What is the difference between passing in generic parameter some generic class with and without his generic parameter?
Example:
Simple Generic class:
public class Foo<T> { /*...*/ }
Simple class that extend simple generic class setting the generic parameter to some irrelevant type:
public class FooFoo extends Foo<Type1> { /*...*/ }
Another generic class
public class Bar<T> extends FooFoo { /*...*/ }
Our base class that as generic parameter need something that extends class Foo
public class TestFooClass<T extends Foo<?>> { /*...*/ }
And the question what is the deference between this two parameters
public class BarTestOne extends TestFooClass<Bar> { /*...*/ }
public class BarTestTwo extends TestFooClass<Bar<?>> { /*...*/ }
Problem
Class<T> class = (Class<T>) ((Foo)getClass().getGenericSuperclass()).getActualTypeArguments()[0];
In the first case code works in the second doesn't.
It looks like you are trying to determine the actual type that the TestFooClass is parameterized with?
In that context, the difference between using generic class with and without its generic parameter is that getActualTypeArguments()[0] will:
In the first case provide an instance of Class representing the raw type
In the second case provide an instance of ParameterizedType (thus one may get ClassCastException). If you call getRawType() on that ParameterizedType, you will get Class representing the raw type.
This:
BarTestOne one = new BarTestOne();
BarTestTwo two = new BarTestTwo();
Class<?> clazz1 = (Class<?>) ((ParameterizedType) one.getClass().getGenericSuperclass()).getActualTypeArguments()[0];
Class<?> clazz2 = (Class<?>) ((ParameterizedType) ((ParameterizedType) two.getClass().getGenericSuperclass()).getActualTypeArguments()[0]).getRawType();
System.out.println(clazz1.equals(clazz2));
This will return true.
Alas, any deeper answer is beyond my knowledge of generics.
Bar means Bar<Object> and Bar<?> doesn't. For example, if you have a List, you can add to it, if you have a List<?> you can't, because the compiler doesn't know if your object is compatible with the "actual generic type" of the object.
As for the reflection code, I don't know. It says getClass(), which depends on the object you call it on; in this case the object is obviously this... From where is this code called?
Related
While sub classing from generic class type/Formal type parameter T/E with valid class type/Actual type parameter say e.g. Type/String there are many combinations occurs and that confusing which one to use and when?
public class SubClass<T> implements SuperIfc<T> <-- It is straight forward to understand
public class SubClass<T> implements SuperIfc<Type>
public class SubClass<Type> implements SuperIfc<T>
public class SubClass<Type> implements SuperIfc<Type>
public class SubClass<Type> implements SuperIfc
public class SubClass implements SuperIfc<Type>
public class SubClass implements SuperIfc<T> <--- Hope we cannot declare <T> in his case while initialising SubClass.
// Bounded type parameter
public class SubClass<T extends Type> implements SuperIfc<Type>
public class SubClass<T extends Type> implements SuperIfc<T> <-- Looks <T> at SuperIfc also refers <T extends Type>, and no need to declare it again at SuperIfc.
// Recursive type bound
public class SubClass<T extends Comparable<T>>> implements SuperIfc<T>
public class SubClass<T extends Comparable<T>>> implements SuperIfc<Type>
So that i can be more clearer on solving incompatible types while subclassing
Case_1:
public class Test {
interface TestIfc {
public static <T extends TestIfc> T of(int choice) {
if(choice == 1) {
return new TestImpl(); <-- PROB_1: incompatible type error
} else {
return new SomeOtherTestImpl(); //incompatible type error
}
}
}
static class TestImpl implements TestIfc {}
static class SomeOtherTestImpl<T extends TestIfc> implements TestIfc {
//The below method also having same error though with declaration
public T of() {
return new TestImpl(); <-- PROB_2: incompatible type error
}
}
}
Case_1: PROB_1: return type is T extends TestIfc and returned TestImpl implements TestIf So what is wrong?
Case_1: PROB_2: Similar to PROB_1, how to rectify without external casting. Please help.
