Preventing subclasses from adding methods - java

This might seem like an odd thing to want, but is there a way in Java to stop subclasses from adding new methods (including constructors) whilst still allowing subclasses to override methods?
The actual situation is where we have an abstract class with some abstract methods and a constructor
abstract class A {
abstract A doX();
abstract boolean isY();
public A(String s){ ... };
}
and we want all concrete subclasses of this class to only override these methods and constructor.
This is about enforcing a certain style in our code i.e. stopping other people working on the code from adding things. We could just tell them not to, but that rarely works, so we wondered if there was a programmatic way of achieving this.
Obviously the class cannot be final. Efficiency isn't paramount - cleaner code is more important.
Update - dynamic approach
As has been pointed out in the answers, there is no way to do this statically as the only way to prevent subclasses being created is using final, which won't work. But I could use a dynamic approach so my current solution is to add this aspect to the project (which already uses AspectJ).
public aspect WatchA{
before() : execute(* A.*()) || execute(* A.*(..)) {
String methodCalled = joinPoint.getSignature().getName();
Class<?> c = Class.forName(args[0])
Method[] allMethods = c.getDeclaredMethods();
boolean found = false;
for(Method m : allMethods)
found |= m.getName().equals(methodCalled);
if(!found)
throw new RuntimeException("Do not add method "+methodCalled+" to A");
}
}
Which will cause their tests to fail if they use any of these new methods.

You cannot do that. Only if classes are final can you ensure that no subclass can be created.
You can also make methods final (even in abstract classes) so that overriding them is forbidden.
Your best bet is to create an interface, with all methods you want visible, and force all users of your API to access the objects via this interface. This way, even if implementations add their own stuff, said stuff won't be visible.
One solution for this is to implement a factory to return the concrete classes; for "added security", you could put all implementations in the same package as this factory and make constructors package local (but this is often not practical):
public final class MyFactory
{
// ....
public MyInterface getConcrete()
{
return new MyInterfaceImpl();
}
// etc etc -- getStones(), getTar(), getFeathers() and so on
}
Note that builders can also be used for that.

If you really wan't to do this.. one way would be to programatically check in the abstract class constructor that the methods defined in the class are those that are allowed.
import java.lang.reflect.Method;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
public abstract class Base {
private static final Set<String> allowedMethodNames = new HashSet<>(Arrays.asList("doThis", "wait", "wait", "wait", "equals", "toString", "hashCode", "getClass", "notify", "notifyAll"));
public Base() {
Set<String> allMethods = new HashSet<>();
for (Method aMethod : getClass().getMethods()) {
allMethods.add(aMethod.getName());
}
if (!allowedMethodNames.equals(allMethods)) {
allMethods.removeAll(allowedMethodNames);
throw new IllegalStateException("Following methods not allowed <" + allMethods + ">");
}
}
public abstract void doThis();
}
public class Disallowed extends Base {
#Override
public void doThis() {
System.out.println("dooooooo");
}
public void doSomethingElse() {
System.out.println("not allowed");
}
public static void main(String[] args) {
new Allowed().doThis();
new Disallowed();
}
}
public class Allowed extends Base {
#Override
public void doThis() {
System.out.println("doing this");
}
}
When someone is trying create an instance of 'Disallowed' it would fail. However 'new Allowed().doThis()' will work fine.
A more graceful way to do this would be to introduce a custom annotation + annotation processor and do the same check during the compilation time.

There is no such way.
Why would you want to enforce such a coding style?
If you really must enforce such a style you could create a "rule enforcer" which checks your classpath and compares the methods of your abstract parent classes with their sub classes.

It is Java which means flexibility. So java gives you more convinient in using the abstract methods and overriding them from your subclasses. Also one should have an idea of adding new methods to these subclasses. Even java can't change this. If it does then the whole Java community crash. It is impossible that you can prevent from adding methods to their subclasses. Only you can stop them extending your classes and overridding your methods.

Related

What pattern should be used, strategy?

