We Keep Coding

Java - Is there a way to declare a generic class type? [duplicate]

How can I achieve this?
public class GenericClass<T>
{
public Type getMyType()
{
//How do I return the type of T?
}
}
Everything I have tried so far always returns type Object rather than the specific type used.

As others mentioned, it's only possible via reflection in certain circumstances.
If you really need the type, this is the usual (type-safe) workaround pattern:
public class GenericClass<T> {
private final Class<T> type;
public GenericClass(Class<T> type) {
this.type = type;
}
public Class<T> getMyType() {
return this.type;
}
}

I have seen something like this
private Class<T> persistentClass;
public Constructor() {
this.persistentClass = (Class<T>) ((ParameterizedType) getClass()
.getGenericSuperclass()).getActualTypeArguments()[0];
}
in the hibernate GenericDataAccessObjects Example

Generics are not reified at run-time. This means the information is not present at run-time.
Adding generics to Java while mantaining backward compatibility was a tour-de-force (you can see the seminal paper about it: Making the future safe for the past: adding genericity to the Java programming language).
There is a rich literature on the subject, and some people are dissatisfied with the current state, some says that actually it's a lure and there is no real need for it. You can read both links, I found them quite interesting.

Use Guava.
import com.google.common.reflect.TypeToken;
import java.lang.reflect.Type;
public abstract class GenericClass<T> {
private final TypeToken<T> typeToken = new TypeToken<T>(getClass()) { };
private final Type type = typeToken.getType(); // or getRawType() to return Class<? super T>
public Type getType() {
return type;
}
public static void main(String[] args) {
GenericClass<String> example = new GenericClass<String>() { };
System.out.println(example.getType()); // => class java.lang.String
}
}
A while back, I posted some full-fledge examples including abstract classes and subclasses here.
Note: this requires that you instantiate a subclass of GenericClass so it can bind the type parameter correctly. Otherwise it'll just return the type as T.

Java generics are mostly compile time, this means that the type information is lost at runtime.
class GenericCls<T>
{
T t;
}
will be compiled to something like
class GenericCls
{
Object o;
}
To get the type information at runtime you have to add it as an argument of the ctor.
class GenericCls<T>
{
private Class<T> type;
public GenericCls(Class<T> cls)
{
type= cls;
}
Class<T> getType(){return type;}
}
Example:
GenericCls<?> instance = new GenericCls<String>(String.class);
assert instance.getType() == String.class;

Sure, you can.
Java does not use the information at run time, for backwards compatibility reasons. But the information is actually present as metadata and can be accessed via reflection (but it is still not used for type-checking).
From the official API:
http://download.oracle.com/javase/6/docs/api/java/lang/reflect/ParameterizedType.html#getActualTypeArguments%28%29
However, for your scenario I would not use reflection. I'm personally more inclined to use that for framework code. In your case I would just add the type as a constructor param.

public abstract class AbstractDao<T>
{
private final Class<T> persistentClass;
public AbstractDao()
{
this.persistentClass = (Class<T>) ((ParameterizedType) this.getClass().getGenericSuperclass())
.getActualTypeArguments()[0];
}
}

I used follow approach:
public class A<T> {
protected Class<T> clazz;
public A() {
this.clazz = (Class<T>) ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0];
}
public Class<T> getClazz() {
return clazz;
}
}
public class B extends A<C> {
/* ... */
public void anything() {
// here I may use getClazz();
}
}

I dont think you can, Java uses type erasure when compiling so your code is compatible with applications and libraries that were created pre-generics.
From the Oracle Docs:
Type Erasure
Generics were introduced to the Java language to provide tighter type
checks at compile time and to support generic programming. To
implement generics, the Java compiler applies type erasure to:
Replace all type parameters in generic types with their bounds or
Object if the type parameters are unbounded. The produced bytecode,
therefore, contains only ordinary classes, interfaces, and methods.
Insert type casts if necessary to preserve type safety. Generate
bridge methods to preserve polymorphism in extended generic types.
Type erasure ensures that no new classes are created for parameterized
types; consequently, generics incur no runtime overhead.
http://docs.oracle.com/javase/tutorial/java/generics/erasure.html

Technique described in this article by Ian Robertson works for me.
In short quick and dirty example:
public abstract class AbstractDAO<T extends EntityInterface, U extends QueryCriteria, V>
{
/**
* Method returns class implementing EntityInterface which was used in class
* extending AbstractDAO
*
* #return Class<T extends EntityInterface>
*/
public Class<T> returnedClass()
{
return (Class<T>) getTypeArguments(AbstractDAO.class, getClass()).get(0);
}
/**
* Get the underlying class for a type, or null if the type is a variable
* type.
*
* #param type the type
* #return the underlying class
*/
public static Class<?> getClass(Type type)
{
if (type instanceof Class) {
return (Class) type;
} else if (type instanceof ParameterizedType) {
return getClass(((ParameterizedType) type).getRawType());
} else if (type instanceof GenericArrayType) {
Type componentType = ((GenericArrayType) type).getGenericComponentType();
Class<?> componentClass = getClass(componentType);
if (componentClass != null) {
return Array.newInstance(componentClass, 0).getClass();
} else {
return null;
}
} else {
return null;
}
}
/**
* Get the actual type arguments a child class has used to extend a generic
* base class.
*
* #param baseClass the base class
* #param childClass the child class
* #return a list of the raw classes for the actual type arguments.
*/
public static <T> List<Class<?>> getTypeArguments(
Class<T> baseClass, Class<? extends T> childClass)
{
Map<Type, Type> resolvedTypes = new HashMap<Type, Type>();
Type type = childClass;
// start walking up the inheritance hierarchy until we hit baseClass
while (!getClass(type).equals(baseClass)) {
if (type instanceof Class) {
// there is no useful information for us in raw types, so just keep going.
type = ((Class) type).getGenericSuperclass();
} else {
ParameterizedType parameterizedType = (ParameterizedType) type;
Class<?> rawType = (Class) parameterizedType.getRawType();
Type[] actualTypeArguments = parameterizedType.getActualTypeArguments();
TypeVariable<?>[] typeParameters = rawType.getTypeParameters();
for (int i = 0; i < actualTypeArguments.length; i++) {
resolvedTypes.put(typeParameters[i], actualTypeArguments[i]);
}
if (!rawType.equals(baseClass)) {
type = rawType.getGenericSuperclass();
}
}
}
// finally, for each actual type argument provided to baseClass, determine (if possible)
// the raw class for that type argument.
Type[] actualTypeArguments;
if (type instanceof Class) {
actualTypeArguments = ((Class) type).getTypeParameters();
} else {
actualTypeArguments = ((ParameterizedType) type).getActualTypeArguments();
}
List<Class<?>> typeArgumentsAsClasses = new ArrayList<Class<?>>();
// resolve types by chasing down type variables.
for (Type baseType : actualTypeArguments) {
while (resolvedTypes.containsKey(baseType)) {
baseType = resolvedTypes.get(baseType);
}
typeArgumentsAsClasses.add(getClass(baseType));
}
return typeArgumentsAsClasses;
}
}

