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Why is a `val` inside an `object` not automatically final?

Tags:

field

scala

final

What is the reason for vals not (?) being automatically final in singleton objects? E.g.

object NonFinal {
   val a = 0
   val b = 1

   def test(i: Int) = (i: @annotation.switch) match {
      case `a` => true
      case `b` => false
   }
}

results in:

<console>:12: error: could not emit switch for @switch annotated match
          def test(i: Int) = (i: @annotation.switch) match {
                                                     ^

Whereas

object Final {
   final val a = 0
   final val b = 1

   def test(i: Int) = (i: @annotation.switch) match {
      case `a` => true
      case `b` => false
   }
}

Compiles without warnings, so presumably generates the faster pattern matching table.

Having to add final seems pure annoying noise to me. Isn't an object final per se, and thus also its members?

like image 706
0__ Avatar asked Sep 06 '12 22:09

0__


3 Answers

This is addressed explicitly in the specification, and they are automatically final:

Members of final classes or objects are implicitly also final, so the final modifier is generally redundant for them, too. Note, however, that constant value definitions (§4.1) do require an explicit final modifier, even if they are defined in a final class or object.

Your final-less example compiles without errors (or warnings) with 2.10-M7, so I'd assume that there's a problem with the @switch checking in earlier versions, and that the members are in fact final.


Update: Actually this is more curious than I expected—if we compile the following with either 2.9.2 or 2.10-M7:

object NonFinal {
  val a = 0
}

object Final {
  final val a = 0
}

javap does show a difference:

public final class NonFinal$ implements scala.ScalaObject {
  public static final NonFinal$ MODULE$;
  public static {};
  public int a();
}

public final class Final$ implements scala.ScalaObject {
  public static final Final$ MODULE$;
  public static {};
  public final int a();
}

You see the same thing even if the right-hand side of the value definitions isn't a constant expression.

So I'll leave my answer, but it's not conclusive.

like image 96
Travis Brown Avatar answered Nov 11 '22 16:11

Travis Brown


You're not asking "why aren't they final", you're asking "why aren't they inlined." It just happens that final is how you cue the compiler that you want them inlined.

The reason they are not automatically inlined is separate compilation.

object A { final val x = 55 }
object B { def f = A.x }

When you compile this, B.f returns 55, literally:

public int f();
  0: bipush        55
  2: ireturn       

That means if you recompile A, B will be oblivious to the change. If x is not marked final in A, B.f looks like this instead:

  0: getstatic     #19                 // Field A$.MODULE$:LA$;
  3: invokevirtual #22                 // Method A$.x:()I
  6: ireturn       

Also, to correct one of the other answers, final does not mean immutable in scala.

like image 34
psp Avatar answered Nov 11 '22 18:11

psp


To address the central question about final on an object, I think this clause from the spec is more relevant:

A constant value definition is of the form final val x = e where e is a constant expression (§6.24). The final modifier must be present and no type annotation may be given. References to the constant value x are themselves treated as constant expressions; in the generated code they are replaced by the definition’s right-hand side e.

Of significance:

  • No type annotation may be given
  • The expression e is used in the generated code (by my reading, as the original unevaluated constant expression)

It sounds to me like the compiler is required by the spec to use these more like macro replacements rather than values that are evaluated in place at compile time, which could have impacts on how the resulting code runs.

I think it is particularly interesting that no type annotation may be given.

This, I think points to our ultimate answer, though I cannot come up with an example that shows the runtime difference for these requirements. In fact, in my 2.9.2 interpreter, I don't even get the enforcement of the first rule.

like image 44
Tony K. Avatar answered Nov 11 '22 16:11

Tony K.