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I have a situation where I have to get the fully qualified name of a class I generate dynamically in Scala. Here's what I have so far.

import scala.reflect.runtime.universe
import scala.tools.reflect.ToolBox

val tb = universe.runtimeMirror(getClass.getClassLoader).mkToolBox()

val generatedClass = "class Foo { def addOne(i: Int) = i + 1 }"
tb.compile(tb.parse(generatedClass))

val fooClass:String = ???

Clearly this is just a toy example, but I just don't know how to get the fully qualified name of Foo. I tried sticking a package declaration into the code but that threw an error when calling tb.compile.

Does anyone know how to get the fully qualified class name or (even better) to specify the package that Foo gets compiled under?

Thanks

EDIT

After using the solution proposed I was able to get the class name. However, the next step is the register this class to take some actions later. Specifically I'm trying to make use of the UDTRegistration within Apache Spark to handle my own custom UserDefinedTypes. This strategy works fine when I manually create all the types, however, I want to use them to extend other types I may not know about.

After reading this it seems like what I'm trying to do might not be possible using code compiled at runtime using reflection. Maybe a better solution is to use Scala macros, but I'm very new to that area.

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Jon
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1 Answers1

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You may use define instead of compile to generate new class and get its package

val cls = tb.define(tb.parse(generatedClass).asInstanceOf[universe.ImplDef])
println(cls.fullName) //__wrapper$1$d1de39015284494799acd2875643f78e.Foo
Odomontois
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  • This is perfect, but I realized that my question was insufficient. I've edited the question to reflect my new needs. – Jon Dec 07 '16 at 16:27