I think there is another elegant solution.
What you want to do is (safely) "pass" the type of the generic type parameter up from the concerete class to the superclass.
If you allow yourself to think of the class type as "metadata" on the class, that suggests the Java method for encoding metadata in at runtime: annotations.
First define a custom annotation along these lines:
import java.lang.annotation.*;
#Target(ElementType.TYPE)
#Retention(RetentionPolicy.RUNTIME)
public #interface EntityAnnotation {
Class entityClass();
}
You can then have to add the annotation to your subclass.
#EntityAnnotation(entityClass = PassedGenericType.class)
public class Subclass<PassedGenericType> {...}
Then you can use this code to get the class type in your base class:
import org.springframework.core.annotation.AnnotationUtils;
.
.
.
private Class getGenericParameterType() {
final Class aClass = this.getClass();
EntityAnnotation ne =
AnnotationUtils.findAnnotation(aClass, EntityAnnotation.class);
return ne.entityClass();
}
Some limitations of this approach are:
You specify the generic type (PassedGenericType) in TWO places rather than one which is non-DRY.
This is only possible if you can modify the concrete subclasses.

Here's one way, which I've had to use once or twice:
public abstract class GenericClass<T>{
public abstract Class<T> getMyType();
}
Along with
public class SpecificClass extends GenericClass<String>{
#Override
public Class<String> getMyType(){
return String.class;
}
}

This is my solution:
import java.lang.reflect.Type;
import java.lang.reflect.TypeVariable;
public class GenericClass<T extends String> {
public static void main(String[] args) {
for (TypeVariable typeParam : GenericClass.class.getTypeParameters()) {
System.out.println(typeParam.getName());
for (Type bound : typeParam.getBounds()) {
System.out.println(bound);
}
}
}
}

Here is working solution!!!
#SuppressWarnings("unchecked")
private Class<T> getGenericTypeClass() {
    try {
        String className = ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0].getTypeName();
        Class<?> clazz = Class.forName(className);
        return (Class<T>) clazz;
    } catch (Exception e) {
        throw new IllegalStateException("Class is not parametrized with generic type!!! Please use extends <> ");
    }
}
NOTES:
Can be used only as superclass
1. Has to be extended with typed class (Child extends Generic<Integer>)
OR
2. Has to be created as anonymous implementation (new Generic<Integer>() {};)

You can't. If you add a member variable of type T to the class (you don't even have to initialise it), you could use that to recover the type.

One simple solution for this cab be like below
public class GenericDemo<T>{
private T type;
GenericDemo(T t)
{
this.type = t;
}
public String getType()
{
return this.type.getClass().getName();
}
public static void main(String[] args)
{
GenericDemo<Integer> obj = new GenericDemo<Integer>(5);
System.out.println("Type: "+ obj.getType());
}
}

To complete some of the answers here, I had to get the ParametrizedType of MyGenericClass, no matter how high is the hierarchy, with the help of recursion:
private Class<T> getGenericTypeClass() {
return (Class<T>) (getParametrizedType(getClass())).getActualTypeArguments()[0];
}
private static ParameterizedType getParametrizedType(Class clazz){
if(clazz.getSuperclass().equals(MyGenericClass.class)){ // check that we are at the top of the hierarchy
return (ParameterizedType) clazz.getGenericSuperclass();
} else {
return getParametrizedType(clazz.getSuperclass());
}
}

Here is my solution
public class GenericClass<T>
{
private Class<T> realType;
public GenericClass() {
findTypeArguments(getClass());
}
private void findTypeArguments(Type t) {
if (t instanceof ParameterizedType) {
Type[] typeArgs = ((ParameterizedType) t).getActualTypeArguments();
realType = (Class<T>) typeArgs[0];
} else {
Class c = (Class) t;
findTypeArguments(c.getGenericSuperclass());
}
}
public Type getMyType()
{
// How do I return the type of T? (your question)
return realType;
}
}
No matter how many level does your class hierarchy has,
this solution still works, for example:
public class FirstLevelChild<T> extends GenericClass<T> {
}
public class SecondLevelChild extends FirstLevelChild<String> {
}
In this case, getMyType() = java.lang.String

Here is my trick:
public class Main {
public static void main(String[] args) throws Exception {
System.out.println(Main.<String> getClazz());
}
static <T> Class getClazz(T... param) {
return param.getClass().getComponentType();
}
}

Just in case you use store a variable using the generic type you can easily solve this problem adding a getClassType method as follows:
public class Constant<T> {
private T value;
#SuppressWarnings("unchecked")
public Class<T> getClassType () {
return ((Class<T>) value.getClass());
}
}
I use the provided class object later to check if it is an instance of a given class, as follows:
Constant<?> constant = ...;
if (constant.getClassType().equals(Integer.class)) {
Constant<Integer> integerConstant = (Constant<Integer>)constant;
Integer value = integerConstant.getValue();
// ...
}

