I know this isn't a good question to ask and I might get cursed to ask it but I cannot find any place to get help on this question
Below is a Generic class that appeared in my interview question (which I have already failed). The question was to tell what this Class declaration is doing and in what circumstances this could be used for ?
I have very limited understanding of Generic programming but I understand that 'T' is Type and 'extends' here means that the Type should have inherited 'SimpleGenericClass' but I do not understand the '?' at the end and in what circumstances this Class could be potentially used for
public abstract class SimpleGenericClass<T extends SimpleGenericClass<?>> {
}
First, because the class SimpleGenericClass is abstract, it is meant to be subclassed.
Second, it is a generic class which means that inside the class somewhere you will almost assuredly be using the generic parameter T as the type of a field.
public abstract class SimpleGenericClass<T...> {
T x;
}
Now the first interesting thing here is that T is bounded. Because it is declared as T extends SimpleGenericClass<?> it can only be SimpleGenericClass<?> or some subclass of SimpleGenericClass<?>. You also asked about thr ?. That's known as a wildcard and there is a pretty good explanation of it at the Java Tutorial on Wildcards. In your case we would say this is a "SimpleGenericClass of unknown." It is needed in Java because SimpleGenericClass<Object> is NOT the superclass of SimpleGenericClass<String>, for example.
The second interesting thing though is that since T is a SimpleGenericClass of some sort, your class is more than likely defining recursive structures. What comes to my mind are trees (think of expression trees) where SimpleGenericClass is the (abstract) node type, designed to be subclassed with all kinds of specialized node types.
UPDATE This SO question on self-bounded generics might be helpful to you.
UPDATE 2
I went ahead and put together some code that illustrates how this can be used. The app doesn't do anything but it does compile and it shows you how the generic bounds can supply some possibly-meaningful constraints.
public abstract class Node<T extends Node<?>> {
public abstract T[] getChildren();
}
class NumberNode extends Node {
int data;
public Node[] getChildren() {return new Node[]{};}
}
class IdentifierNode extends Node {
int data;
public Node[] getChildren() {return new Node[]{};}
}
class PlusNode extends Node {
NumberNode left;
NumberNode right;
public NumberNode[] getChildren() {return new NumberNode[]{};}
}
The nice thing here is that NumberNode[] is a valid return type for PlusNode.getChildren! Does that matter in practice? No idea, but it is pretty cool. :)
It's not the greatest example, but the question was rather open ended ("what might such a thing be used for?"). There are other ways to define trees, of course.
This really only means that you allow the user of class SimpleGenericClass to parametrize instances of the class with the type T. However, T cannot be any type, but must be a subtype of SampleGenericClass (or SampleGenericClass itself).
In the remainder of the code of class SimpleGenericClass you may use type T in method signatures.
Let's assume for a second that SimpleGenericClass is not abstract. When using it, you could then write:
new SimpleGenericClass<SampleGenericClass<String>>();
I.e. you parametrize SimpleGenericClass with SampleGenericClass and SampleGenericClass with String.
This basically sais: in this class you have a Type placeholder called T, and a restriction on that placeholder, it must be of type SimpleGenericClass or something that extends it. Once you obey that rule you can create instances of your class and give an actual type to T, that later on can be used in methods of that class, something like this:
public class C <T extends Number>{
public void doSomething(T t) {
}
public static void main(String... args) {
//works:
C<Number> c = new C<Number>();
c.doSomething(new Number() {
//Aonimous implementation of number
});
//won't work
//C<Object> c = new C<Object>();
C<Integer> c2 = new C<Integer>();
c2.doSomething(new Integer(1));
//won't work
//c2.doSomething(new Number() {
//Aonimous implementation of number
//});
}
}
The SimpleGenericClass<?> is pretty redundant at this point. If another generic type is needed on this class, you can have more than one (SimpleGenericClass<T extends SimpleGenericClass, T2 extends Whatever>)
By definition it says that the SimpleGenericClass can work on a type <T> which is subclass of SimpleGenericClass.
So I assume there will be some operations which will work on <T>.
Now to see why one would define a template like this - (not much I can think of , really ) may be a scenario where the SimpleGenericClass is an abstract class (just realized it is as per OP :P) and expects that it can work on any concrete classes ?
Guys what do you think ?
