hibernate的lazy初始化结果

package com.ehcache;import java.io.Serializable;public class User implements Serializable{ ???private int id; ???private String name; ???public User() { ???????????} ???public int getId() { ???????return id; ???} ???public void setId(int id) { ???????this.id = id; ???} ???public String getName() { ???????return name; ???} ???public void setName(String name) { ???????this.name = name; ???} ???@Override ???public String toString() { ???????return "User [id=" + id + ", name=" + name + "]"; ???} ???}

<?xml version="1.0"?><!DOCTYPE hibernate-mapping PUBLIC ???????"-//Hibernate/Hibernate Mapping DTD 3.0//EN" ???????"http://www.hibernate.org/dtd/hibernate-mapping-3.0.dtd"><hibernate-mapping package="com.ehcache"> ???<!-- name属性：哪个类 table属性：对应哪个表，如果不写，默认的表名就是类的简单名称 --> ???<class name="User" table="tt_user" > ???????<cache usage="read-write" /> ???????<id name="id"> ???????????<generator class="native" /> ???????</id> ???????<property name="name" not-null="true" /> ???</class></hibernate-mapping>

 ???????@Test ???????public void test2() { ???????????Session session = sessionFactory.openSession(); ???????????session.beginTransaction(); ???????????User u1 = session.load(User.class, 1);
 ???????????//这边注释session关闭后打印u1会报错 no session ???????????// Hibernate.initialize(u1); ???????????session.getTransaction().commit(); ???????????session.close(); ???????????System.out.println(u1); ???????????// return c1; ???????}