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ClassNotFoundException when deserializing a binary class file's contents

I don't know much about Java. I'm trying to read a file containing an int and various instances of a class called "Automobile". When I deserialize it, though, the program throws a ClassNotFoundException and I can't seem to understand why.

Here's the code:

        try {
        FileInputStream fin = new FileInputStream(inputFile);
        ObjectInputStream input = new ObjectInputStream(fin);

        conto = input.readInt();

        Automobile[] macchine = new Automobile[conto];

        for(int i = 0; i < conto; i++) {
            macchine[i] = (Automobile)input.readObject();
        }

        String targa;
        System.out.print("\nInserire le cifre di una targa per rintracciare l'automobile: ");
        targa = sc1.nextLine();

        for(int i = 0; i < conto; i++) {
            if(macchine[i].getTarga().equals(targa))
                System.out.println(macchine[i]);
        }

    } catch(IOException e) {
        System.out.println("Errore nella lettura del file "+inputFile);
    } catch(java.lang.ClassNotFoundException e) {
        System.out.println("Class not found");
    }

Thanks in advance.

EDIT: here's the stacktrace

java.lang.ClassNotFoundException: es4.Automobile
    at java.net.URLClassLoader$1.run(URLClassLoader.java:202)
    at java.security.AccessController.doPrivileged(Native Method)
    at java.net.URLClassLoader.findClass(URLClassLoader.java:190)
    at java.lang.ClassLoader.loadClass(ClassLoader.java:307)
    at sun.misc.Launcher$AppClassLoader.loadClass(Launcher.java:301)
    at java.lang.ClassLoader.loadClass(ClassLoader.java:248)
    at java.lang.Class.forName0(Native Method)
    at java.lang.Class.forName(Class.java:247)
    at java.io.ObjectInputStream.resolveClass(ObjectInputStream.java:604)
    at java.io.ObjectInputStream.readNonProxyDesc(ObjectInputStream.java:1575)
    at java.io.ObjectInputStream.readClassDesc(ObjectInputStream.java:1496)
    at java.io.ObjectInputStream.readOrdinaryObject(ObjectInputStream.java:1732)
    at java.io.ObjectInputStream.readObject0(ObjectInputStream.java:1329)
    at java.io.ObjectInputStream.readObject(ObjectInputStream.java:351)
    at es4p2.Main.main(Main.java:35)
like image 439
Gabriele Cirulli Avatar asked Jun 09 '11 22:06

Gabriele Cirulli


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3 Answers

When you deserialize a serialized object tree, the classes of all the objects have to be on the classpath. In this context, a ClassNotFoundException most likely means that one of the classes required is not on the classpath. You have to address this for deserialization to work.

In this case, the es4.Automobile is missing.


Could the problem be caused by a custom exception I made which is fired by Automobile?

The only other possibilities I can think of are:

  • es4.Automobile has a direct or indirect dependency on some other class that is missing
  • the static initialization of es4.Automobile or a dependent class has thrown an exception that has not been caught internally to the class.

But both of those should (I think) have resulted in a different stack trace.


I just noticed the package name is es4p2, not es4. Why does it say es4? Could it be because the program which saves the file uses another package name?

I've no idea why they are different. You'd need to talk to whoever wrote the code / produced the serialized objects. However, this is most likely the cause of your problem. A class with a different package name is a different class. Period.


You should always output (or better, log) the stacktrace when an unexpected exception is caught. That will tell you (and us) more about what has gone wrong, and in this case the name of the class that is missing.

like image 104
Stephen C Avatar answered Oct 04 '22 08:10

Stephen C


This generally happens if your class Automobile is not in the runtime classpath.

like image 41
adarshr Avatar answered Oct 02 '22 08:10

adarshr


This is and old question but this may help someone else. I faced the same issue and the problem was that I was not using the current thread class loader. You will find below the serializer class that I used in a grails project, should be quite straightforward use this in java Hope this helps

public final class Serializer<T> {

/**
 * Converts an Object to a byte array.
 *
 * @param object, the Object to serialize.
 * @return, the byte array that stores the serialized object.
 */

public static byte[] serialize(T object) {

    ByteArrayOutputStream bos = new ByteArrayOutputStream()
    ObjectOutput out = null
    try {
        out = new ObjectOutputStream(bos)
        out.writeObject(object)

        byte[] byteArray = bos.toByteArray()
        return byteArray

    } catch (IOException e) {
        e.printStackTrace()
        return null

    } finally {
        try {
            if (out != null)
                out.close()
        } catch (IOException ex) {
            ex.printStackTrace()
            return null
        }
        try {
            bos.close()
        } catch (IOException ex) {
            ex.printStackTrace()
            return null
        }
    }

}

/**
 * Converts a byte array to an Object.
 *
 * @param byteArray, a byte array that represents a serialized Object.
 * @return, an instance of the Object class.
 */
public static Object deserialize(byte[] byteArray) {
    ByteArrayInputStream bis = new ByteArrayInputStream(byteArray)
    ObjectInput input = null
    try {
        input = new ObjectInputStream(bis){
            @Override protected Class<?> resolveClass(final ObjectStreamClass desc) throws IOException, ClassNotFoundException {
                ClassLoader cl = Thread.currentThread().getContextClassLoader();
                if (cl == null)  return super.resolveClass(desc);
                return Class.forName(desc.getName(), false, cl);
            }
        };
        Object o = input.readObject()
        return o

    } catch (ClassNotFoundException | IOException e) {
        e.printStackTrace()
        return null
    } finally {
        try {
            bis.close()
        } catch (IOException ex) {
        }
        try {
            if (input != null)
                input.close()
        } catch (IOException ex) {
            ex.printStackTrace()
            return null
        }
    }
}
like image 31
Maximiliano Sibert Avatar answered Oct 04 '22 08:10

Maximiliano Sibert