I guess you have got the question in this form (T instead of ?):
public abstract class SimpleGenericClass<T extends SimpleGenericClass<T>>
Take a look at this code:
abstract class Foo<SubClassOfFoo extends Foo<SubClassOfFoo>>
{
/** subclasses are forced to return themselves from this method */
public abstract SubClassOfFoo subclassAwareDeepCopy();
}
class Bar extends Foo<Bar> {
public Bar subclassAwareDeepCopy() {
Bar b = new Bar();
// ...
return b;
}
}
Bar b = new Bar();
Foo<Bar> f = b;
Bar b2 = b.subclassAwareDeepCopy();
Bar b3 = f.subclassAwareDeepCopy(); // no need to cast, return type is Bar
The trick going on with Foo<SubClassOfFoo extends Foo<SubClassOfFoo>> is:
Any subclass of Foo must supply a type argument to Foo.
That type argument must actually be a subclass of Foo.
Subclasses of Foo (like Bar) follow the idiom that the type
argument they supply to Foo is themselves.
Foo has a method that returns SubClassOfFoo. Combined
with the above idiom, this allows Foo to formulate a contract that
says “any subclass of me must implement subclassAwareDeepCopy() and
they must declare that it returns that actual subclass“.
To say that another way: this idiom allows a superclass (such as an Abstract Factory) to define methods whose argument types and return types are in terms of the subclass type, not the superclass type.
The trick is done for example in Enum JDK class:
public abstract class Enum<E extends Enum<E>>
Refer here for more details.
Related
Ok, I'll try to explain this as cleanly as I can.
I've created a generic abstract controller class that has a method hasCreatePermissions that looks something like this:
public abstract class ApplicationController<
AppEntity extends ApplicationEntity,
AppService extends ApplicationService<AppEntity>,
DTOManager extends ApplicationDTOManager
> {
// Other methods, properties, etc...
public boolean hasCreatePermissions(DTOManager.CreationRequest requestBody, Optional<UUID> requestingUser) {
return false;
}
}
Essentially, I want any class that overrides this method to be able to use its own DTOManager class as the parameter when it overrides this method.
The generic ApplicationDTOManager class looks like
public abstract class ApplicationDTOManager {
public abstract class CreationRequest {}
public abstract class CreationResponse {}
}
and any class that inherits ApplicationDTOManager can add classes that extend CreationRequest and CreationResponse for their own implementation of respective DTOs.
However, lets say I try to extend it with a UserResource class (assume UserDTOManager exists with an implementation for CreationRequest):
#RestController
public class UserResource extends ApplicationController<
User,
UserService<User>,
UserDTOManager
> {
#Override
public boolean hasCreatePermissions(UserDTOManager.CreationRequest requestBody, Optional<UUID> requestingUser) {
// Stuff
}
}
I'm told that this does not override any super class methods. Why? Is there any way to achieve this as I did not want to pass too many generics to my ApplicationController class, but also cannot have a constructor.
class ApplicationController<
AppEntity extends ApplicationEntity,
No, stop right there. This is declaring a type variable with the bound rules: "Must be either ApplicationEntity or any subtype thereof" and you named it AppEntity. This is going to make your head go in circles when you read the code later, 'I keep confusing reified types with type variables' comprises 95% of all confusion about generics. I know it seems unreadable, but there really is just only one way to go about it, and that is to use single capital letters for all your type vars. So let's fix that right now:
public abstract class ApplicationController<
E extends ApplicationEntity,
S extends ApplicationService<E>,
M extends ApplicationDTOManager> {
Which then immediately lets us see a big problem in the next line:
public boolean hasCreatePermissions(M.CreationRequest requestBody) {}
Of course that can't work. M is a type variable, at compile time the compiler has no idea what type it is. It could be bound to some type that doesn't even exist yet right now. You can't ask for a reified inner type on a type variable. You can of course simply talk about ApplicationDTOManager.CreationRequest and that's presumably what you want.