Case_2:
public interface SuperIfc<T> {
public T create(Object label);
}
class Type {
public static Type of(){
return new Type();
}
}
------
public class SubClass<Type> implements SuperIfc<Type>{
#Override
public Type create() {
return Type.of(); <---- PROB_1: cannot resolve method
}
}
-------
public class SubClass<T extends Type> implements SuperIfc<Type>{
#Override
public Type create() {
return Type.of(); <---- PROB_1: is resolved
}
}
SuperIfc<Type> object = new SubClass(); <-- PROB_2 Unchecked assignement warning
SuperIfc<Type> object = new SubClass<TypeImpl>(); <-- PROB_3: bound should extend Type
I would like to know how to resolve Case_2, PROB_1 and PROB_2 together?
How to write subclass for generic super class with class types and what are the rules?
What should be taken care when changing generic T to class Type while subclassing? may be the difference between below and when to use?
public class SubClass<Type> implements SuperIfc<Type>
public class SubClass<Type> implements SuperIfc
public class SubClass implements SuperIfc<Type>
public class SubClass<T extends Type> implements SuperIfc<Type>
public class SubClass<T extends Type> implements SuperIfc<T>
public class SubClass<T> implements SuperIfc<Type>
In the first of() method, the method can return any type that implements InformationIfc, but your method always returns a specific implementation - InformationImpl - which is not acceptable.
For example, if you had some other class SomeOtherInformationImpl that implements that interface, the caller of that method would be allowed to write:
SomeOtherInformationImpl i = InformationImpl.of();
but your method doesn't return a SomeOtherInformationImpl.
The second of() method has the same issue as the first method.
If you instantiated your class with:
InformationImpl i = new InformationImpl<SomeOtherInformationImpl>();
the of() method would have to return a SomeOtherInformationImpl, not a InformationImpl.
Problems with case one.
PROB_1: return type is T extends TestIfc
Why do you have a generic here at all?. Since you have a static method I can do.
TestIfc broken = TestIfc<SomeOtherImplementation>.of(0);
SomeOtherImplementation is not a TestImpl. This is broken by design. What you really want is.
public static TestIfc of(int choice)
Next.
static class SomeOtherTestImpl<T extends TestIfc> implements TestIfc {
TestIfc is not parameterized, SomeOtherTestImp is, but it is completely unrelated to the interface you're implementing. Not to mention, TestIfc has a static method of that has nothing to do with the interface.
If I had to guess, I would think you want.
interface TestIfc<T>{}
static class TestImpl implements TestIfc<TestImpl> {}
static class SomeOtherTestImpl<T extends TestIfc> implements TestIfc<T>{}
That is the best I could come up with, because it is unclear what you actually want to happen.
Your examples for question 3
public class SubClass<Type> implements SuperIfc<Type>
This is broken, because SubClass<Type> declares Type to be the name of a generic parameter. It puts no restriction on the type, hence you get the method not found error.
public class SubClass<Type> implements SuperIfc
Broken, makes a generic parameter named Type, has nothing to do with your raw type version of SuperIfc
public SubClass implements SuperIfc<Type>
This is good.
public class SubClass<T extends Type> implements SuperIfc<Type>
public class SubClass<T> implements SuperIfc<Type>
These are both good, but the T has no relation to the SuperIfc parameter, hence your implementation would be.
public Type create(Object label);
The first generic parameter says the name you're going to use through the class.
This is a long answer, but please read through it (at least case 1 and the very end, you can skip over problem 2's solution in case 2 if you wish)
Case 1
Problem 1:
Your problem here is that the compiler cannot prove that T is the same as TestImpl or SomeOtherTestImpl. What if there was another class, call it TestFoo, that implemented TestIfc? Then, if you called the method as TestIfc.<TestFoo>of(1), you would expect an object of type TestFoo, but you would instead get a TestImpl, which is wrong. The compiler doesn't want you doing that, so it throws an error. What you should do is just remove the generics, like so:
public static TestImpl of(int choice) {
if(choice == 1) {
return new TestImpl();
} else {
return new SomeOtherTestImpl();
}
}
And then you can safely call TestIfc.of.
Problem 2:
This is basically the same thing. The compiler can't prove that the type parameter T of SomeOtherTestImpl is the same as TestImpl. What if you have an object like this (where TestFoo does not extend TestImpl but implements TestIfc)?
SomeOtherTestImpl<TestFoo> testImpl = ...;
And then you try to call the of method like this:
TestFoo testFoo = testImpl.of();
Can you see the problem here? Your of method returns a TestImpl, but you expect it to return a TestFoo because T is TestFoo. The compiler stops you from doing that. Again, you should just get rid of the generics:
public TestImpl of() {
return new TestImpl(); //This works now
}
Case 2
Problem 1
This problem is caused simply because you named your type parameter the same thing as your class - Type. When you changed the name of your type parameter to, say, T, it'll work because now Type is the name of a class. Before, your type parameter Type was hiding the class Type.
