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How to prevent Serialization to break a Singleton Class Pattern?
A Singleton pattern states that a class can have a single instance and multiple instances are not permitted to be created. For this purpose, we make the constructor of the class a private and return a instance via a static method. But using serialization, we can still create multiple instance of a class. See the example below −
Example - Breaking Singleton
import java.io.ByteArrayInputStream; import java.io.ByteArrayOutputStream; import java.io.IOException; import java.io.ObjectInputStream; import java.io.ObjectOutputStream; import java.io.Serializable; public class Tester{ public static void main(String[] args) throws ClassNotFoundException, IOException{ A a = A.getInstance(); A b = (A) getSerializedCopy(a); System.out.println(a.hashCode()); System.out.println(b.hashCode()); } public static Object getSerializedCopy(Object sourceObject) throws IOException, ClassNotFoundException { ObjectOutputStream objectOutputStream = null; ObjectInputStream objectInputStream = null; ByteArrayOutputStream byteArrayOutputStream = new ByteArrayOutputStream(); objectOutputStream = new ObjectOutputStream(byteArrayOutputStream); objectOutputStream.writeObject(sourceObject); objectOutputStream.flush(); objectInputStream = new ObjectInputStream( new ByteArrayInputStream(byteArrayOutputStream.toByteArray())); return objectInputStream.readObject(); } } class A implements Serializable { private static A a; private A(){} public static A getInstance(){ if(a == null){ a = new A(); } return a; } }
Output
1550089733 865113938
Here you can see, we've created another object of a Singleton class. Let's see how to prevent such a situation −
Override readResolve() method in the singleton class.
Example - Protecting Singleton
// implement readResolve method protected Object readResolve() { return a; }
Output
1550089733 1550089733
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