I do have a service which needs to handle two types of meal.
#Service
class MealService {
private final List<MealStrategy> strategies;
MealService(…) {
this.strategies = strategies;
}
void handle() {
var foo = …;
var bar = …;
strategies.forEach(s -> s.remove(foo, bar));
}
}
There are two strategies, ‘BurgerStrategy’ and ‘PastaStrategy’. Both implements Strategy interface with one method called remove which takes two parameters.
BurgerStrategy class retrieves meals of enum type burger from the database and iterate over them and perform some operations. Similar stuff does the PastaStrategy.
The question is, does it make sense to call it Strategy and implement it this way or not?
Also, how to handle duplications of the code in those two services, let’s say both share the same private methods. Does it make sense to create a Helper class or something?
does it make sense to call it Strategy and implement it this way or not
I think these classes ‘BurgerStrategy’ and ‘PastaStrategy’ have common behaviour. Strategy pattern is used when you want to inject one strategy and use it. However, you are iterating through all behaviors. You did not set behaviour by getting one strategy and stick with it. So, in my honour opinion, I think it is better to avoid Strategy word here.
So strategy pattern would look like this. I am sorry, I am not Java guy. Let me show via C#. But I've provided comments of how code could look in Java.
This is our abstraction of strategy:
public interface ISoundBehaviour
{
void Make();
}
and its concrete implementation:
public class DogSound : ISoundBehaviour // implements in Java
{
public void Make()
{
Console.WriteLine("Woof");
}
}
public class CatSound : ISoundBehaviour
{
public void Make()
{
Console.WriteLine("Meow");
}
}
And then we stick with one behaviour that can also be replaced:
public class Dog
{
ISoundBehaviour _soundBehaviour;
public Dog(ISoundBehaviour soundBehaviour)
{
_soundBehaviour = soundBehaviour;
}
public void Bark()
{
_soundBehaviour.Make();
}
public void SetAnotherSound(ISoundBehaviour anotherSoundBehaviour)
{
_soundBehaviour = anotherSoundBehaviour;
}
}
how to handle duplications of the code in those two services, let’s say both share the same private methods.
You can create one base, abstract class. So basic idea is to put common logic into some base common class. Then we should create abstract method in abstract class. Why? By doing this, subclasses will have particular logic for concrete case. Let me show an example.
An abstract class which has common behaviour:
public abstract class BaseMeal
{
// I am not Java guy, but if I am not mistaken, in Java,
// if you do not want method to be overriden, you shoud use `final` keyword
public void CommonBehaviourHere()
{
// put here code that can be shared among subclasses to avoid code duplication
}
public abstract void UnCommonBehaviourShouldBeImplementedBySubclass();
}
And its concrete implementations:
public class BurgerSubclass : BaseMeal // extends in Java
{
public override void UnCommonBehaviourShouldBeImplementedBySubclass()
{
throw new NotImplementedException();
}
}
public class PastaSubclass : BaseMeal // extends in Java
{
public override void UnCommonBehaviourShouldBeImplementedBySubclass()
{
throw new NotImplementedException();
}
}

How to prevent sublasses from default implementing a method of super class?

I have a method which adds Objects to an static list like this:
#PostConstruct
protected void registerToTransactionList() {
TransactionValidator.registerTransactionList(this);
}
registerTransactionList method just adds "this" to the static list, this method is in BalanceTransactionValidator class which extends TransactionValidator (owner of static list),the problem is all subclasses of BalanceTransactionValidator class are added to static list either,and if I override registerToTransactionList method in them like this:
#Override
#PostConstruct
protected void registerToTransactionList() {
}
It doesn't add subclasses but doesn't add BalanceTransactionValidator either. Can anybody help me on this? Please notice sublasses are overriding this method by default.
make the method private to block the visibility
private void registerToTransactionList() {
}
or make the method final to block it from been override
protected final void registerToTransactionList() {
}
There are two ways of achieving that:
Keep your method as it is; but then you have to actively check for the type of your objects before externally calling that method
Change your whole logic and make that method private
It won't help to make the method final as suggested in one of the comments - your problem is not that subclasses are overwriting that method; in essence, you have a design problem: you wish that subclasses should not invoke that method at all.
So, the only real option that makes sense here is "2.". You see, by having public method on a class that you want to be extended you are implicitly saying: it is perfectly fine to call that method; on any object that is instance of the base class (or child class!).
And in your case, that is not true: you actually do not want that the code behind this method runs for child classes. Then you shouldn't put that method in the list of public/protected methods of your base class!
Finally: you might want to step back and do some reading about good OO design. Class hierarchies do not fall from the sky: you willfully design them for a certain purpose. In other words: there is more to inheritance than just putting some "A extends B" on your class declaration. You have to understand each and every method on your B class; and how your child classes should deal with them!
EDIT: after some more thinking, I guess you are doing things "the wrong way", like:
class BaseClass {
public final void doRegistration() {
BaseClass toRegister = getObjectForRegistration();
if (toRegister != null) { ... register toRegister ...
}
protected BaseClass getObjectForRegistration() {
return null;
}
With that code, you could then put
protected BaseClass getObjectForRegistration() {
if (this instanceof ClassThatShouldBeRegistered) {
return this;
}
return null;
}
into that one class that wants to be registered. Probably there could be even nicer ways of doing so; but after some thinking I don't see how we could avoid the instanceof. But the above code should work; and it only requires specific code only in your base class and in that one class that wants to register something.