However, again, there is no need for a type parameter, since, you can just do this, and it'll satisfy all constraints.
public class SubClass implements SuperIfc<Type> {
#Override
public Type create() {
return Type.of();
}
}
Problem 2:
The unchecked assignment is because you didn't provide type arguments to SubClass when you did
SuperIfc<Type> object = new SubClass();
This can be fixed by either explicitly giving Type to SubClass:
SuperIfc<Type> object = new SubClass<Type>();
or by putting in an empty diamond (which I much prefer). This second approach means that the compiler can infer by itself that by new Subclass<>(), you mean new Subclass<Type>().
SuperIfc<Type> object = new SubClass<>();
Why the compiler complains:
That constructor call (new Subclass()) is basically like calling a method that looks like public SubClass makeNewSubClass() {...}. In fact, let's replace that statement above with this one, which will cause the same warning.
SuperIfc<Type> object = makeNewSubClass();
SubClass normally takes 1 type parameter (call it T), but here, it's not given any type parameters. That means that T could be anything, any class that extends Type (because of the constraint SubClass<T extends Type>).
You might thinking that if T is always going to be a subclass of Type, it should work all right, because the type of object above is SuperIfc<Type>. Assume there's a class like this - class TypeFoo extends Type - and that the method makeNewSubClass actually returns an object of type SubClass<TypeFoo> (new SubClass<TypeFoo>()).
Because of this, you expect object to be SuperIfc<Type> but it's actually SubClass<TypeFoo>. You probably think that that's all right, because after all, TypeFoo is a subclass of Type, right? Well, Java's generics are actually invariant, which means that for the compiler, SubClass<TypeFoo> is not a subclass of SuperIfc<Type> - it thinks they're 2 completely unrelated types. Don't ask me - I have no idea why :). Even SubClass<TypeFoo> isn't considered the same as or a subclass of SubClass<Type>!
That's why the compiler emits a warning - because raw types (that's what you call it when you have new SubClass() without giving type arguments) could represent anything. As for why it emits a warning and not an error - generics were introduced in Java 5, so for code before that to compile, the compiler lets you off with just a warning.
Problem 3:
According to the error, the type argument you gave (TypeImpl) should extend Type. This is because you defined SubClass as class SubClass<T extends Type> .... Here, you are giving TypeImpl as an argument for T, and so TypeImpl must extend Type, so all you need to do is put do class TypeImpl extends Type to solve that particular error.
However, even after you do that, you'll get an "incompatible types" error because, as I said earlier when talking about Problem 2, SuperIfc<Type> is not considered a supertype of SubClass<TypeImpl> even if TypeImpl extends Type.
You can do this
SuperIfc<TypeImpl> object = new SubClass<TypeImpl>();
or you can do this
SuperIfc<? extends Type> object = new SubClass<TypeImpl>();
In the second solution, we lose the information of what exactly T is in SuperIfc<T>; all we know is that it extends Type. The benefit of the second one is that you can later reassign any SubClass object to it (which doesn't work with the first version because it only allows SubClass<TypeImpl>:
SuperIfc<? extends Type> object = new SubClass<TypeImpl>();
object = new SubClass<Type>(); //works great
Alternative solution for Case 2
What it seems like you want to do is return objects of different type depending on type parameters. That's not really possible to do, since type parameters are erased at runtime. There are workarounds, though.
Here's one way to do it with a functional interface. It involves no casting and ensures type safety.
//This is the functional interface. You can also use Supplier here, of course
interface Constructor<T> {
T create();
}
class SubClass<T extends Type> implements SuperIfc<T> {
public Constructor<T> ctor;
public SubClass(Constructor<T> ctor) {
this.ctor = ctor;
}
#Override
public T create(Object label) {
return ctor.create();
}
}
class Type {}
class TypeImpl extends Type{}
By using a Constructor object, you will know that the returned Type object will always be of the right type. This is very easy to apply to your existing code - you don't need to write your own ConstructorImpl classes or anything.
You can now write this, and all you need to add is a method reference to the constructor (Type::new).