Alternatively, you're thinking that subtypes of ApplicationDTOManager may also want to subclass ApplicationDTOManager.CreationRequest. This is possible, but then all ApplicationDTOManager types need to carry their associated CreationRequest type as a type variable. We thus fix a few things:
public class ApplicationDTOManager<R extends ApplicationDTOManager.CreationRequest> {
public static class CreationRequest {}
}
You may have a non-static inner class named CreationRequest. I'm going to stop you again on that - non-static inners have an invisible inner field of their outer's type, and combining that invisible voodoo magic with generics just doesn't work. Don't do it. You can explicitly make that field if you must have it, and make a constructor that takes it. This is what javac generates for you if you don't add static to your inner classes. But by making it explicit, you take control of the generics, which you have to here, and avoid confusion which given the nature of the question seems pertinent.
I'm told that this does not override any super class methods. Why?
Java's method names include all their erased types. The name of this method:
class List<T extends Number> {
int hello(String name, boolean[] hi, T arg) throws SQLException {}
}
is, as far as the JVM is concerned, hello(Ljava/lang/String;[ZLjava/lang/Number;)I.
Yeah, no, really. javap -c -v a class file and you'll see it. (I is integer, [ is array, Z is boolean, and Ltxt; encodes ref type names in JVM style, e.g. with slashes and dollars instead of dots). It's written name(params)ret.
If you then subtype something and introduce a method whose erased JVM name is identical, you're overriding. If you don't, it is not an override. Merely an overload. Overrides are dynamically dispatched. But overloads are not - The names are all linked up at compile time. However, for any given 'JVM method name', the lookup is done dynamically based on the receiver type. In other words:
class Fruit {
void hi(Fruit f) { System.out.println("Fruit Fruit"); }
void hi(Apple a) { System.out.println("Fruit Apple"); }
}
class Apple extends Fruit {
void hi(Fruit f) { System.out.println("Apple Fruit"); }
void hi(Apple a) { System.out.println("Apple Apple"); }
}
Fruit f = new Fruit();
Fruit a = new Apple();
a.hi(a);
Will print Apple Fruit. You'd think it should print Apple Apple perhaps - we are calling apple's hi passing an apple, no? But the invocation a.hi(a) is invoking the method named hi(Lfruit;)V (because the type of a is Fruit). The receiver variable (a) has compile time type Fruit, but its real type is Apple. So, which of the hi(Lfruit;)V methods is chosen is done with dynamic dispatch - you get apple's hi(Fruit). Deciding between going with hi(Fruit) and hi(Apple) is done by the compiler only. Given that the type of the expression a is Fruit, you get hi(Fruit). The fact that if you resolve this expression, you get an object whose .getClass() returns Apple.class, doesn't change this.
Hence, what you wrote, does not override. Different name, even if you erase.
Toss generics in the mix and it gets funky. But you can do this.
public abstract class ApplicationDTOManager<I extends CreationRequest, O extends CreationResponse> {
public abstract static class CreationRequest {}
public abstract static class CreationResponse {}
}
public abstract class ApplicationController<
E extends ApplicationEntity,
S extends ApplicationService<AppEntity>,
I extends CreationRequest,
O extends CreationResponse,
M extends ApplicationDTOManager<I, O>
>
// heck that's probably too many, at some point generics aren't worth it
{
public abstract boolean hasCreatePermissions(I requestBody);
}
#RestController
public class UserResource extends ApplicationController<
User,
// UserService<User>, // this seems wrong!
UserService, // UserService should extends ApplicationService<User>
UserDTOManager.CreationRequest,
UserDTOManager.CreationResponse,
UserDTOManager> {
#Override
public boolean hasCreatePermissions(UserDTOManager.CreationRequest requestBody, Optional<UUID> requestingUser) {
// Stuff
}
}
Not sure all this is worth the pain, but, if you insist on linking all this together with generics, the above is the only way. You cannot express the notion 'has an inner type that is a subtype of CreationRequest' is a generics bound.
If you override a method you cannot have a different signature, If the method you override requires a DTOManager.CreateRequest you cannot use a child class within the override method.
You have to "support" all types of input that the parent method could take.
I believe that you need this because the permission validation relies on methods or fields of the child class. If so you should implement it inside the child class.
For example, I have a some class:
public class Test<T> {
public T setField() {
return this;
}
}
Of course, it's an incorrect class. But can I write it some else? (to it hasn't errors).
P.S. I know that I can write an abstract method to override it in the child class. But I ask about the generic class only (in the its class body only).
If you want to return an instance of your class (this), it does simply by using the class declaration (Test<T>):
public Test<T> setField() {
return this;
}
If you want to return a type of a generic parameter, look at #wero's answer.