Here is my solution. The examples should explain it. The only requirement is that a subclass must set the generic type, not an object.
import java.lang.reflect.AccessibleObject;
import java.lang.reflect.Field;
import java.lang.reflect.Method;
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;
import java.lang.reflect.TypeVariable;
import java.util.HashMap;
import java.util.Map;
public class TypeUtils {
/*** EXAMPLES ***/
public static class Class1<A, B, C> {
public A someA;
public B someB;
public C someC;
public Class<?> getAType() {
return getTypeParameterType(this.getClass(), Class1.class, 0);
}
public Class<?> getCType() {
return getTypeParameterType(this.getClass(), Class1.class, 2);
}
}
public static class Class2<D, A, B, E, C> extends Class1<A, B, C> {
public B someB;
public D someD;
public E someE;
}
public static class Class3<E, C> extends Class2<String, Integer, Double, E, C> {
public E someE;
}
public static class Class4 extends Class3<Boolean, Long> {
}
public static void test() throws NoSuchFieldException {
Class4 class4 = new Class4();
Class<?> typeA = class4.getAType(); // typeA = Integer
Class<?> typeC = class4.getCType(); // typeC = Long
Field fieldSomeA = class4.getClass().getField("someA");
Class<?> typeSomeA = TypeUtils.getFieldType(class4.getClass(), fieldSomeA); // typeSomeA = Integer
Field fieldSomeE = class4.getClass().getField("someE");
Class<?> typeSomeE = TypeUtils.getFieldType(class4.getClass(), fieldSomeE); // typeSomeE = Boolean
}
/*** UTILS ***/
public static Class<?> getTypeVariableType(Class<?> subClass, TypeVariable<?> typeVariable) {
Map<TypeVariable<?>, Type> subMap = new HashMap<>();
Class<?> superClass;
while ((superClass = subClass.getSuperclass()) != null) {
Map<TypeVariable<?>, Type> superMap = new HashMap<>();
Type superGeneric = subClass.getGenericSuperclass();
if (superGeneric instanceof ParameterizedType) {
TypeVariable<?>[] typeParams = superClass.getTypeParameters();
Type[] actualTypeArgs = ((ParameterizedType) superGeneric).getActualTypeArguments();
for (int i = 0; i < typeParams.length; i++) {
Type actualType = actualTypeArgs[i];
if (actualType instanceof TypeVariable) {
actualType = subMap.get(actualType);
}
if (typeVariable == typeParams[i]) return (Class<?>) actualType;
superMap.put(typeParams[i], actualType);
}
}
subClass = superClass;
subMap = superMap;
}
return null;
}
public static Class<?> getTypeParameterType(Class<?> subClass, Class<?> superClass, int typeParameterIndex) {
return TypeUtils.getTypeVariableType(subClass, superClass.getTypeParameters()[typeParameterIndex]);
}
public static Class<?> getFieldType(Class<?> clazz, AccessibleObject element) {
Class<?> type = null;
Type genericType = null;
if (element instanceof Field) {
type = ((Field) element).getType();
genericType = ((Field) element).getGenericType();
} else if (element instanceof Method) {
type = ((Method) element).getReturnType();
genericType = ((Method) element).getGenericReturnType();
}
if (genericType instanceof TypeVariable) {
Class<?> typeVariableType = TypeUtils.getTypeVariableType(clazz, (TypeVariable) genericType);
if (typeVariableType != null) {
type = typeVariableType;
}
}
return type;
}
}

If you have a class like:
public class GenericClass<T> {
private T data;
}
with T variable, then you can print T name:
System.out.println(data.getClass().getSimpleName()); // "String", "Integer", etc.

Use an abstract method that returns the class type then use it in that class and wherever you extend generic class you will have to implement that abstract method to return the required class type
public class AbsractService<T>{
public abstract Class<T> getClassType ();
.......
}
at runtime
class AnimalService extends AbstractService<Animal>{
#Override
public Class<Animal> getClassType (){
return Animal.class;
}
.....
}

public static final Class<?> getGenericArgument(final Class<?> clazz)
{
return (Class<?>) ((ParameterizedType) clazz.getGenericSuperclass()).getActualTypeArguments()[0];
}

If you are working with spring:
public static Class<?>[] resolveTypeArguments(Class<?> parentClass, Class<?> subClass) {
if (subClass.isSynthetic()) {
return null;
}
return GenericTypeResolver.resolveTypeArguments(subClass, parentClass);
}
By the way, GenericTypeResolver will still get null for the non-subclasses class like the question mentioned, because the generic info of such class was completely erased after compilation.
The only way to solve this question may be:
public class GenericClass<T>
{
private final Class<T> clazz;
public Foo(Class<T> clazz) {
this.clazz= clazz;
}
public Type getMyType()
{
return clazz;
}
}

If you cannot change the generic class and use one of the method already explained on this page, then simple approach would be to get the type class based on the runtime instance class name.
Class getType(GenericType runtimeClassMember){
if (ClassA.class.equals(runtimeClassMember.getClass()){
return TypeForClassA.class;
} else if (ClassB.class.equals(runtimeClassMember.getClass()){
return TypeForClassB.class;
}
//throw an expectation or do whatever you want for the cases not described in the if section.
}

I did the same as #Moesio Above but in Kotlin it could be done this way:
class A<T : SomeClass>() {
var someClassType : T
init(){
this.someClassType = (javaClass.genericSuperclass as ParameterizedType).actualTypeArguments[0] as Class<T>
}
}

This was inspired by Pablo's and CoolMind's answers.
Occasionally I have also used the technique from kayz1's answer (expressed in many other answers as well), and I believe it is a decent and reliable way to do what the OP asked.
I chose to define this as an interface (similar to PJWeisberg) first because I have existing types that would benefit from this functionality, particularly a heterogeneous generic union type:
public interface IGenericType<T>
{
Class<T> getGenericTypeParameterType();
}
Where my simple implementation in a generic anonymous interface implementation looks like the following:
//Passed into the generic value generator function: toStore
//This value name is a field in the enclosing class.
//IUnionTypeValue<T> is a generic interface that extends IGenericType<T>
value = new IUnionTypeValue<T>() {
...
private T storedValue = toStore;
...
#SuppressWarnings("unchecked")
#Override
public Class<T> getGenericTypeParameterType()
{
return (Class<T>) storedValue.getClass();
}
}
I imagine this could be also implemented by being built with a class definition object as the source, that's just a separate use-case.
I think the key is as many other answers have stated, in one way or another, you need to get the type information at runtime to have it available at runtime; the objects themselves maintain their type, but erasure (also as others have said, with appropriate references) causes any enclosing/container types to lose that type information.

