hmm, I dunno why IntelliJ is unhappy, but you can always fall back to
getting a class from the String:
Class.forName(scala.reflect.ClassTag$$anon$1)
perhaps the class is package private or something, and the repl somehow
subverts it ...
On Tue, Apr 14, 2015 at 5:44 PM, Arun Lists
Wow, it all works now! Thanks, Imran!
In case someone else finds this useful, here are the additional classes
that I had to register (in addition to my application specific classes):
val tuple3ArrayClass = classOf[Array[Tuple3[Any, Any, Any]]]
val anonClass =
Hi Arun,
It can be hard to use kryo with required registration because of issues
like this -- there isn't a good way to register all the classes that you
need transitively. In this case, it looks like one of your classes has a
reference to a ClassTag, which in turn has a reference to some
Hi Imran,
Thanks for the response! However, I am still not there yet.
In the Scala interpreter, I can do:
scala classOf[scala.reflect.ClassTag$$anon$1]
but when I try to do this in my program in IntelliJ, it indicates an error:
Cannot resolve symbol ClassTag$$anon$1
Hence I am not any closer
Those funny class names come from scala's specialization -- its compiling a
different version of OpenHashMap for each primitive you stick in the type
parameter. Here's a super simple example:
*➜ **~ * more Foo.scala
class Foo[@specialized X]
*➜ **~ * scalac Foo.scala
*➜ **~ * ls
Hi,
I am trying to register classes with KryoSerializer. This has worked with
other programs. Usually the error messages are helpful in indicating which
classes need to be registered. But with my current program, I get the
following cryptic error message:
*Caused by:
I am trying to register classes with KryoSerializer. I get the following
error message:
How do I find out what class is being referred to by: *OpenHashMap$mcI$sp ?*
*com.esotericsoftware.kryo.KryoException:
java.lang.IllegalArgumentException: Class is not registered:
