Why is this class not Serializable? - java

I was using the Mersenne-Twister implementation at http://www.math.sci.hiroshima-u.ac.jp/~m-mat/MT/VERSIONS/JAVA/MTRandom.java as a drop-in replacement for the default java.util.Random class. However, four fields (an int, a boolean and two byte[]) are marked as transient. This means that I can't serialize an object of this class without implementing custom functionality.
The question is, is there any reason that these fields are marked transient? Is there any code in there that holds information that won't make any sense when the object is read in from a file? I removed the transient modifier from the fields and it seems to work fine, but I haven't tested it intensively and so might there be cases where it breaks?
Personally, I can't see why, since all that's done in the class is arithmetic.

From the comment on serialVersionUID, it looks like the author didn't want to consider serialisation. Adding transient may have suppressed some compiler/IDE warnings.

Most likely the reasoning behind making all of the non-static fields of the class transient was so that the MTRandom class stays binary compatible with java.util.Random, from which it is extended.
So theoretically, you could serialize an MTRandom instance, and deserialize it as a Random instance and everything would work.
If those fields aren't transient, then they would be serialized and become incompatible.
However, as far as I can tell, removing the transients shouldn't cause a problem for you.

Related

How are incompatible classes with same serialVersionUID handled? [duplicate]

I have a class that is serialised. Now I need to add a new variable into the class, with setter and getter methods. This class is sent over wire in RMI.
Without changing the UID, can I add new parameters and getter and setter methods for it? I tried to write an example class that is sent over wire, and did not change the UID, and added new parameters and getter and setter methods for it. On the other end, I tested it and I still got the values properly. I had assumed, if I add new parameters, getter and setter methods, I need to change the UID. Am I wrong?
If you hard-code the SerialVersionUID of a class, (to 1L, usually), store some instances, and then re-define the class, you basically get this behavior (which is more or less common sense):
New fields (present in class definition, not present in the serialized instance) are assigned a default value, which is null for objects, or the same value as an uninitialized field for primitives.
Removed fields (not present in class definition but present in the serialized instance) are simply ignored.
So the general rule of thumb is, if you simply add fields and methods, and don't change any of the existing stuff, AND if you're OK with default values for these new fields, you're generally OK.
Wow, a lot of bad information.
Java serialization is +very+ robust. There are a very well defined set of rules governing backwards compatibility of objects with the same uid and different data. the basic idea is that as long as you don't change the the type of an existing member, you can maintain the same uid without data issues.
that said, your code still needs to be smart about handling classes with potentially missing data. the object may deserialize correctly, but there may not be data in certain fields (e.g. if you added a field to the class and are deserializing an old version of the class). if your code can handle this, than you can probably keep the current uid. if not, then you should probably change it.
in addition to the pre-defined rules, there are advanced usage scenarios where you could even change the type of existing fields and still manage to deserialize the data, but that generally only necessary in extreme situations.
java serialization is very well documented online, you should be able to find all this information in the relevant sun/oracle tutorials/docs.
This only matters if you let Java generate a default UID for your class. It uses the actual members and methods of the class to generate it, thus making it invalid once you change the class structure. If you provide an UID for your class then this only matters if you need to deserialize older versions of your class from a file and such.
Want to define few point to highlight the changes which impacts serialization.
Below you will find the link to Oracle Java Docs for more details.
Incompatible Changes
Incompatible changes to classes are those changes for which the guarantee of interoperability cannot be maintained. The incompatible changes that may occur while evolving a class are:
Deleting fields
Moving classes up or down the hierarchy
Changing a nonstatic field to static or a nontransient field to transient
Changing the declared type of a primitive field
Changing the writeObject or readObject method so that it no longer writes or reads the default field data or changing it so that it attempts to write it or read it when the previous version did not.
Changing a class from Serializable to Externalizable or vice versa.
Changing a class from a non-enum type to an enum type or vice versa.
Removing either Serializable or Externalizable.
Adding the writeReplace or readResolve method to a class, if the behavior would produce an object that is incompatible with any older version of the class.
Link from where the above information is taken
http://docs.oracle.com/javase/7/docs/platform/serialization/spec/version.html#6678