Casting this to T makes only sense if you use a recursive type bound. Still you need to add a cast and suppress the warning:
abstract class Test<T extends Test<T>> {
#SuppressWarnings("unchecked")
public T setField()
{
return (T)this;
}
}
Define derived classes like:
public class DerivedTest extends Test<DerivedTest> { ... }
and now this works without any casts needed in the client code:
DerivedTest d = new DerivedTest().setField();
It is not possible for something like this to work safely in general in Java. this is only known to be an instance of Test<T>, which is not known to be a subtype of T, the type you want to return. There is no bound on T that can guarantee that this is an instance of T. No matter what bounds you give T, for any type that you can use for T that satisfies those bounds (call it X), I can just define a new class (unrelated to X) that extends Test<X>, and you cannot return an instance of this class as X, because it is not an X.
Assuming that I have a basic enum like:
public enum Color { Red, Green, Blue}
How can one write a generic class which only accepts "enum classes" so that a concrete instantiation of that generic class might look like MyClass<Color>?
Edit:
What a actually want to do is to write a generic abstract class containing a function returning all enum "entries" as list:
public abstract class EnumListBean<E extends Enum<E>> {
public List<E> getEnumList() {
return Arrays.asList(E.values());
}
}
While Day.values() is available E.values() is not. What i am doing wrong here?
public class EnumAcceptor<E extends Enum<E>> {
...
}
Use E as a type inside your class.
See Istvan Devai for answer to original question.
For the follow up, methods like values() are static methods, so you're out of luck trying to get that from a generic parameter. As a poor solution, you can pass the enum's Class object into the constructor. and use Class.getEnumConstants. But you might as well pass MyEnum.values() into the constructor rather than the class, and so avoid reflection altogether. It's a real shame there isn't a sensible enum metaclass.
An enum really declares a class derived from Enum. As such, you can use:
public class MyClass<T extends Enum> { }
Note that #Istvan's solution can only accept elements of the enum, which is fine if that is all you want.
Although you cannot pass the enum itself as a parameter (because it does not actually have an object equivalent) you can specify that you must receive the class of the enum in your constructor and derive the enum's details from that:
public class EnumAcceptor<E extends Enum<E>> {
public EnumAcceptor(Class<E> c) {
// Can get at the enum constants through the class.
E[] es = c.getEnumConstants();
}
enum ABC {
A, B, C;
}
public static void main(String args[]) {
EnumAcceptor<ABC> abcAcceptor = new EnumAcceptor<ABC>(ABC.class);
}
}
You can't use E.values() due to type erasure -- the type of E is not available at run-time.
For the specific case in your question, you're probably better off using Guava's Lists.newArrayList:
List<Color> days = Lists.newArrayList(Color.values());
Is there a way in java to specify, that the type parameter of a generic class must be an interface (not just extending it!)
What I want to do is the following:
public class MyClass<X extends SomeInterface, Y extends SomeOtherClass & X>
Meaning that Y must be a subclass of SomeOtherClass AND implement X.
What I currently get by the compiler is
The type X is not an interface; it cannot be specified as a bounded parameter
So, how can I tell the compiler that X must always be an interface?
Edit:
OK, I guess I oversimplified my problem a bit. Let's use my actual application domain to make it more clear:
I have an API for representing diagrams. A Diagram contains Node and Edge objects. All these three classes implement the Shape interface. Shapes may have child shapes, a parent shape and belong to a diagram.
The thing is, that I need to make two versions of this API: one open-source with just basic functionality and an extended one with more features. However, the extended API must only provide methods, which return the extended types (ExtendedDiagram, ExtendedNode, ExtendedEdge and (here comes the problem) ExtendedShape).
So I have something like this:
/* BASIC CLASSES */
public interface Shape<X extends Shape<X,Y>, Y extends Diagram<X,Y>>{
public List<X> getChildShapes();
public X getParent();
public Y getDiagram();
...
}
public class Diagram<X extends Shape<X,Y>, Y extends Diagram<X,Y>> implements Shape<X,Y>{...}
public class Edge<X extends Shape<X,Y>, Y extends Diagram<X,Y>> implements Shape<X,Y>{...}
...
