How are Scala closures transformed to Java objects?

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被撕碎了的回忆 2021-02-01 20:45

I\'m currently looking at closure implementations in different languages. When it comes to Scala, however, I\'m unable to find any documentation on how a closure is mapped to Ja

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  •  孤城傲影
    2021-02-01 21:18

    Let's take apart a set of examples so we can see how they differ. (If using RC1, compile with -no-specialization to keep things easier to understand.)

    class Close {
      var n = 5
      def method(i: Int) = i+n
      def function = (i: Int) => i+5
      def closure = (i: Int) => i+n
      def mixed(m: Int) = (i: Int) => i+m
    }
    

    First, let's see what method does:

    public int method(int);
      Code:
       0:   iload_1
       1:   aload_0
       2:   invokevirtual   #17; //Method n:()I
       5:   iadd
       6:   ireturn
    

    Pretty straightforward. It's a method. Load the parameter, invoke the getter for n, add, return. Looks just like Java.

    How about function? It doesn't actually close any data, but it is an anonymous function (called Close$$anonfun$function$1). If we ignore any specialization, the constructor and apply are of most interest:

    public scala.Function1 function();
      Code:
       0:   new #34; //class Close$$anonfun$function$1
       3:   dup
       4:   aload_0
       5:   invokespecial   #35; //Method Close$$anonfun$function$1."":(LClose;)V
       8:   areturn
    
    public Close$$anonfun$function$1(Close);
      Code:
       0:   aload_0
       1:   invokespecial   #43; //Method scala/runtime/AbstractFunction1."":()V
       4:   return
    
    public final java.lang.Object apply(java.lang.Object);
      Code:
       0:   aload_0
       1:   aload_1
       2:   invokestatic    #26; //Method scala/runtime/BoxesRunTime.unboxToInt:(Ljava/lang/Object;)I
       5:   invokevirtual   #28; //Method apply:(I)I
       8:   invokestatic    #32; //Method scala/runtime/BoxesRunTime.boxToInteger:(I)Ljava/lang/Integer;
       11:  areturn
    
    public final int apply(int);
      Code:
       0:   iload_1
       1:   iconst_5
       2:   iadd
       3:   ireturn
    

    So, you load a "this" pointer and create a new object that takes the enclosing class as its argument. This is standard for any inner class, really. The function doesn't need to do anything with the outer class so it just calls the super's constructor. Then, when calling apply, you do the box/unbox tricks and then call the actual math--that is, just add 5.

    But what if we use a closure of the variable inside Close? Setup is exactly the same, but now the constructor Close$$anonfun$closure$1 looks like this:

    public Close$$anonfun$closure$1(Close);
      Code:
       0:   aload_1
       1:   ifnonnull   12
       4:   new #48; //class java/lang/NullPointerException
       7:   dup
       8:   invokespecial   #50; //Method java/lang/NullPointerException."":()V
       11:  athrow
       12:  aload_0
       13:  aload_1
       14:  putfield    #18; //Field $outer:LClose;
       17:  aload_0
       18:  invokespecial   #53; //Method scala/runtime/AbstractFunction1."":()V
       21:  return
    

    That is, it checks to make sure that the input is non-null (i.e. the outer class is non-null) and saves it in a field. Now when it comes time to apply it, after the boxing/unboxing wrapper:

    public final int apply(int);
      Code:
       0:   iload_1
       1:   aload_0
       2:   getfield    #18; //Field $outer:LClose;
       5:   invokevirtual   #24; //Method Close.n:()I
       8:   iadd
       9:   ireturn
    

    you see that it uses that field to refer to the parent class, and invokes the getter for n. Add, return, done. So, closures are easy enough: the anonymous function constructor just saves the enclosing class in a private field.

    Now, what about if we close not an internal variable, but a method argument? That's what Close$$anonfun$mixed$1 does. First, look at what the mixed method does:

    public scala.Function1 mixed(int);
      Code:
       0:   new #39; //class Close$$anonfun$mixed$1
       3:   dup
       4:   aload_0
       5:   iload_1
       6:   invokespecial   #42; //Method Close$$anonfun$mixed$1."":(LClose;I)V
       9:   areturn
    

    It loads the parameter m before calling the constructor! So it's no surprise that the constructor looks like this:

    public Close$$anonfun$mixed$1(Close, int);
      Code:
       0:   aload_0
       1:   iload_2
       2:   putfield    #18; //Field m$1:I
       5:   aload_0
       6:   invokespecial   #43; //Method scala/runtime/AbstractFunction1."":()V
       9:   return
    

    where that parameter is saved in a private field. No reference to the outer class is kept because we don't need it. And you ought not be surprised by apply either:

    public final int apply(int);
      Code:
       0:   iload_1
       1:   aload_0
       2:   getfield    #18; //Field m$1:I
       5:   iadd
       6:   ireturn
    

    Yes, we just load that stored field and do our math.

    I'm not sure what you were doing to not see this with your example--objects are a little tricky because they have both MyObject and MyObject$ classes and the methods get split between the two in a way that may not be intuitive. But apply definitely applies things, and overall the whole system works pretty much the way you'd expect it to (after you sit down and think about it really hard for a really long time).

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