Enumerated(EnumType.STRING)在hibernate中怎样映射?

Question

大家知道jdb5.0中增加了annotation的功能，有些项目hibernate已经不用**.hbm.xml了。
在类里可以定义枚举类，比如：
@Entity
@Table(name = "contact")
@AttributeOverride(name = "id", column = @Column(name = "contact_id"))
@org.hibernate.annotations.Entity(selectBeforeUpdate=true)
public class ContactVO extends VersionedVO implements EntityAccess
{
    private static final long serialVersionUID = 3834315033474906417L;    public static enum ContactType {
        TELEPHONE,
        EMAIL,
        LETTER,
        FAX,
        PERSONAL;
    }    public static enum BoundType {
        IN,
        OUT;
    }    public static enum ImportType {
        NONE,
        NEW,
        ACCEPTED;
    }    public static enum ContentType {
        PLAINTEXT,
        HTML;
    }    private Date contactTimestamp;    private String subject;    private String content;    private String contactName;    private ContactType contactType;    private BoundType boundType;    private CompanyVO company;    private ContentType contentType = ContentType.PLAINTEXT;    private ImportType importType = ImportType.NONE;    private Set<PersonContactVO> personContacts = new HashSet<PersonContactVO>();
    
    private JobVO job;
    
    private Integer ownerUser;    private Integer ownerGroup;    private Access accessUser;    private Access accessGroup;    private Access accessGlobal;
    
    @Column(name = "bound_type")
    @Enumerated(EnumType.STRING)
    public BoundType getBoundType()
    {
        return boundType;
    }    public void setBoundType(BoundType boundType)
    {
        this.boundType = boundType;
    }    @Column(name = "contact_name")
    public String getContactName()
    {
        return contactName;
    }    public void setContactName(String contactName)
    {
        this.contactName = contactName;
    }    @Column(name = "contact_timestamp")
    public Date getContactTimestamp()
    {
        return contactTimestamp;
    }    public void setContactTimestamp(Date contactTimestamp)
    {
        this.contactTimestamp = contactTimestamp;
    }    @Column(name = "contact_type")
    @Enumerated(EnumType.STRING)
    public ContactType getContactType()
    {
        return contactType;
    }    public void setContactType(ContactType contactType)
    {
        this.contactType = contactType;
    }    @Column(name = "content")
    public String getContent()
    {
        return content;
    }    public void setContent(String content)
    {
        this.content = content;
    }    @Column(name = "subject")
    public String getSubject()
    {
        return subject;
    }    public void setSubject(String subject)
    {
        this.subject = subject;
    }    @Column(name = "content_type")
    @Enumerated(EnumType.STRING)
    public ContentType getContentType()
    {
        return contentType;
    }    public void setContentType(ContentType contentType)
    {
        this.contentType = contentType;
    }    @Column(name = "import_type")
    @Enumerated(EnumType.STRING)
    public ImportType getImportType()
    {
        return importType;
    }    public void setImportType(ImportType importType)
    {
        this.importType = importType;
    }    @ManyToOne
    @JoinColumn(name = "company_id")
    public CompanyVO getCompany()
    {
        return company;
    }    public void setCompany(CompanyVO company)
    {
        this.company = company;
    }    @OneToMany(mappedBy = "contact", cascade=CascadeType.ALL)
    @Cascade({ org.hibernate.annotations.CascadeType.ALL, org.hibernate.annotations.CascadeType.DELETE_ORPHAN }) 
    public Set<PersonContactVO> getPersonContacts()
    {
        return personContacts;
    }    public void setPersonContacts(Set<PersonContactVO> personContacts)
    {
        this.personContacts = personContacts;
    }    @ManyToOne
    @JoinColumn(name = "job_id")
    public JobVO getJob()
    {
        return job;
    }
    
    public void setJob(JobVO job)
    {
        this.job = job;
    }
    
    @Column(name = "owner_user")
    public Integer getOwnerUser()
    {
        return ownerUser;
    }    public void setOwnerUser(Integer ownerUser)
    {
        this.ownerUser = ownerUser;
    }    @Column(name = "owner_group")
    public Integer getOwnerGroup()
    {
        return ownerGroup;
    }    public void setOwnerGroup(Integer ownerGroup)
    {
        this.ownerGroup = ownerGroup;
    }    @Column(name = "access_user")
    @Enumerated(EnumType.STRING)
    public Access getAccessUser()
    {
        return accessUser;
    }    public void setAccessUser(Access accessUser)
    {
        this.accessUser = accessUser;
    }    @Column(name = "access_group")
    @Enumerated(EnumType.STRING)
    public Access getAccessGroup()
    {
        return accessGroup;
    }    public void setAccessGroup(Access accessGroup)
    {
        this.accessGroup = accessGroup;
    }
    
    @Column(name = "access_global")
    @Enumerated(EnumType.STRING)
    public Access getAccessGlobal()
    {
        return accessGlobal;
    }    public void setAccessGlobal(Access accessGlobal)
    {
        this.accessGlobal = accessGlobal;
    } 
    
    protected void toString(ToStringBuilder builder)
    {
        builder.append("company", company != null ? company.getId() : "null");
        builder.append("job", job != null ? job.getId() : "null");
        builder.append("contactTimestamp", contactTimestamp);
        builder.append("subject", subject);
        builder.append("content", content);
        builder.append("contentType", contentType);
        builder.append("contactName", contactName);
        builder.append("contactType", contactType);
        builder.append("boundType", boundType);
        builder.append("importType", importType);
        builder.append("ownerUser", ownerUser);
        builder.append("ownerGroup", ownerGroup);
        builder.append("accessUser", accessUser);
        builder.append("accessGroup", accessGroup);
        builder.append("accessGlobal", accessGlobal);
    }}请问@Enumerated(EnumType.STRING)
在原先的***.hbm.xml中怎样映射啊？
方便点说我想放弃用annotation的方法，还原到以前老的hibernate映射方法，不知道hibernate有没有支持@Enumerated(EnumType.STRING)
的type.