How do I represent the boxed Double in pure Scala?

Question

In Scala there are 2 representations of double-precision numbers, one is an AnyVal, the other AnyRef. On the JVM they are mapped to the primitive double and the class java.lang.Double respectively.

Now what happens on platforms other than the JVM? I can use Scala.Double for the primitive, but how do I specify that I want a reference to the boxed Double without specifying java.lang.Double?

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[Context - left to make sense of Thomasz' answer, but not the fundamental issue.

I have a Double that I want to inject with Spring into a Wicket component:

class MyPanel(id: String) extends Panel(id) {
  @SpringBean(name="rxAlarmLimitDb") var alarmLimitDb: Double = _

If I specify the type as scala.Double as above, the injector fails, as it can only inject Objects.

If I specify java.lang.Double as the type of the field, all is well

class MyPanel(id: String) extends Panel(id) {
  @SpringBean(name="rxAlarmLimitDb") var alarmLimitDb: java.lang.Double = _

But I'm trying to reduce my dependance on falling back on the Java API, so how do I represent the boxed Double without it? ]

Answer 1

scala.Double == double in Java. When you box a scala.Double, it becomes a java.lang.Double.

scala> val f = 45d;
f: Double = 45.0

scala> println("f=" + f.getClass)
f=double

scala> def foo(d: Any) = println("d=" + d.getClass)
foo: (d: Any)Unit

scala> foo(f)
d=class java.lang.Double

There isn't any way of creating an object of type scala.Double. It's just an alias for double. So for your problem, you need to use a java.lang.Double, or enclose it in your own type and provide implicit conversions.

This definition makes sense if you think about it. All of the interaction between java & scala code which requires autoboxing & unboxing will work as expected.

If it makes a difference, you can always do:

type BoxedDouble = java.lang.Double

then you won't have to see the java.lang :-)