Creating Immutable Objects in Java

I'd like to create a few immutable objects for my codebase. What's the best way to really deliver the message that a given class is intended to be immutable? Should I make all of my fields final, and initialize during object construction? (This seems really awkward...) Should I create some Immutable interface, and have objects implement it? (Since Java doesn't have some standard interface behind this, I thought they had some other way of dealing with it.) What's the standard way this is dealt with? (If it's simply done by adding a bunch of comments around the fields exclaiming that they shouldn't be modified once initialized, that's fine too.)
Should I make all of my fields final, and initialize during object construction?
Yes. And ensure that those types are themselves immutable, or that you create copies when you return values from getter methods. And make the class itself final. (Otherwise your class on its own may be immutable, but that doesn't mean that any instance of your class would be immutable - because it could be an instance of a mutable subclass.)
(This seems really awkward...)
It's hard to know what to suggest without knowing how you find it to be awkward - but the builder pattern can often be useful. I usually use a nested static class for that, often with a static factory method. So you end up with:
Foo foo = Foo.newBuilder()
.setName("asd")
.setPoints(10)
.setOtherThings("whatever")
.build();
Yes and no. Making all fields final is not a guarantee in and of itself. If you'd like to get really in-depth with this there are a number of chapters in Effective Java by Joshua Bloch dealing with immutability and the considerations involved. Item 15 in Effective Java covers the bulk of it and references the other items in question.
He offers these five steps:
Don’t provide any methods that modify the object’s state (known as muta-
tors).
Ensure that the class can’t be extended.
Make all fields final.
Make all fields private.
Ensure exclusive access to any mutable components.
One way to learn how to do all of this is to see how the language designers make classes immutable by reviewing the source for classes like String which are immutable (for example see http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/lang/String.java).
Write a unit test that will fail if your coworkers make the class mutable.
Using Mutability Detector, you can write a test like this:
import static org.mutabilitydetector.unittesting.MutabilityAssert.assertImmutable;
#Test public void isImmutable() {
assertImmutable(MyImmutableThing.class)
}
If a coworker comes along, and, for example, adds a setter method to your class, the test will fail. Your use case is one of the core purposes of Mutability Detector.
Disclaimer: I wrote it.

ObjectOutputStream doesn't correctly save fields of superclasses unless they're Serializable, why no error or warning?

Say I have an ArrayList<B> array of objects from a certain class B, which extends A. B has an instance field bb and A a field aa. I know that saving array to a .dat-file using ObjectOutputStream requires that B (not just ArrayList!) implement Serializable. I've found, however, that when loading the object back from the file (using an ObjectInputStream):
arrayLoaded = (ArrayList<B>)myObjIn.readObject();
the loaded array isn't identical to the original array: In the particular case, arrayLoaded.get(0).bb has the same value as in array, but arrayLoaded.get(0).aa is "zeroed". It has a default initialization value, regardless of its value when array was saved to file. However, this problem is solved by letting also A implement Serializable.
What bothers me is that this error is so subtle: no exception, no warning (in eclipse), nothing. Is there a reason for this or is this simply an oversight by the java developers? Do I just have to accept it and think hard about which classes in the hierarchy implement Serializable every time I want to use object IO streams?
Just because B implements Serializable, that does not retroactively include the fields of the non-serializable superclass in what gets serialized. (This makes sense, especially when you consider that being able to serialize private and package-private fields of any class just by extending it and implementing Serializable would violate its encapsulation.)
A field declared in A will behave the same as a field declared as transient in B. There is a workaround however. From the documentation for Serializable:
To allow subtypes of non-serializable classes to be serialized, the
subtype may assume responsibility for saving and restoring the state
of the supertype's public, protected, and (if accessible) package
fields. The subtype may assume this responsibility only if the class
it extends has an accessible no-arg constructor to initialize the
class's state.
So you will need to implement writeObject and readObject in B to handle the serialization/deserialization of A.aa.
What bothers me is that this error is so subtle: no exception, no warning (in eclipse), nothing. Is there a reason for this or is this simply an oversight by the java developers?
It is by design. (See #Paul Bellora's answer). The alternatives would be to:
Make it illegal to declare a class Serializable unless its superclass is Serializable. That's obviously unworkable.
Automatically serialize the superclasses fields which breaks if the should or can't be serialized. (Note that we can't rely on transient here, because if the designer of the superclass didn't intend the it to be serializable, he/she won't have labelled the fields.)
Do I just have to accept it and think hard about which classes in the hierarchy implement Serializable every time I want to use object IO streams?
Basically, yes. In particular, you need to think hard when you write a Serializable subclass of an existing non-Serializable class.
I guess it is possible to write FindBugs / PMD / etc rules to flag this particular usage as potentially problematic.