SuperIfc<Type> object1 = new SubClass<>(Type::new);
SuperIfc<? extends Type> object2 = new SubClass<>(TypeImpl::new);
With lambdas, you can also write this in a slightly more verbose way:
SuperIfc<TypeImpl> object = new SubClass<>(() -> new TypeImpl());
---
SuperIfc<? extends Type> object = new SubClass<>(() -> new TypeImpl());
I have some interface with generic type interface DummyInterface<T> and i have class to which i have to set interface
class DummyClass<T>{
public void setDummyInterface(DummyInterface<T> dummyInterface){
//set interface
}
}
Imagine the situation where i store DummyClass with different generics in ArrayList. When i get class from list i don't know the generic of class, and i want to check if interface has the same generic or not;
For example i want to check like this
if(dummyInterface<SomeTypeIKnow>.getGenericType() == dummyClass.getGenericType()){
dummyClass.setDummyInterface(dummyInterface);
}
You can use something like TypeTools (a library that I authored) to resolve type arguments so long as the argument is captured in a type definition. For example:
interface DummyStringInterface extends DummyInterface<String> {}
Class<?> t = TypeResolver.resolveRawArgument(DummyInterface.class, DummyStringInterface.class);
assert t == String.class;
Not entirely sure I understand your question, but this code will allow you to get the Generic types for an object. They allow you to perform reflection on the generic parameters of an object.
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;
public class GenericClassUtils {
public static <T> Class<T> getClassFromGeneric(
Object parentObj,
int oridnalParamterizedTypeIndex) throws Exception{
Type[] typeArray = getParameterizedTypeListAsArray(parentObj);
return (Class<T>)typeArray[oridnalParamterizedTypeIndex];
}
public static <T> Type[] getParameterizedTypeListAsArray(Object parentObj){
return ((ParameterizedType) parentObj.getClass()
.getGenericSuperclass())
.getActualTypeArguments();
}
}
Perhaps by calling something like this:
Class clazz = getClassFromGeneric(dummyClass, 0);
if (clazz.isInstance(SomeTypeIKnow.class)){
dummyClass.setDummyInterface(dummyInterface);
}
I've long used an idiom in Java for using the class information of a (non-abstract) class in the methods of its (generally abstract) ancestor class(es) (unfortunately I can't find the name of this pattern):
public abstract class Abstract<T extends Abstract<T>> {
private final Class<T> subClass;
protected Abstract(Class<T> subClass) {
this.subClass = subClass;
}
protected T getSomethingElseWithSameType() {
....
}
}
An example of a subclass thereof:
public class NonGeneric extends Abstract<NonGeneric> {
public NonGeneric() {
super(NonGeneric.class);
}
}
However, I'm having trouble defining a subclass of Abstract which has its own generic parameters:
public class Generic<T> extends Abstract<Generic<T>> {
public Generic() {
super(Generic.class);
}
}
This example is not compilable; likewise, it is not possible to specify the generic types using e.g. Generic<T>.class or even to use a wildcard like Generic<?>.
I also tried replacing the declaration of generic type T in the superclass to ? extends T, but that isn't compilable either.
Is there any way I can get this pattern to work with generic base classes?
The "pattern" (idiom) of passing an instance of Class<T> (typically to the constructor) is using Class Literals as Runtime-Type Tokens, and is used to keep a runtime reference to the generic type, which is otherwise erased.
The solution is firstly to change the token class bound to:
Class<? extends T>
and then to put a similar requirement on your generic subclass as you did with your super class; have the concrete class pass a type token, but you can type it properly as a parameter:
These classes compile without casts or warnings:
public abstract class Abstract<T extends Abstract<T>> {
private final Class<? extends T> subClass;
protected Abstract(Class<? extends T> subClass) {
this.subClass = subClass;
}
}
public class NonGeneric extends Abstract<NonGeneric> {
public NonGeneric() {
super(NonGeneric.class);
}
}
public class Generic<T> extends Abstract<Generic<T>> {
public Generic(Class<? extends Generic<T>> clazz) {
super(clazz);
}
}
And finally at the concrete class, if you declare the usage as its own class, it doesn't require a cast anywhere:
public class IntegerGeneric extends Generic<Integer> {
public IntegerGeneric() {
super(IntegerGeneric.class);
}
}
I haven't figured out how to create an instance of Generic (anonymous or not) without a cast:
// can someone fill in the parameters without a cast?
new Generic<Integer>(???); // typed direct instance
new Generic<Integer>(???) { }; // anonymous
I don't think it's possible, but I welcome being shown otherwise.