It might be useful to someone. You can Use java.lang.ref.WeakReference;
this way:
class SomeClass<N>{
WeakReference<N> variableToGetTypeFrom;
N getType(){
return variableToGetTypeFrom.get();
}
}

I found this to be a simple understandable and easily explainable solution
public class GenericClass<T> {
private Class classForT(T...t) {
return t.getClass().getComponentType();
}
public static void main(String[] args) {
GenericClass<String> g = new GenericClass<String>();
System.out.println(g.classForT());
System.out.println(String.class);
}
}

Get parameter class in generic java method [duplicate]

I have a generics class, Foo<T>. In a method of Foo, I want to get the class instance of type T, but I just can't call T.class.
What is the preferred way to get around it using T.class?

The short answer is, that there is no way to find out the runtime type of generic type parameters in Java. I suggest reading the chapter about type erasure in the Java Tutorial for more details.
A popular solution to this is to pass the Class of the type parameter into the constructor of the generic type, e.g.
class Foo<T> {
final Class<T> typeParameterClass;
public Foo(Class<T> typeParameterClass) {
this.typeParameterClass = typeParameterClass;
}
public void bar() {
// you can access the typeParameterClass here and do whatever you like
}
}

I was looking for a way to do this myself without adding an extra dependency to the classpath. After some investigation I found that it is possible as long as you have a generic supertype. This was OK for me as I was working with a DAO layer with a generic layer supertype. If this fits your scenario then it's the neatest approach IMHO.
Most generics use cases I've come across have some kind of generic supertype e.g. List<T> for ArrayList<T> or GenericDAO<T> for DAO<T>, etc.
Pure Java solution
The article Accessing generic types at runtime in Java explains how you can do it using pure Java.
#SuppressWarnings("unchecked")
public GenericJpaDao() {
this.entityBeanType = ((Class) ((ParameterizedType) getClass()
.getGenericSuperclass()).getActualTypeArguments()[0]);
}
Spring solution
My project was using Spring which is even better as Spring has a handy utility method for finding the type. This is the best approach for me as it looks neatest. I guess if you weren't using Spring you could write your own utility method.
import org.springframework.core.GenericTypeResolver;
public abstract class AbstractHibernateDao<T extends DomainObject> implements DataAccessObject<T>
{
#Autowired
private SessionFactory sessionFactory;
private final Class<T> genericType;
private final String RECORD_COUNT_HQL;
private final String FIND_ALL_HQL;
#SuppressWarnings("unchecked")
public AbstractHibernateDao()
{
this.genericType = (Class<T>) GenericTypeResolver.resolveTypeArgument(getClass(), AbstractHibernateDao.class);
this.RECORD_COUNT_HQL = "select count(*) from " + this.genericType.getName();
this.FIND_ALL_HQL = "from " + this.genericType.getName() + " t ";
}
Full code example
Some people are struggling in the comments to get this working so I wrote a small application to show both approaches in action.
https://github.com/benthurley82/generic-type-resolver-test

There is a small loophole however: if you define your Foo class as abstract.
That would mean you have to instantiate you class as:
Foo<MyType> myFoo = new Foo<MyType>(){};
(Note the double braces at the end.)
Now you can retrieve the type of T at runtime:
Type mySuperclass = myFoo.getClass().getGenericSuperclass();
Type tType = ((ParameterizedType)mySuperclass).getActualTypeArguments()[0];
Note however that mySuperclass has to be the superclass of the class definition actually defining the final type for T.
It is also not very elegant, but you have to decide whether you prefer new Foo<MyType>(){} or new Foo<MyType>(MyType.class); in your code.
For example:
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;
import java.util.ArrayDeque;
import java.util.Deque;
import java.util.NoSuchElementException;
/**
* Captures and silently ignores stack exceptions upon popping.
*/
public abstract class SilentStack<E> extends ArrayDeque<E> {
public E pop() {
try {
return super.pop();
}
catch( NoSuchElementException nsee ) {
return create();
}
}
public E create() {
try {
Type sooper = getClass().getGenericSuperclass();
Type t = ((ParameterizedType)sooper).getActualTypeArguments()[ 0 ];
return (E)(Class.forName( t.toString() ).newInstance());
}
catch( Exception e ) {
return null;
}
}
}
Then:
public class Main {
// Note the braces...
private Deque<String> stack = new SilentStack<String>(){};
public static void main( String args[] ) {
// Returns a new instance of String.
String s = stack.pop();
System.out.printf( "s = '%s'\n", s );
}
}

A standard approach/workaround/solution is to add a class object to the constructor(s), like:
public class Foo<T> {
private Class<T> type;
public Foo(Class<T> type) {
this.type = type;
}
public Class<T> getType() {
return type;
}
public T newInstance() {
return type.newInstance();
}
}

Here is a working solution:
#SuppressWarnings("unchecked")
private Class<T> getGenericTypeClass() {
    try {
        String className = ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0].getTypeName();
        Class<?> clazz = Class.forName(className);
        return (Class<T>) clazz;
    } catch (Exception e) {
        throw new IllegalStateException("Class is not parametrized with generic type!!! Please use extends <> ");
    }
}
NOTES:
Can be used only as superclass
Has to be extended with typed class (Child extends Generic<Integer>)
OR
Has to be created as anonymous implementation (new Generic<Integer>() {};)

Imagine you have an abstract superclass that is generic:
public abstract class Foo<? extends T> {}
And then you have a second class that extends Foo with a generic Bar that extends T:
public class Second extends Foo<Bar> {}
You can get the class Bar.class in the Foo class by selecting the Type (from bert bruynooghe answer) and infering it using Class instance:
Type mySuperclass = myFoo.getClass().getGenericSuperclass();
Type tType = ((ParameterizedType)mySuperclass).getActualTypeArguments()[0];
//Parse it as String
String className = tType.toString().split(" ")[1];
Class clazz = Class.forName(className);
You have to note this operation is not ideal, so it is a good idea to cache the computed value to avoid multiple calculations on this. One of the typical uses is in generic DAO implementation.
The final implementation:
public abstract class Foo<T> {
private Class<T> inferedClass;
public Class<T> getGenericClass(){
if(inferedClass == null){
Type mySuperclass = getClass().getGenericSuperclass();
Type tType = ((ParameterizedType)mySuperclass).getActualTypeArguments()[0];
String className = tType.toString().split(" ")[1];
inferedClass = Class.forName(className);
}
return inferedClass;
}
}
The value returned is Bar.class when invoked from Foo class in other function or from Bar class.

