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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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