Using methods from a subclass on an object that is an instance of the superclass

Let's say there's a class that I use extensively and is returned by a method.
CommonClass obj = getCommonObject();
Now I want to extend this class to create some utility method to avoid repeating myself.
public CommonClassPlus extends CommonClass {
public String dontRepeatYourself() {
// the reason I'm creating a subclass
}
}
Of course I would like to use my improved class for the method above, however, downcasting isn't allowed.
CommonClassPlus obj = getCommonObject();
//Cannot cast to CommonClassPlus
How can I use the method dontRepeatYourself() if I can only work with the object that is an instance of the superclass?
CommonClass and getCommonObject() are from an external library and I cannot change them.
You cannot add behavior to an existing instance in Java (like you could in JavaScript, for example).
The closest you can get in Java is the Decorator pattern:
CommonClassPlus obj = decorate(getCommonObject());
where decorate() is
public CommonClassPlus decorate(CommonClass x) {
return new CommonClassPlus(x);
}
This approach creates a potentially huge amount of boilerplate because it must delegate each method call to the wrapped instance. If a method in CommonClass is final and there is no interface you can reimplement, then this approach fails altogether.
In most cases you will be able to get along with a simple static helper method:
public static String dontRepeatYourself(CommonClass x) {
...
}
If CommonClass is from an external library, you probably want to wrap it in an Adapter Pattern anyway, using the principle of Composition over Inheritance.
This gives you complete control if you want to, say, change the library you're using, and allows you to add functionality like dontRepeatYourself().
public class CommonClassAdapter implements MyAdapter {
private final CommonClass common;
private final String cachedResult;
// Note that I'm doing dependency injection here
public CommonClassAdapter(CommonClass common) {
this.common = common;
// Don't expose these because they shouldn't be called more than once
common.methodIOnlyCallOnce();
cachedResult = common.anotherMethodIOnlyCallOnce();
}
#Override
public void someMethod() {
common.someMethodWithDifferentName();
}
#Override
public String dontRepeatYourself() {
return cachedResult;
}
}
Note also that most modern IDEs have things like Eclipse's Source -> Generate Delegate Methods to make this process faster.