/* EXTENDED CLASSES */
public interface ExtendedShape extends Shape<ExtendedShape,ExtendedDiagram> { ... }
public class ExtendedDiagram extends Diagram<ExtendedShape,ExtenedDiagram> implements ExtendedShape { ... }
public class ExtendedEdge extends Edge<ExtendedShape,ExtenedDiagram> implements ExtendedShape { ... }
...
The extended API works fine and the basic API code gives some warnings, but the main problem occurs when using the basic API:
public class SomeImporter<X extends Shape<X,Y>, Y extends Diagram<X,Y>, E extends Edge<X,Y>>{
private Y diagram;
public void addNewEdge(E newEdge){
diagram.addChildShape(newEdge);
...
That last line gives me the following warning:
The method addChildShape(X) in the type Diagram is not applicable for the arguments (E)
So now, I would just like to specify that E also needs to implement X and all would be fine - I hope ;)
Does all that make sense? Do you guys know a way to do that? Or is there even a better way to get the extended API with the said restrictions?
Thanks for sticking with me, any help is greatly appreciated!
You can use:
class Foo<T extends Number & Comparable> {...}
A class Foo with one type parameter, T. Foo must be instantiated with a type that is a subtype of Number and that implements Comparable.
In the generics context, <Type extends IInterface> handles both extends and implements. Here's an example:
public class GenericsTest<S extends Runnable> {
public static void main(String[] args) {
GenericsTest<GT> t = new GenericsTest<GT>();
GenericsTest<GT2> t2 = new GenericsTest<GT>();
}
}
class GT implements Runnable{
public void run() {
}
}
class GT2 {
}
GenericsTest will accept GT because it implements Runnable. GT2 does not, therefore it fails when trying to compile that second GenericsTest instantiation.
Maybe you can simplify your model a bit: too much generics become quickly a real pain in terms of readability, and that's quite an issue if you define a public API. Usually, if you can't understand anymore what should be inside the brackets, then you're going too far for your need - and you can't expect users to understand it better than yourself...
Anyway, in order to make your code compile, you may try defining something like this, in the Shape type:
public <S extends Shape<?,?>> void addChildShape(S shape);
That should do it.
HTH
You wrote the following:
public interface Shape<X extends Shape<X,Y>, Y extends Diagram<X,Y>>{
public List<X> getChildShapes();
public X getParent();
public Y getDiagram();
...
}
I would advise, at the minimum, getting rid of the X type variable, as follows:
public interface Shape<Y>{
public List<Shape<Y>> getChildShapes();
public Shape<Y> getParent();
public Diagram<Y> getDiagram();
...
}
The reason being is that what you originally wrote suffers from a potentially unbounded recursive nesting of the type parameters. A shape may be nested within a parent shape, which may be nested within another, all of which must be accounted for in the type signature... not a good recipe for readability. Well, it doesn't happen quite that way in your example, in which you declare "Shape<X>" instead of "Shape<Shape<X>>" but that's the direction you're going in, if you ever wanted to actually use Shape on its own...
I would probably also recommend going one step further and getting rid of the Y variable for similar reasons. Java generics don't cope very well with this sort of composition. When attempting to enforce static types for this type of modelling via generics, I've found that the type system starts to break down when you start to extend things later on.
It's typical of the Animal/Dog problem... Animal has a getChildren(), but Dog's children must also be Dogs... Java doesn't cope with this well because (in part due to the lack of abstract types as in languages like Scala, but I'm not saying you should rush off and use Scala for this problem either) the type variables have to start being declared in all sorts of places where they don't really belong.
Use a pre-processor to generate the "reduced" version of your code. Using apt and annotations might be a nice way to do it.
I might be WAY off base here, but my understanding of generics is a little different.
I am asking someone to correct me if I am wrong.
IMO -
This is a very confusing structure that you have. You have SubClasses of Shape being referenced infinitely it looks like.
Your Shape interface is utilized in the same manner as a HashMap, but I have never seen a HashMap do what you are trying to do, eventually you have to have X be a class in Shape. Otherwise you are doing HashMap
If you always want X to be a "IS A" relationship to an interface it won't happen. That is not what generics are for. Generics are used to apply methods to multiple Objects, and interfaces cannot be Objects. Interfaces define a contract between a client and a class. All you can do with the is say that you will accept any Object that implements Runnable, because all or some of your methods are required to utilize the Runnable interface methods. Otherwise if you don't specify and you define as , then your contract between your class with the client can produce unexpected behavior and cause either the wrong return value or an exception to be thrown.
