How to get effective default value of XmlType#name?

Question

In my test utility code, I print XML content like this.

public static <T> void printXml(final Class<T> type, final T instance)
    throws JAXBException {

    final JAXBContext context = JAXBContext.newInstance(type);

    final Marshaller marshaller = context.createMarshaller();
    marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, Boolean.TRUE);

    if (type.getAnnotation(XmlRootElement.class) == null) {
        marshaller.setProperty(Marshaller.JAXB_FRAGMENT, Boolean.TRUE);
        final XmlType xmlType = type.getAnnotation(XmlType.class);
        // xmlType.name() -> "##default"
        marshaller.marshal(
            new JAXBElement<>(new QName(xmlType.namespace(),
                                        xmlType.name()), type, instance),
            System.out);
        return;
    }

    marshaller.marshal(instance, System.out);
}

Types with @XmlRootElement work just fine.

But when I tried to print a type which has not an @XmlRootElement I got this.

<##default xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="...">
...
</##default>

The xmlns part is fine. The marshaller found the namespaceUri I want.

The question is, see that ##default?, how can I find the name part?

I want this.

<someName xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns="...">
...
</someName>

Answer 1

If a name isn't specified on the annotation it defaults to ##default. To handle this case you would need to do what JAXB impls do and apply the algorithm defined in the JSR-222 specification on the short class name to get the name.