Unit testing for object immutability

I want to make sure that a given group of objects is immutable.
I was thinking about something along the lines of:
check if every field is private final
check if class is final
check for mutable members
So I guess my question is: is 3. possible ?
I can check recursively whether every member of a class has its fields private final, but this is not enough since a class can have e method named getHaha(param) which adds the given param to an array for instance.
So is there a good way to check if an object is immutable or is it even possible ?
Thanks,
You may want to check out this project:
Mutability Detector
This library attempts to analyse the bytecode of a particular class, to discover if it is immutable or not. It allows testing for this condition in a unit test, as demonstrated in a video available here. It is certainly not perfect (a String field will be considered mutable, and your array example is not handled well) but it's more sophisticated than what FindBugs offers (i.e. only checking that every field is final).
Disclaimer: I wrote it ;-)
If you generate your data model and all its code, you can ensure the possible Data Value objects you create will be immutable to meet your needs.
The problem you have is that there is different forms of immutability. Even String would fail your test Are String, Date, Method immutable? You can prove that a class is strictly immutable this way, but you are likely to be better off generating your data model.
Yes, you can write an immutability detector.
First of all, you are not going to be just writing a method which determines whether a class is immutable; instead, you will need to write an immutability detector class, because it is going to have to maintain some state. The state of the detector will be the detected immutability of all classes which it has examined so far. This is not only useful for performance, but it is actually necessary because a class may contain a circular reference, which would cause a simplistic immutability detector to fall into infinite recursion.
The immutability of a class has four possible values: Unknown, Mutable, Immutable, and Calculating. You will probably want to have a map which associates each class that you have encountered so far to an immutability value. Of course, Unknown does not actually need to be implemented, since it will be the implied state of any class which is not yet in the map.
So, when you begin examining a class, you associate it with a Calculating value in the map, and when you are done, you replace Calculating with either Immutable or Mutable.
For each class, you only need to check the field members, not the code. The idea of checking bytecode is rather misguided.
First of all, you should not check whether a class is final; The finality of a class does not affect its immutability. Instead, a method which expects an immutable parameter should first of all invoke the immutability detector to assert the immutability of the class of the actual object that was passed. This test can be omitted if the type of the parameter is a final class, so finality is good for performance, but strictly speaking not necessary. Also, as you will see further down, a field whose type is of a non-final class will cause the declaring class to be considered as mutable, but still, that's a problem of the declaring class, not the problem of the non-final immutable member class. It is perfectly fine to have a tall hierarchy of immutable classes, in which all the non-leaf nodes must of course be non-final.
You should not check whether a field is private; it is perfectly fine for a class to have a public field, and the visibility of the field does not affect the immutability of the declaring class in any way, shape, or form. You only need to check whether the field is final and its type is immutable.
When examining a class, what you want to do first of all is to recurse to determine the immutability of its super class. If the super is mutable, then the descendant is by definition mutable too.
Then, you only need to check the declared fields of the class, not all fields.
If a field is non-final, then your class is mutable.
If a field is final, but the type of the field is mutable, then your class is mutable. (Arrays are by definition mutable.)
If a field is final, and the type of the field is Calculating, then ignore it and proceed to the next field. If all fields are either immutable or Calculating, then your class is immutable.
If the type of the field is an interface, or an abstract class, or a non-final class, then it is to be considered as mutable, since you have absolutely no control over what the actual implementation may do. This might seem like an insurmountable problem, because it means that wrapping a modifiable collection inside an UnmodifiableCollection will still fail the immutability test, but it is actually fine, and it can be handled with the following workaround.
Some classes may contain non-final fields and still be effectively immutable. An example of this is the String class. Other classes which fall into this category are classes which contain non-final members purely for performance monitoring purposes (invocation counters, etc.), classes which implement popsicle immutability (look it up), and classes which contain members that are interfaces which are known to not cause any side effects. Also, if a class contains bona fide mutable fields but promises not to take them into account when computing hashCode() and equals(), then the class is of course unsafe when it comes to multi-threading, but it can still be considered as immutable for the purpose of using it as a key in a map. So, all these cases can be handled in one of two ways:
Manually adding classes (and interfaces) to your immutability detector. If you know that a certain class is effectively immutable despite the fact that the immutability test for it fails, you can manually add an entry to your detector which associates it with Immutable. This way, the detector will never attempt to check whether it is immutable, it will always just say 'yes, it is.'
Introducing an #ImmutabilityOverride annotation. Your immutability detector can check for the presence of this annotation on a field, and if present, it may treat the field as immutable despite the fact that the field may be non-final or its type may be mutable. The detector may also check for the presence of this annotation on the class, thus treating the class as immutable without even bothering to check its fields.
I hope this helps future generations.
I doubt you can do this with unit tests. The best way would be to be careful during writing the class or looking into the code. Precisely because of the problem that methods on the object can mutate its state which you might not see from the outside. Just because it's discouraged doesn't mean it doesn't happen :-)
Pretty sure it is impossible. Consider this function:
public void doSomething() {
if (System.currentTimeMillis() % 100000 == 0) {
this.innerMember.changeState();
}
}
First, you won't be able to detect it by running every class function, as this function changes the state of object precisely only once in 100 seconds.
Second, you won't be able to detect it by parsing code, as you do not know if changeState() function changes the state of innerMember or not.
This thread can help How do I identify immutable objects in Java. Take a look at the second popular answer, it might be possible to check for any immutability problems with FindBugs. If you run it on every commit then you can call it a unit test :)
EDIT
It seems that FindBugs only check for final, that's not much. You could implement your own rule according to you patterns and classes which you use in the code.

Java: Static transient fields

I just found out in Java you can declare a field 'static transient' - the compiler doesn't complain. This doesn't seem to be useful in any way since static fields are not serialized, as we all know.
But I wonder, is there actually a case where 'static transient' fields are useful?
Nope - you said it yourself, static fields aren't serialized.
Kinda weird that the compiler lets you do that though.
In most cases, it is not useful. Static fields are indeed not serialized by the default serializer.
However, static transient fields can be detected via reflection. If someone writes its own serializer and he wants to also serialize static fields, then he might take the transient keyword in consideration and skip the serialization of that particular field.
PS: This answer is posted for the sake of completeness, and is based on Peter Lawrey's comment. Credits to him.

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