refine Range serialization

apache · Oct 5, 2024 · 382a693 · 382a693
1 parent ea961fc
commit 382a693
Showing 1 changed file with 9 additions and 0 deletions.
diff --git a/docs/guide/scala_guide.md b/docs/guide/scala_guide.md
@@ -35,14 +35,23 @@ When using fury for scala serialization, you should create fury at least with fo
 ```scala
 import org.apache.fury.Fury
 import org.apache.fury.serializer.scala.ScalaSerializers
+import org.apache.fury.serializer.collection.CollectionSerializers.DefaultJavaCollectionSerializer
 
 val fury = Fury.builder()
   .withScalaOptimizationEnabled(true)
   .requireClassRegistration(true)
   .withRefTracking(true)
   .build()
 
+// Register optimized fury serializers for scala
 ScalaSerializers.registerSerializers(fury)
+// serialize range as (start, step, end) instead of collection
+// this will be handled in next version automatically.
+fury.registerSerializer(classOf[Range.Inclusive], classOf[DefaultJavaCollectionSerializer])
+fury.registerSerializer(classOf[Range.Exclusive], classOf[DefaultJavaCollectionSerializer])
+fury.registerSerializer(classOf[NumericRange], classOf[DefaultJavaCollectionSerializer])
+fury.registerSerializer(classOf[NumericRange.Inclusive], classOf[DefaultJavaCollectionSerializer])
+fury.registerSerializer(classOf[NumericRange.Exclusive], classOf[DefaultJavaCollectionSerializer])
 ```
 
 Depending on the object types you serialize, you may need to register some scala internal types: