serialization in java

by Geethalakshmi 2011-01-04 10:09:35

Serialization in Java


Java object serialization is used to persist Java objects to a file, database, network, process or any other system. Serialization flattens objects into an ordered, or serialized stream of bytes. The ordered stream of bytes can then be read at a later time, or in another environment, to recreate the original objects.

Java serialization does not cannot occur for transient or static fields. Marking the field transient prevents the state from being written to the stream and from being restored during deserialization. Java provides classes to support writing objects to streams and restoring objects from streams. Only objects that support the java.io.Serializable interface or the java.io.Externalizable interface can be written to streams.
public interface Serializable

The Serializable interface has no methods or fields. (Marker Interface)
Only objects of classes that implement java.io.Serializable interface can be serialized or deserialized.

Serializing an Object:

The ObjectOutputStream class is used to serialize an Object. The following SerializeDemo program instantiates an Employee object and serializes it to a file.

When the program is done executing, a file named employee.ser is created. The program does not generate any output, but study the code and try to determine what the program is doing.

Note: When serializing an object to a file, the standard convention in Java is to give the file a .ser extension.

import java.io.*;

public class SerializeDemo
{
public static void main(String [] args)
{
Employee e = new Employee();
e.name = "Reyan Ali";
e.address = "Phokka Kuan, Ambehta Peer";
e.SSN = 11122333;
e.number = 101;
try
{
FileOutputStream fileOut =
new FileOutputStream("employee.ser");
ObjectOutputStream out =
new ObjectOutputStream(fileOut);
out.writeObject(e);
out.close();
fileOut.close();
}catch(IOException i)
{
i.printStackTrace();
}
}
}


Deserializing an Object:

The following DeserializeDemo program deserializes the Employee object created in the SerializeDemo program. Study the program and try to determine its output:

import java.io.*;
public class DeserializeDemo
{
public static void main(String [] args)
{
Employee e = null;
try
{
FileInputStream fileIn =
new FileInputStream("employee.ser");
ObjectInputStream in = new ObjectInputStream(fileIn);
e = (Employee) in.readObject();
in.close();
fileIn.close();
}catch(IOException i)
{
i.printStackTrace();
return;
}catch(ClassNotFoundException c)
{
System.out.println(.Employee class not found.);
c.printStackTrace();
return;
}
System.out.println("Deserialized Employee...");
System.out.println("Name: " + e.name);
System.out.println("Address: " + e.address);
System.out.println("SSN: " + e.SSN);
System.out.println("Number: " + e.number);
}
}


This would produce following result:

Deserialized Employee...
Name: Reyan Ali
Address:Phokka Kuan, Ambehta Peer
SSN: 0
Number:101
Here are following important points to be noted:

The try/catch block tries to catch a ClassNotFoundException, which is declared by the readObject() method. For a JVM to be able to deserialize an object, it must be able to find the bytecode for the class. If the JVM can't find a class during the deserialization of an object, it throws a ClassNotFoundException.

Notice that the return value of readObject() is cast to an Employee reference.

The value of the SSN field was 11122333 when the object was serialized, but because the field is transient, this value was not sent to the output stream. The SSN field of the deserialized Employee object is 0.

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