Created attachment 841 [details]
junit test

As I understand the test source, you have some existing inner class coded as

public class OuterClass {
	public class InnerClass {
	}
}

then theInnerClassElement.getName() has sourceName property = "InnerClass", and
its fullName property should be set to "whatever.package.OuterClass.InnerClass".

The superclass property is just an identifier, nothing more - you, as the
caller, are responsible for translating the name into appropriate target context
(e.g. prefix it with outerclass/package). If you want to be sure, the best is
maybe create Identifier from the inner class' full name.

Resolved for 3.3.x or earlier, no new info since then -> closing.

Resolved for 3.4.x or earlier, no new info since then -> closing.