如何在没有任何相关信息的情况下通过JAXB封送对象？

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;

public class JAXBExample {
    public static void main(String[] args) {
        try {
            // 创建要封送的Java对象
            Person person = new Person();
            person.setName("John");
            person.setAge(30);

            // 创建JAXBContext
            JAXBContext jaxbContext = JAXBContext.newInstance(Person.class);

            // 创建Marshaller
            Marshaller marshaller = jaxbContext.createMarshaller();
            marshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);

            // 封送对象到XML文档
            marshaller.marshal(person, System.out);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

// 要封送的Java对象
class Person {
    private String name;
    private int age;

    // 省略构造方法、getter和setter

    // 添加JAXB注解
    @XmlElement
    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    @XmlElement
    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }
}