For example:
public interface Animal {
void eat();
void speak();
}
public interface Dog extends Animal {
void scratch();
void sniff();
}
public interface Cat extends Animal {
void sleep();
void stretch();
}
public GoldenRetriever implements Dog {
public GoldenRetriever() { }
void eat() {
System.out.println("I like my Kibbles!");
}
void speak() {
System.out.println("Rufff!");
}
void scratch() {
System.out.println("I hate this collar.");
}
void sniff() {
System.out.println("Ummmm?");
}
}
public Tiger implements Cat {
public Tiger() { }
void eat() {
System.out.println("This meat is tasty.");
}
void speak() {
System.out.println("Roar!");
}
void sleep() {
System.out.println("Yawn.");
}
void stretch() {
System.out.println("Mmmmmm.");
}
}
Now if you did this class you can expect that you CAN always call 'speak()' & 'sniff()'
public class Kingdom<X extends Dog> {
public Kingdom(X dog) {
dog.toString();
dog.speak();
dog.sniff();
}
}
However, if you did this you CANNOT ALWAYS call 'speak()' & 'sniff()'
public class Kingdom<X> {
public Kingdom(X object) {
object.toString();
object.speak();
object.sniff();
}
}
CONCLUSION:
Generics give you the ability to utilize methods on a wide range of objects, not interfaces. Your final entry into a generic MUST be a type of Object.
The reserved word “extends” as along with a type parameter T is used to specify a bound.
‘…in this context, extends is used in a general sense to mean either "extends" (as in classes) or "implements" (as in interfaces).’
[ https://docs.oracle.com/javase/tutorial/java/generics/bounded.html ]
In short, “extends” can only be used to specify a bound (whether a class or an interface) for some class type parameter T and not any interface type parameter T.
In your case,
public class MyClass<X extends SomeInterface, Y extends SomeOtherClass & X>
The compiler resolves X to be a class. For the second occurrence of X along with the type parameter Y (which clearly needs to be a class anyway), it requires X to be an interface. Since it has already resolved X to be class, it signals the error for the second occurrence of X,
The type X is not an interface;
Further, had X been specified in the first occurrence as an unbounded parameter, the compiler would have resolved it to be either a class or an interface and it would’ve considered the second occurrence of X to be a possible interface and thus allowed compilation. Since it was not so, the compiler clarifies,
it cannot be specified as a bounded parameter
I want to have a function defined in a superclass that returns an instance of the subclass that is used to invoke the function. That is, say I have class A with a function plugh. Then I create subclasses B and C that extend A. I want B.plugh to return a B and C.plugh to return a C. Yes, they could return an A, but then the caller would have to either cast it to the right subtype, which is a pain when used a lot, or declare the receiving variable to be of the supertype, which loses type safety.
So I was trying to do this with generics, writing something like this:
class A<T extends A>
{
private T foo;
public T getFoo()
{
return foo;
}
}
class B extends A<B>
{
public void calcFoo()
{
foo=... whatever ...
}
}
class C extends A<C>
{
public void calcFoo()
{
foo=... whatever ...
}
}
This appears to work but it looks pretty ugly.
For one thing, I get warnings on class A<T extends A>. The compiler says that A is generic and I should specify the type. I guess it wants me to say class A<T extends A<x>. But what would I put in for x? I think I could get stuck in an infinite loop here.
It seems weird to write class B extends A<B>, but this causes no complaints, so maybe that's just fine.
Is this the right way to do it? Is there a better way?
class A<T extends A<T>> {
Or, I prefer:
class A<THIS extends A<THIS>> {
Like java.lang.Enum. It uglifies client code that just wants A, but has to write A<?>.
Note this does not in any way break LSP, and is GOOD OO.
Or you could just use covariant return types.
Doing this is, from a strictly OO perspective, a Very Bad Thing™. It violates the Liskov Substitution Principle for one - if B & C are both A, then B.getFoo() and C.getFoo() should return the same thing - which in this case they won't. If the caller needs to know what it's getting when it calls the method, there's a good chance your inheritance hierarchy is wrong.