I had this problem in an abstract generic class. In this particular case, the solution is simpler:
abstract class Foo<T> {
abstract Class<T> getTClass();
//...
}
and later on the derived class:
class Bar extends Foo<Whatever> {
#Override
Class<T> getTClass() {
return Whatever.class;
}
}

Actually, it is possible (without external libraries!)
The following is my (ugly, yet effective) solution for this problem:
import java.lang.reflect.TypeVariable;
public static <T> Class<T> getGenericClass() {
__<T> instance = new __<T>();
TypeVariable<?>[] parameters = instance.getClass().getTypeParameters();
return (Class<T>)parameters[0].getClass();
}
// Generic helper class which (only) provides type information. This avoids the
// usage of a local variable of type T, which would have to be initialized.
private final class __<T> {
private __() { }
}

You can't do it because of type erasure. See also Stack Overflow question Java generics - type erasure - when and what happens.

A better route than the Class the others suggested is to pass in an object that can do what you would have done with the Class, e.g., create a new instance.
interface Factory<T> {
T apply();
}
<T> void List<T> make10(Factory<T> factory) {
List<T> result = new ArrayList<T>();
for (int a = 0; a < 10; a++)
result.add(factory.apply());
return result;
}
class FooFactory<T> implements Factory<Foo<T>> {
public Foo<T> apply() {
return new Foo<T>();
}
}
List<Foo<Integer>> foos = make10(new FooFactory<Integer>());

I assume that, since you have a generic class, you would have a variable like that:
private T t;
(this variable needs to take a value at the constructor)
In that case you can simply create the following method:
Class<T> getClassOfInstance()
{
return (Class<T>) t.getClass();
}
Hope it helps!

It's possible:
class Foo<T> {
Class<T> clazz = (Class<T>) DAOUtil.getTypeArguments(Foo.class, this.getClass()).get(0);
}
You need two functions from hibernate-generic-dao/blob/master/dao/src/main/java/com/googlecode/genericdao/dao/DAOUtil.java.
For more explanations, see Reflecting generics.

I found a generic and simple way to do that. In my class I created a method that returns the generic type according to it's position in the class definition. Let's assume a class definition like this:
public class MyClass<A, B, C> {
}
Now let's create some attributes to persist the types:
public class MyClass<A, B, C> {
private Class<A> aType;
private Class<B> bType;
private Class<C> cType;
// Getters and setters (not necessary if you are going to use them internally)
}
Then you can create a generic method that returns the type based on the index of the generic definition:
/**
* Returns a {#link Type} object to identify generic types
* #return type
*/
private Type getGenericClassType(int index) {
// To make it use generics without supplying the class type
Type type = getClass().getGenericSuperclass();
while (!(type instanceof ParameterizedType)) {
if (type instanceof ParameterizedType) {
type = ((Class<?>) ((ParameterizedType) type).getRawType()).getGenericSuperclass();
} else {
type = ((Class<?>) type).getGenericSuperclass();
}
}
return ((ParameterizedType) type).getActualTypeArguments()[index];
}
Finally, in the constructor just call the method and send the index for each type. The complete code should look like:
public class MyClass<A, B, C> {
private Class<A> aType;
private Class<B> bType;
private Class<C> cType;
public MyClass() {
this.aType = (Class<A>) getGenericClassType(0);
this.bType = (Class<B>) getGenericClassType(1);
this.cType = (Class<C>) getGenericClassType(2);
}
/**
* Returns a {#link Type} object to identify generic types
* #return type
*/
private Type getGenericClassType(int index) {
Type type = getClass().getGenericSuperclass();
while (!(type instanceof ParameterizedType)) {
if (type instanceof ParameterizedType) {
type = ((Class<?>) ((ParameterizedType) type).getRawType()).getGenericSuperclass();
} else {
type = ((Class<?>) type).getGenericSuperclass();
}
}
return ((ParameterizedType) type).getActualTypeArguments()[index];
}
}

That is pretty straight forward.
If you need from within the same class:
Class clazz = this.getClass();
ParameterizedType parameterizedType = (ParameterizedType) clazz.getGenericSuperclass();
try {
Class typeClass = Class.forName( parameterizedType.getActualTypeArguments()[0].getTypeName() );
// You have the instance of type 'T' in typeClass variable
System.out.println( "Class instance name: "+ typeClass.getName() );
} catch (ClassNotFoundException e) {
System.out.println( "ClassNotFound!! Something wrong! "+ e.getMessage() );
}

As explained in other answers, to use this ParameterizedType approach, you need to extend the class, but that seems like extra work to make a whole new class that extends it...
So, making the class abstract it forces you to extend it, thus satisfying the subclassing requirement. (using lombok's #Getter).
#Getter
public abstract class ConfigurationDefinition<T> {
private Class<T> type;
...
public ConfigurationDefinition(...) {
this.type = (Class<T>) ((ParameterizedType) this.getClass().getGenericSuperclass()).getActualTypeArguments()[0];
...
}
}
Now to extend it without defining a new class. (Note the {} on the end... extended, but don't overwrite anything - unless you want to).
private ConfigurationDefinition<String> myConfigA = new ConfigurationDefinition<String>(...){};
private ConfigurationDefinition<File> myConfigB = new ConfigurationDefinition<File>(...){};
...
Class stringType = myConfigA.getType();
Class fileType = myConfigB.getType();

Many people don't know this trick! Actually, I just found it today! It works like a dream! Just check this example out:
public static void main(String[] args) {
Date d=new Date(); //Or anything you want!
printMethods(d);
}
public static <T> void printMethods(T t){
Class<T> clazz= (Class<T>) t.getClass(); // There you go!
for ( Method m : clazz.getMethods()){
System.out.println( m.getName() );
}
}