The major problem you have got here is, there is no class literal for concrete parameterized type. And that makes sense, since parameterized types don't have any runtime type information. So, you can only have class literal with raw types, in this case Generic.class.
Reference:
Java Generics FAQs
Why is there no class literal for concrete parameterized types? :
Well, that's fine, but Generic.class gives you a Class<Generic> which is not compatible with Class<Generic<T>>. A workaround is to find a way to convert it to Class<Generic<T>>, but that too you can't do directly. You would have to add an intermediate cast to Class<?>, which represents the family of all the instantiation of Class. And then downcast to Class<Generic<T>>, which will remove the compiler error, though you will an unchecked cast warning. You can annotate the constructor with #SuppressWarnings("unchecked") to remove the warning.
class Generic<T> extends Abstract<Generic<T>> {
public Generic() {
super((Class<Generic<T>>)(Class<?>)Generic.class);
}
}
There is no need in Class<T> subClass argument. Change:
protected Abstract(Class<T> subClass) {
this.subClass = subClass;
}
to:
protected Abstract(Class subClass) {
this.subClass = subClass;
}
and everything will compile.
I'd like to make a class that looks basically like this:
public class MyClass<T implements Serializable) {
void function() {
Class c = T.class;
}
}
Two errors:
- I cannot call T.class, even though I can do that with any other object type
- I cannot enforce that T implements Serializable in this way
How do I solve my two generics problems?
Cheers
Nik
You can't get the type.
Generics are implemented using something called type-erasure.
When a generic type is instantiated,
the compiler translates those types by
a technique called type erasure — a
process where the compiler removes all
information related to type parameters
and type arguments within a class or
method. Type erasure enables Java
applications that use generics to
maintain binary compatibility with
Java libraries and applications that
were created before generics.
The essence of this is that the type information is used by the compiler and discarded, hence not available at runtime.
With regards to the enforcing T implements Serializable, you just need the following:
public class MyClass<T extends Serializable>)
{
public void function(T obj)
{
...
}
}
This is simply referring to the is a relationship, so an class that implements Serializable, is a Serializable and can be passed to function.
you do this:
public class MyClass<T implements Serializable) {
void function(Class<T> tc) {
...
}
}
Basically, you have to pass in the class at run time in order to see it. You could also do something like this:
public class MyClass<T implements Serializable) {
Class<T> ct;
public MyClass(Class<T> ct){this.ct = ct;}
void function() {
... //you know what the class is here
}
}
It's kind of annoying, but not really that big of a hassle overall.
Something along these lines should do it.
private Class<T> dataType;
Type type = getClass().getGenericSuperclass();
if (type instanceof ParameterizedType) {
ParameterizedType paramType = (ParameterizedType) type;
dataType = (Class<T>) paramType.getActualTypeArguments()[0];
} else if (type instanceof Class) {
dataType = (Class<T>) type;
}
You can't do T.class because java does not actually know which class T is at runtime.
All that information is lost at compilation.
To get the class object for T you can either call getClass() on an instance of T (if you have access to one) or require the user to pass the class object as an argument to function, like:
void function(Class<T> c)
This is not possible without tricks.
The Java Generics FAQ provides an idea for a workaround.
I have a generic type that is parameterized on some Enum, declared like this:
public class FlagsField<T extends Enum<T>> {
private EnumSet<T> _flagSet;
public FlagsField() {
_flagSet = EnumSet.<T>noneOf( /* what goes here? */ );
}
...
}
I want to initialize _flagsField in the constructor as above, but can't figure out for the life of me what the right parameter to the noneOf method is. It needs to be of type Class<T>. If this weren't a generic, you'd use MyFooEnumType.class here, but T.class is not valid.
Thanks!
You've run into type erasure. Your constructor is going to need to look like:
public FlagsField(Class<T> enumClass) {
_flagSet = EnumSet.<T>noneOf(enumClass);
}
You could use this trick in your constructor: (see Generic Data Access Objects, section "Preparing DAOs with lookup")
enumClass = (Class<T>) ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0];
But I believe this code only works when the class is sub-classed and an instance of the sub-class executes it.