aload_1
   4:   invokevirtual   #16; //Method scala/Predef$.longArrayOps:([J)Lscala/collection/mutable/ArrayOps;
   7:   new     #18; //class X$$anonfun$foo$1
   10:  dup
   11:  aload_0
   12:  invokespecial   #22; //Method X$$anonfun$foo$1."<init>":(LX;)V
   15:  getstatic       #27; //Field scala/Array$.MODULE$:Lscala/Array$;
   18:  getstatic       #32; //Field scala/reflect/Manifest$.MODULE$:Lscala/reflect/Manifest$;
   21:  invokevirtual   #36; //Method scala/reflect/Manifest$.Double:()Lscala/reflect/AnyValManifest;
   24:  invokevirtual   #40; //Method scala/Array$.canBuildFrom:(Lscala/reflect/ClassManifest;)Lscala/collection/generic/CanBuildFrom;
   27:  invokeinterface #46,  3; //InterfaceMethod scala/collection/TraversableLike.map:(Lscala/Function1;Lscala/collection/generic/CanBuildFrom;)Ljava/lan                                   g/Object;
   32:  checkcast       #48; //class "[D"
   35:  areturn
  LineNumberTable:
   line 7: 0

那里没有盒子/拆箱的迹象.但我仍然怀疑,所以我用-print()编译它:

[[syntax trees at end of cleanup]]// Scala source: X.scala
package <empty> {
  class X extends java.lang.Object with ScalaObject {
    def foo(ls: Array[Long]): Array[Double] = scala.this.Predef.longArrayOps(ls).map({
(new anonymous class X$$anonfun$foo$1(X.this): Function1)
}, scala.this.Array.canBuildFrom(reflect.this.Manifest.Double())).$asInstanceOf[Array[Double]]();
  def this(): X = {
    X.super.this();
    ()
  }
};
@SerialVersionUID(0) final <synthetic> class X$$anonfun$foo$1 extends scala.runtime.AbstractFunction1$mcDJ$sp with Serializable {
  final def apply(x$1: Long): Double = X$$anonfun$foo$1.this.apply$mcDJ$sp(x$1);
  <specialized> def apply$mcDJ$sp(v1: Long): Double = v1.toDouble();
  final <bridge> def apply(v1: java.lang.Object): java.lang.Object = scala.Double.box(X$$anonfun$foo$1.this.apply(scala.Long.unbox(v1)));
    def this($outer: X): anonymous class X$$anonfun$foo$1 = {
      X$$anonfun$foo$1.super.this();
      ()
    }
  }
}

这段代码的主要意见是,创建的匿名功能已被专门用于Long => Double与该map功能是由提供longArrayOps(ls).map(ArrayOps是不是专业).

问题是:"在这个例子中发生了装箱/拆箱吗？"

Answer 1

可能有拳击正在进行中.您必须查看实际map调用的位置:

27:  invokeinterface #46,  3; //InterfaceMethod scala/collection/TraversableLike.map:(Lscala/Function1;Lscala/collection/generic/CanBuildFrom;)Ljava/lang/Object;

在运行时,这应该去ArrayOps#ofLong.由于ArrayOps不是专门的,所以你的map-call 不可能在没有拳击的情况下完成.

要确定你可以尝试通过字节码跟踪调用(可能很难,因为在你的函数中你可能会遇到运行时调度)或者使用调试器(由于大多数调试器没有显示字节码,但很难您可以设置断点到Scala或Java的拳击方法).

由于实际原因,我们发现如果装箱是在字节码中完成的话可能无关紧要,因为对于热方法,Hotspot编译器有时会设法完全内联标准的高阶函数调用链,在这种情况下可以消除装箱.