I've created an example based on one of two most promising solutions here from this question.
The result is however not so promising, at least for my use case.
Only one approach is working, but you need a super class containing the method and the generic has to be set in the child class and cannot be assigned dynamically (which my use case sadly is)
import org.junit.jupiter.api.Test;
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;
import static org.junit.jupiter.api.Assertions.assertEquals;
import static org.junit.jupiter.api.Assertions.assertThrows;
public class GenericTest {
/**
* only this will work!
*/
#Test
void testGetGenericTypeClassFromChildClassWithSpecifiedType() {
TestClassWithSpecifiedType parent = new TestClassWithSpecifiedType();
assertEquals(SomeGenericType.class, parent.getGenericTypeClass());
}
/**
* won't work!
*/
#Test
void testGetGenericTypeClassFromChildClassWithUnspecifiedType() {
TestClassWithUnspecifiedType<SomeGenericType> parent = new TestClassWithUnspecifiedType<>();
assertThrows(IllegalStateException.class, parent::getGenericTypeClass);
}
/**
* won't work
*/
#Test
void testGetGenericTypeClassWithUnspecifiedType() {
SomeGenericTypedClass<SomeGenericType> parent = new SomeGenericTypedClass<>();
assertThrows(IllegalStateException.class, parent::getGenericTypeClass);
}
/**
* won't work
* returns object instead!
*/
#Test
void testGetLoadedClassFromObject() {
Foo<SomeGenericType> foo = new Foo<>();
Class<?> barClass = foo.getBarClass();
assertEquals(SomeGenericType.class, barClass);
}
/**
* A class that has specified the type parameter
*/
public static class TestClassWithSpecifiedType extends AbstractGenericTypedClass<SomeGenericType> {
}
/**
* A class where the type parameter will be specified on demand
*
* #param <T>
*/
public static class TestClassWithUnspecifiedType<T> extends AbstractGenericTypedClass<T> {
}
/**
* An abstract class, because otherwise finding the parameter will not work
*/
#SuppressWarnings("unchecked")
public static abstract class AbstractGenericTypedClass<T> {
#SuppressWarnings("unchecked")
public Class<T> getGenericTypeClass() {
try {
String className = ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0].getTypeName();
Class<?> clazz = Class.forName(className);
return (Class<T>) clazz;
} catch (Exception e) {
throw new IllegalStateException("Class is not parametrized with generic type!!! Please use extends <> ");
}
}
}
/**
* A typed class without abstract super class
*
* #param <T>
*/
public static class SomeGenericTypedClass<T> {
#SuppressWarnings("unchecked")
public Class<T> getGenericTypeClass() {
try {
String className = ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0].getTypeName();
Class<?> clazz = Class.forName(className);
return (Class<T>) clazz;
} catch (Exception e) {
throw new IllegalStateException("Class is not parametrized with generic type!!! Please use extends <> ");
}
}
}
/**
* Some generic type - won't work with primitives such as String, Integer, Double!
*/
public static class SomeGenericType {
}
public static class Foo<T> {
// The class:
private final Class<?> barClass;
public Foo() {
try {
// Im giving it [0] cuz Bar is the first TypeParam
Type[] bounds = getClass().getTypeParameters()[0].getBounds();
// Here, we get the class now:
barClass = Class.forName(bounds[0].getTypeName());
} catch (ClassNotFoundException e) {
// will never happen!
throw new Error("Something impossible happened!", e);
}
}
public Class<?> getBarClass() {
return barClass;
}
}
}
I do not really understand why this has to be so complicated, but I bet there have to be some technical limitations for the dynamically setting of parameters.

public <T> T yourMethodSignature(Class<T> type) {
// get some object and check the type match the given type
Object result = ...
if (type.isAssignableFrom(result.getClass())) {
return (T)result;
} else {
// handle the error
}
}

If you are extending or implementing any class/interface that are using generics , you may get the Generic Type of parent class/interface, without modifying any existing class/interface at all.
There could be three possibilities,
Case 1
When your class is extending a class that is using Generics
public class TestGenerics {
public static void main(String[] args) {
Type type = TestMySuperGenericType.class.getGenericSuperclass();
Type[] gTypes = ((ParameterizedType)type).getActualTypeArguments();
for(Type gType : gTypes){
System.out.println("Generic type:"+gType.toString());
}
}
}
class GenericClass<T> {
public void print(T obj){};
}
class TestMySuperGenericType extends GenericClass<Integer> {
}
Case 2
When your class is implementing an interface that is using Generics
public class TestGenerics {
public static void main(String[] args) {
Type[] interfaces = TestMySuperGenericType.class.getGenericInterfaces();
for(Type type : interfaces){
Type[] gTypes = ((ParameterizedType)type).getActualTypeArguments();
for(Type gType : gTypes){
System.out.println("Generic type:"+gType.toString());
}
}
}
}
interface GenericClass<T> {
public void print(T obj);
}
class TestMySuperGenericType implements GenericClass<Integer> {
public void print(Integer obj){}
}
Case 3
When your interface is extending an interface that is using Generics
public class TestGenerics {
public static void main(String[] args) {
Type[] interfaces = TestMySuperGenericType.class.getGenericInterfaces();
for(Type type : interfaces){
Type[] gTypes = ((ParameterizedType)type).getActualTypeArguments();
for(Type gType : gTypes){
System.out.println("Generic type:"+gType.toString());
}
}
}
}
interface GenericClass<T> {
public void print(T obj);
}
interface TestMySuperGenericType extends GenericClass<Integer> {
}

Actually, I suppose you have a field in your class of type T. If there's no field of type T, what's the point of having a generic Type? So, you can simply do an instanceof on that field.
In my case, I have a List<T> items; in my class, and I check if the class type is "Locality" by
if (items.get(0) instanceof Locality) ...
Of course, this only works if the total number of possible classes is limited.