Default method in interface in Java 8 and Bean Info Introspector

I have a little problem with default methods in Interface and BeanInfo Introspector.
In this example, there is interface: Interface
public static interface Interface {
default public String getLetter() {
return "A";
}
}
and two classes ClassA and ClassB:
public static class ClassA implements Interface {
}
public static class ClassB implements Interface {
public String getLetter() {
return "B";
}
}
In main method app prints PropertyDescriptors from BeanInfo:
public static String formatData(PropertyDescriptor[] pds) {
return Arrays.asList(pds).stream()
.map((pd) -> pd.getName()).collect(Collectors.joining(", "));
}
public static void main(String[] args) {
try {
System.out.println(
formatData(Introspector.getBeanInfo(ClassA.class)
.getPropertyDescriptors()));
System.out.println(
formatData(Introspector.getBeanInfo(ClassB.class)
.getPropertyDescriptors()));
} catch (IntrospectionException e) {
e.printStackTrace();
}
}
And the result is:
class
class, letter
Why default method "letter" is not visible as property in ClassA? Is it bug or feature?
I guess, Introspector does not process interface hierarchy chains, even though with Java 8 virtual extention methods (aka defenders, default methods) interfaces can have something that kinda sorta looks like property methods. Here's a rather simplistic introspector that claims it does: BeanIntrospector
Whether this can be considered a bug is somewhat of a gray area, here's why I think so.
Obviously, now a class can "inherit" from an interface a method that has all the qualities of what's oficially considered a getter/setter/mutator. But at the same time, this whole thing is against interface's purpose -- an interface can not possibly provide anything that can be considered a property, since it's stateless and behaviorless, it's only meant to describe behavior. Even defender methods are basically static unless they access real properties of a concrete implementation.
On the other hand, if we assume defenders are officially inherited (as opposed to providing default implementation which is a rather ambiguous definition), they should result in synthetic methods being created in the implementing class, and those belong to the class and are traversed as part of PropertyDescriptor lookup. Obviously this is not the way it is though, otherwise the whole thing would be working. :) It seems that defender methods are getting some kind of special treatment here.
Debugging reveals that this method is filtered out at Introspector#getPublicDeclaredMethods():
if (!method.getDeclaringClass().equals(clz)) {
result[i] = null; // ignore methods declared elsewhere
}
where clz is a fully-qualified name of the class in question.
Since ClassB has custom implementation of this method, it passes the check successfully while ClassA doesn't.
I think also that it is a bug.
You can solve this using a dedicated BeanInfo for your class, and by providing somthing like that :
/* (non-Javadoc)
* #see java.beans.SimpleBeanInfo#getAdditionalBeanInfo()
*/
#Override
public BeanInfo[] getAdditionalBeanInfo()
{
Class<?> superclass = Interface.class;
BeanInfo info = null;
try
{
info = Introspector.getBeanInfo(superclass);
}
catch (IntrospectionException e)
{
//nothing to do
}
if (info != null)
return new BeanInfo[] { info };
return null;
}
This is because you only have your method on Interface and ClassB, not on ClassA directly. However it sounds to me like a bug since I'd expect that property to showup on the list. I suspect Inrospector did not catch up with Java 8 features yet.

