Convert java.util.IdentityHashMap to scala.immutable.Map - scala

Question

What is the simplest way to convert a java.util.IdentityHashMap[A,B] into a subtype of scala.immutable.Map[A,B]? I need to keep keys separate unless they are eq.
Here's what I've tried so far:
scala> case class Example()
scala> val m = new java.util.IdentityHashMap[Example, String]()
scala> m.put(Example(), "first!")
scala> m.put(Example(), "second!")
scala> m.asScala // got a mutable Scala equivalent OK
res14: scala.collection.mutable.Map[Example,String] = Map(Example() -> first!, Example() -> second!)
scala> m.asScala.toMap // doesn't work, since toMap() removes duplicate keys (testing with ==)
res15: scala.collection.immutable.Map[Example,String] = Map(Example() -> second!)

Answer 1

Here's a simple implementation of identity map in Scala. In usage, it should be similar to standard immutable map.
Example usage:
val im = IdentityMap(
new String("stuff") -> 5,
new String("stuff") -> 10)
println(im) // IdentityMap(stuff -> 5, stuff -> 10)
Your case:
import scala.collection.JavaConverters._
import java.{util => ju}
val javaIdentityMap: ju.IdentityHashMap = ???
val scalaIdentityMap = IdentityMap.empty[String,Int] ++ javaIdentityMap.asScala
Implementation itself (for performance reasons, there may be some more methods that need to be overridden):
import scala.collection.generic.ImmutableMapFactory
import scala.collection.immutable.MapLike
import IdentityMap.{Wrapper, wrap}
class IdentityMap[A, +B] private(underlying: Map[Wrapper[A], B])
extends Map[A, B] with MapLike[A, B, IdentityMap[A, B]] {
def +[B1 >: B](kv: (A, B1)) =
new IdentityMap(underlying + ((wrap(kv._1), kv._2)))
def -(key: A) =
new IdentityMap(underlying - wrap(key))
def iterator =
underlying.iterator.map {
case (kw, v) => (kw.value, v)
}
def get(key: A) =
underlying.get(wrap(key))
override def size: Int =
underlying.size
override def empty =
new IdentityMap(underlying.empty)
override def stringPrefix =
"IdentityMap"
}
object IdentityMap extends ImmutableMapFactory[IdentityMap] {
def empty[A, B] =
new IdentityMap(Map.empty)
private class Wrapper[A](val value: A) {
override def toString: String =
value.toString
override def equals(other: Any) = other match {
case otherWrapper: Wrapper[_] =>
value.asInstanceOf[AnyRef] eq otherWrapper.value.asInstanceOf[AnyRef]
case _ => false
}
override def hashCode =
System.identityHashCode(value)
}
private def wrap[A](key: A) =
new Wrapper(key)
}

Answer 2

One way to handle this would be change what equality means for the class, e.g.
scala> case class Example() {
override def equals( that:Any ) = that match {
case that:AnyRef => this eq that
case _ => false
}
}
defined class Example
scala> val m = new java.util.IdentityHashMap[Example, String]()
m: java.util.IdentityHashMap[Example,String] = {}
scala> m.put(Example(), "first!")
res1: String = null
scala> m.put(Example(), "second!")
res2: String = null
scala> import scala.collection.JavaConverters._
import scala.collection.JavaConverters._
scala> m.asScala
res3: scala.collection.mutable.Map[Example,String] = Map(Example() -> second!, Example() -> first!)
scala> m.asScala.toMap
res4: scala.collection.immutable.Map[Example,String] = Map(Example() -> second!, Example() -> first!)
Or if you don't want to change equality for the class, you could make a wrapper.
Of course, this won't perform as well as a Map that uses eq instead of ==; it might be worth asking for one....