This question is old, but now the best is use google Gson.
An example to get custom viewModel.
Class<CustomViewModel<String>> clazz = new GenericClass<CustomViewModel<String>>().getRawType();
CustomViewModel<String> viewModel = viewModelProvider.get(clazz);
Generic type class
class GenericClass<T>(private val rawType: Class<*>) {
constructor():this(`$Gson$Types`.getRawType(object : TypeToken<T>() {}.getType()))
fun getRawType(): Class<T> {
return rawType as Class<T>
}
}

I wanted to pass T.class to a method which make use of Generics
The method readFile reads a .csv file specified by the fileName with fullpath. There can be csv files with different contents hence i need to pass the model file class so that i can get the appropriate objects. Since this is reading csv file i wanted to do in a generic way. For some reason or other none of the above solutions worked for me. I need to use
Class<? extends T> type to make it work. I use opencsv library for parsing the CSV files.
private <T>List<T> readFile(String fileName, Class<? extends T> type) {
List<T> dataList = new ArrayList<T>();
try {
File file = new File(fileName);
Reader reader = new BufferedReader(new InputStreamReader(new FileInputStream(file)));
Reader headerReader = new BufferedReader(new InputStreamReader(new FileInputStream(file)));
CSVReader csvReader = new CSVReader(headerReader);
// create csv bean reader
CsvToBean<T> csvToBean = new CsvToBeanBuilder(reader)
.withType(type)
.withIgnoreLeadingWhiteSpace(true)
.build();
dataList = csvToBean.parse();
}
catch (Exception ex) {
logger.error("Error: ", ex);
}
return dataList;
}
This is how the readFile method is called
List<RigSurfaceCSV> rigSurfaceCSVDataList = readSurfaceFile(surfaceFileName, RigSurfaceCSV.class);

I'm using workaround for this:
class MyClass extends Foo<T> {
....
}
MyClass myClassInstance = MyClass.class.newInstance();

I have a custom Array class and I want to know the data type [duplicate]

How can I achieve this?
public class GenericClass<T>
{
public Type getMyType()
{
//How do I return the type of T?
}
}
Everything I have tried so far always returns type Object rather than the specific type used.

As others mentioned, it's only possible via reflection in certain circumstances.
If you really need the type, this is the usual (type-safe) workaround pattern:
public class GenericClass<T> {
private final Class<T> type;
public GenericClass(Class<T> type) {
this.type = type;
}
public Class<T> getMyType() {
return this.type;
}
}

I have seen something like this
private Class<T> persistentClass;
public Constructor() {
this.persistentClass = (Class<T>) ((ParameterizedType) getClass()
.getGenericSuperclass()).getActualTypeArguments()[0];
}
in the hibernate GenericDataAccessObjects Example

Generics are not reified at run-time. This means the information is not present at run-time.
Adding generics to Java while mantaining backward compatibility was a tour-de-force (you can see the seminal paper about it: Making the future safe for the past: adding genericity to the Java programming language).
There is a rich literature on the subject, and some people are dissatisfied with the current state, some says that actually it's a lure and there is no real need for it. You can read both links, I found them quite interesting.

Use Guava.
import com.google.common.reflect.TypeToken;
import java.lang.reflect.Type;
public abstract class GenericClass<T> {
private final TypeToken<T> typeToken = new TypeToken<T>(getClass()) { };
private final Type type = typeToken.getType(); // or getRawType() to return Class<? super T>
public Type getType() {
return type;
}
public static void main(String[] args) {
GenericClass<String> example = new GenericClass<String>() { };
System.out.println(example.getType()); // => class java.lang.String
}
}
A while back, I posted some full-fledge examples including abstract classes and subclasses here.
Note: this requires that you instantiate a subclass of GenericClass so it can bind the type parameter correctly. Otherwise it'll just return the type as T.

Java generics are mostly compile time, this means that the type information is lost at runtime.
class GenericCls<T>
{
T t;
}
will be compiled to something like
class GenericCls
{
Object o;
}
To get the type information at runtime you have to add it as an argument of the ctor.
class GenericCls<T>
{
private Class<T> type;
public GenericCls(Class<T> cls)
{
type= cls;
}
Class<T> getType(){return type;}
}
Example:
GenericCls<?> instance = new GenericCls<String>(String.class);
assert instance.getType() == String.class;

Sure, you can.
Java does not use the information at run time, for backwards compatibility reasons. But the information is actually present as metadata and can be accessed via reflection (but it is still not used for type-checking).
From the official API:
http://download.oracle.com/javase/6/docs/api/java/lang/reflect/ParameterizedType.html#getActualTypeArguments%28%29
However, for your scenario I would not use reflection. I'm personally more inclined to use that for framework code. In your case I would just add the type as a constructor param.

public abstract class AbstractDao<T>
{
private final Class<T> persistentClass;
public AbstractDao()
{
this.persistentClass = (Class<T>) ((ParameterizedType) this.getClass().getGenericSuperclass())
.getActualTypeArguments()[0];
}
}

I used follow approach:
public class A<T> {
protected Class<T> clazz;
public A() {
this.clazz = (Class<T>) ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0];
}
public Class<T> getClazz() {
return clazz;
}
}
public class B extends A<C> {
/* ... */
public void anything() {
// here I may use getClazz();
}
}

I dont think you can, Java uses type erasure when compiling so your code is compatible with applications and libraries that were created pre-generics.
From the Oracle Docs:
Type Erasure
Generics were introduced to the Java language to provide tighter type
checks at compile time and to support generic programming. To
implement generics, the Java compiler applies type erasure to:
Replace all type parameters in generic types with their bounds or
Object if the type parameters are unbounded. The produced bytecode,
therefore, contains only ordinary classes, interfaces, and methods.
Insert type casts if necessary to preserve type safety. Generate
bridge methods to preserve polymorphism in extended generic types.
Type erasure ensures that no new classes are created for parameterized
types; consequently, generics incur no runtime overhead.
http://docs.oracle.com/javase/tutorial/java/generics/erasure.html