Interfaces in Java: cannot make implemented methods protected or private

I know that an interface must be public. However, I don't want that.
I want my implemented methods to only be accessible from their own package, so I want my implemented methods to be protected.
The problem is I can't make the interface or the implemented methods protected.
What is a work around? Is there a design pattern that pertains to this problem?
From the Java guide, an abstract class wouldn't do the job either.
read this.
"The public access specifier indicates that the interface can be used by any class in any package. If you do not specify that the interface is public, your interface will be accessible only to classes defined in the same package as the interface."
Is that what you want?
You class can use package protection and still implement an interface:
class Foo implements Runnable
{
public void run()
{
}
}
If you want some methods to be protected / package and others not, it sounds like your classes have more than one responsibility, and should be split into multiple.
Edit after reading comments to this and other responses:
If your are somehow thinking that the visibility of a method affects the ability to invoke that method, think again. Without going to extremes, you cannot prevent someone from using reflection to identify your class' methods and invoke them. However, this is a non-issue: unless someone is trying to crack your code, they're not going to invoke random methods.
Instead, think of private / protected methods as defining a contract for subclasses, and use interfaces to define the contract with the outside world.
Oh, and to the person who decided my example should use K&R bracing: if it's specified in the Terms of Service, sure. Otherwise, can't you find anything better to do with your time?
When I have butted up against this I use a package accessible inner or nested class to implement the interface, pushing the implemented method out of the public class.
Usually it's because I have a class with a specific public API which must implement something else to get it's job done (quite often because the something else was a callback disguised as an interface <grin>) - this happens a lot with things like Comparable. I don't want the public API polluted with the (forced public) interface implementation.
Hope this helps.
Also, if you truly want the methods accessed only by the package, you don't want the protected scope specifier, you want the default (omitted) scope specifier. Using protected will, of course, allow subclasses to see the methods.
BTW, I think that the reason interface methods are inferred to be public is because it is very much the exception to have an interface which is only implemented by classes in the same package; they are very much most often invoked by something in another package, which means they need to be public.
This question is based on a wrong statement:
I know that an interface must be public
Not really, you can have interfaces with default access modifier.
The problem is I can't make the interface or the implemented methods protected
Here it is:
C:\oreyes\cosas\java\interfaces>type a\*.java
a\Inter.java
package a;
interface Inter {
public void face();
}
a\Face.java
package a;
class Face implements Inter {
public void face() {
System.out.println( "face" );
}
}
C:\oreyes\cosas\java\interfaces>type b\*.java
b\Test.java
package b;
import a.Inter;
import a.Face;
public class Test {
public static void main( String [] args ) {
Inter inter = new Face();
inter.face();
}
}
C:\oreyes\cosas\java\interfaces>javac -d . a\*.java b\Test.java
b\Test.java:2: a.Inter is not public in a; cannot be accessed from outside package
import a.Inter;
^
b\Test.java:3: a.Face is not public in a; cannot be accessed from outside package
import a.Face;
^
b\Test.java:7: cannot find symbol
symbol : class Inter
location: class b.Test
Inter inter = new Face();
^
b\Test.java:7: cannot find symbol
symbol : class Face
location: class b.Test
Inter inter = new Face();
^
4 errors
C:\oreyes\cosas\java\interfaces>
Hence, achieving what you wanted, prevent interface and class usage outside of the package.
Here's how it could be done using abstract classes.
The only inconvenient is that it makes you "subclass".
As per the java guide, you should follow that advice "most" of the times, but I think in this situation it will be ok.
public abstract class Ab {
protected abstract void method();
abstract void otherMethod();
public static void main( String [] args ) {
Ab a = new AbImpl();
a.method();
a.otherMethod();
}
}
class AbImpl extends Ab {
protected void method(){
System.out.println( "method invoked from: " + this.getClass().getName() );
}
void otherMethod(){
System.out.println("This time \"default\" access from: " + this.getClass().getName() );
}
}
Here's another solution, inspired by the C++ Pimpl idiom.
If you want to implement an interface, but don't want that implementation to be public, you can create a composed object of an anonymous inner class that implements the interface.
Here's an example. Let's say you have this interface:
public interface Iface {
public void doSomething();
}
You create an object of the Iface type, and put your implementation in there:
public class IfaceUser {
private int someValue;
// Here's our implementor
private Iface impl = new Iface() {
public void doSomething() {
someValue++;
}
};
}
Whenever you need to invoke doSomething(), you invoke it on your composed impl object.
I just came across this trying to build a protected method with the intention of it only being used in a test case. I wanted to delete test data that I had stuffed into a DB table. In any case I was inspired by #Karl Giesing's post. Unfortunately it did not work. I did figure a way to make it work using a protected inner class.
The interface:
package foo;
interface SomeProtectedFoo {
int doSomeFoo();
}
Then the inner class defined as protected in public class:
package foo;
public class MyFoo implements SomePublicFoo {
// public stuff
protected class ProtectedFoo implements SomeProtectedFoo {
public int doSomeFoo() { ... }
}
protected ProtectedFoo pFoo;
protected ProtectedFoo gimmeFoo() {
return new ProtectedFoo();
}
}
You can then access the protected method only from other classes in the same package, as my test code was as show:
package foo;
public class FooTest {
MyFoo myFoo = new MyFoo();
void doProtectedFoo() {
myFoo.pFoo = myFoo.gimmeFoo();
myFoo.pFoo.doSomeFoo();
}
}
A little late for the original poster, but hey, I just found it. :D
You can go with encapsulation instead of inheritance.
That is, create your class (which won't inherit anything) and in it, have an instance of the object you want to extend.
Then you can expose only what you want.
The obvious disadvantage of this is that you must explicitly pass-through methods for everything you want exposed. And it won't be a subclass...
I would just create an abstract class. There is no harm in it.
With an interface you want to define methods that can be exposed by a variety of implementing classes.
Having an interface with protected methods just wouldn't serve that purpose.
I am guessing your problem can be solved by redesigning your class hierarchy.
One way to get around this is (depending on the situation) to just make an anonymous inner class that implements the interface that has protected or private scope. For example:
public class Foo {
interface Callback {
void hiddenMethod();
}
public Foo(Callback callback) {
}
}
Then in the user of Foo:
public class Bar {
private Foo.Callback callback = new Foo.Callback() {
#Override public void hiddenMethod() { ... }
};
private Foo foo = new Foo(callback);
}
This saves you from having the following:
public class Bar implements Foo.Callback {
private Foo foo = new Foo(this);
// uh-oh! the method is public!
#Override public void hiddenMethod() { ... }
}
I think u can use it now with Java 9 release. From the openJdk notes for Java 9,
Support for private methods in interfaces was briefly in consideration
for inclusion in Java SE 8 as part of the effort to add support for
Lambda Expressions, but was withdrawn to enable better focus on higher
priority tasks for Java SE 8. It is now proposed that support for
private interface methods be undertaken thereby enabling non abstract
methods of an interface to share code between them.
refer https://bugs.openjdk.java.net/browse/JDK-8071453

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