静态成员变量能不能被序列化????
程序如下:
import java.io.*;
class Student implements Serializable
{
private String name;
private transient String password;
private static int count=0;
public Student(String name,String password)
{
System.out.println("调用Student的带参的构造方法");
this.name=name;
this.password=password;
count++;
}
public String toString()
{
return "人数: "+count+" 姓名: "+name+" 密码: "+password;
}
}
public class ObjectSerTest1
{
public static void main(String args[])
{
try{
FileOutputStream fos=new FileOutputStream("test.obj");
ObjectOutputStream oos=new ObjectOutputStream(fos);
Student s1=new Student("张三","12345");
Student s2=new Student("王五","54321");
oos.writeObject(s1);
oos.writeObject(s2);
oos.close();
FileInputStream fis=new FileInputStream("test.obj");
ObjectInputStream ois=new ObjectInputStream(fis);
Student s3=(Student)ois.readObject();
Student s4=(Student)ois.readObject();
System.out.println(s3);
System.out.println(s4);
ois.close();
}
catch(IOException e)
{
e.printStackTrace();
}
catch(ClassNotFoundException e1)
{
e1.printStackTrace();
}
}
}
count为Student的静态成员变量,书上说静态成员不能被序列化,那为什么此程序运行的结果:输出s3和s4时 count为2呢,按理说应该为0才对啊?还是我哪里理解错了,请高手指定,谢谢了
[解决办法]
静态成员属于类级别的,所以不能序列化
这里的不能序列化的意思,是序列化信息中不包含这个静态成员域
你这个测试成功,是因为你都在同一个机器(而且是同一个进程),因为你这个jvm已经把count加载进来了,所以你获取的是加载好的count,如果你是传到另一台机器或者你关掉程序重写写个程序读入test.obj,此时因为别的机器或新的进程是重新加载count的,所以count信息就是初始时的信息。
[解决办法]
同意1楼的
[解决办法]
静态变量不属于对象,属于类。不能被序列化。还有瞬态的变量也不能被序列化 。
test.obj ,你打开看看 ,里面是没有
private transient String password;
private static int count=0;
这两个属性的 序列化信息的 ,输出2是因为 在当前jvm实例中count = 2 ;
你可以这样测试,写两个类来测试,一个类写入序列化文件, 另外一个类读出序列化文件
如 :
import java.io.*;class Student1 implements Serializable { private static final long serialVersionUID = 1L; private String name; private transient String password; private static int count = 0; public Student1(String name, String password) { System.out.println("调用Student的带参的构造方法"); this.name = name; this.password = password; count++; } public String toString() { return "人数: " + count + " 姓名: " + name + " 密码: " + password; }}public class ObjectSerTest1 { public static void main(String args[]) { try { FileOutputStream fos = new FileOutputStream("test.obj"); ObjectOutputStream oos = new ObjectOutputStream(fos); Student1 s1 = new Student1("张三", "12345"); Student1 s2 = new Student1("王五", "54321"); oos.writeObject(s1); oos.writeObject(s2); oos.close(); FileInputStream fis = new FileInputStream("test.obj"); ObjectInputStream ois = new ObjectInputStream(fis); Student1 s3 = (Student1) ois.readObject(); Student1 s4 = (Student1) ois.readObject(); System.out.println(s3); System.out.println(s4); ois.close(); } catch (IOException e) { e.printStackTrace(); } catch (ClassNotFoundException e1) { e1.printStackTrace(); } }}
[解决办法]
执行下下面的代码就可以推翻你的想法了,static不序列化的
import java.io.*;class Student implements Serializable{ private String name; private transient String password; private static int count=0; public Student(String name,String password) { System.out.println("调用Student的带参的构造方法"); this.name=name; this.password=password; count++; } public String toString() { return "人数: "+count+" 姓名: "+name+" 密码: "+password; }}public class ObjectSerTest1{public static void main(String args[]){try{FileOutputStream fos=new FileOutputStream("test.obj");ObjectOutputStream oos=new ObjectOutputStream(fos);Student s1=new Student("张三","12345");Student s2=new Student("王五","54321");System.out.println(s1);System.out.println(s2);oos.writeObject(s1);oos.writeObject(s2);oos.close();new Student("xyq","i love you");FileInputStream fis=new FileInputStream("test.obj");ObjectInputStream ois=new ObjectInputStream(fis); Student s3=(Student)ois.readObject();Student s4=(Student)ois.readObject(); System.out.println(s3);System.out.println(s4); ois.close();}catch(IOException e){e.printStackTrace();}catch(ClassNotFoundException e1){e1.printStackTrace();}}}