Technique described in this article by Ian Robertson works for me.
In short quick and dirty example:
public abstract class AbstractDAO<T extends EntityInterface, U extends QueryCriteria, V>
{
/**
* Method returns class implementing EntityInterface which was used in class
* extending AbstractDAO
*
* #return Class<T extends EntityInterface>
*/
public Class<T> returnedClass()
{
return (Class<T>) getTypeArguments(AbstractDAO.class, getClass()).get(0);
}
/**
* Get the underlying class for a type, or null if the type is a variable
* type.
*
* #param type the type
* #return the underlying class
*/
public static Class<?> getClass(Type type)
{
if (type instanceof Class) {
return (Class) type;
} else if (type instanceof ParameterizedType) {
return getClass(((ParameterizedType) type).getRawType());
} else if (type instanceof GenericArrayType) {
Type componentType = ((GenericArrayType) type).getGenericComponentType();
Class<?> componentClass = getClass(componentType);
if (componentClass != null) {
return Array.newInstance(componentClass, 0).getClass();
} else {
return null;
}
} else {
return null;
}
}
/**
* Get the actual type arguments a child class has used to extend a generic
* base class.
*
* #param baseClass the base class
* #param childClass the child class
* #return a list of the raw classes for the actual type arguments.
*/
public static <T> List<Class<?>> getTypeArguments(
Class<T> baseClass, Class<? extends T> childClass)
{
Map<Type, Type> resolvedTypes = new HashMap<Type, Type>();
Type type = childClass;
// start walking up the inheritance hierarchy until we hit baseClass
while (!getClass(type).equals(baseClass)) {
if (type instanceof Class) {
// there is no useful information for us in raw types, so just keep going.
type = ((Class) type).getGenericSuperclass();
} else {
ParameterizedType parameterizedType = (ParameterizedType) type;
Class<?> rawType = (Class) parameterizedType.getRawType();
Type[] actualTypeArguments = parameterizedType.getActualTypeArguments();
TypeVariable<?>[] typeParameters = rawType.getTypeParameters();
for (int i = 0; i < actualTypeArguments.length; i++) {
resolvedTypes.put(typeParameters[i], actualTypeArguments[i]);
}
if (!rawType.equals(baseClass)) {
type = rawType.getGenericSuperclass();
}
}
}
// finally, for each actual type argument provided to baseClass, determine (if possible)
// the raw class for that type argument.
Type[] actualTypeArguments;
if (type instanceof Class) {
actualTypeArguments = ((Class) type).getTypeParameters();
} else {
actualTypeArguments = ((ParameterizedType) type).getActualTypeArguments();
}
List<Class<?>> typeArgumentsAsClasses = new ArrayList<Class<?>>();
// resolve types by chasing down type variables.
for (Type baseType : actualTypeArguments) {
while (resolvedTypes.containsKey(baseType)) {
baseType = resolvedTypes.get(baseType);
}
typeArgumentsAsClasses.add(getClass(baseType));
}
return typeArgumentsAsClasses;
}
}

I think there is another elegant solution.
What you want to do is (safely) "pass" the type of the generic type parameter up from the concerete class to the superclass.
If you allow yourself to think of the class type as "metadata" on the class, that suggests the Java method for encoding metadata in at runtime: annotations.
First define a custom annotation along these lines:
import java.lang.annotation.*;
#Target(ElementType.TYPE)
#Retention(RetentionPolicy.RUNTIME)
public #interface EntityAnnotation {
Class entityClass();
}
You can then have to add the annotation to your subclass.
#EntityAnnotation(entityClass = PassedGenericType.class)
public class Subclass<PassedGenericType> {...}
Then you can use this code to get the class type in your base class:
import org.springframework.core.annotation.AnnotationUtils;
.
.
.
private Class getGenericParameterType() {
final Class aClass = this.getClass();
EntityAnnotation ne =
AnnotationUtils.findAnnotation(aClass, EntityAnnotation.class);
return ne.entityClass();
}
Some limitations of this approach are:
You specify the generic type (PassedGenericType) in TWO places rather than one which is non-DRY.
This is only possible if you can modify the concrete subclasses.

Here's one way, which I've had to use once or twice:
public abstract class GenericClass<T>{
public abstract Class<T> getMyType();
}
Along with
public class SpecificClass extends GenericClass<String>{
#Override
public Class<String> getMyType(){
return String.class;
}
}

This is my solution:
import java.lang.reflect.Type;
import java.lang.reflect.TypeVariable;
public class GenericClass<T extends String> {
public static void main(String[] args) {
for (TypeVariable typeParam : GenericClass.class.getTypeParameters()) {
System.out.println(typeParam.getName());
for (Type bound : typeParam.getBounds()) {
System.out.println(bound);
}
}
}
}

Here is working solution!!!
#SuppressWarnings("unchecked")
private Class<T> getGenericTypeClass() {
    try {
        String className = ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0].getTypeName();
        Class<?> clazz = Class.forName(className);
        return (Class<T>) clazz;
    } catch (Exception e) {
        throw new IllegalStateException("Class is not parametrized with generic type!!! Please use extends <> ");
    }
}
NOTES:
Can be used only as superclass
1. Has to be extended with typed class (Child extends Generic<Integer>)
OR
2. Has to be created as anonymous implementation (new Generic<Integer>() {};)

You can't. If you add a member variable of type T to the class (you don't even have to initialise it), you could use that to recover the type.

One simple solution for this cab be like below
public class GenericDemo<T>{
private T type;
GenericDemo(T t)
{
this.type = t;
}
public String getType()
{
return this.type.getClass().getName();
}
public static void main(String[] args)
{
GenericDemo<Integer> obj = new GenericDemo<Integer>(5);
System.out.println("Type: "+ obj.getType());
}
}

To complete some of the answers here, I had to get the ParametrizedType of MyGenericClass, no matter how high is the hierarchy, with the help of recursion:
private Class<T> getGenericTypeClass() {
return (Class<T>) (getParametrizedType(getClass())).getActualTypeArguments()[0];
}
private static ParameterizedType getParametrizedType(Class clazz){
if(clazz.getSuperclass().equals(MyGenericClass.class)){ // check that we are at the top of the hierarchy
return (ParameterizedType) clazz.getGenericSuperclass();
} else {
return getParametrizedType(clazz.getSuperclass());
}
}

Here is my solution
public class GenericClass<T>
{
private Class<T> realType;
public GenericClass() {
findTypeArguments(getClass());
}
private void findTypeArguments(Type t) {
if (t instanceof ParameterizedType) {
Type[] typeArgs = ((ParameterizedType) t).getActualTypeArguments();
realType = (Class<T>) typeArgs[0];
} else {
Class c = (Class) t;
findTypeArguments(c.getGenericSuperclass());
}
}
public Type getMyType()
{
// How do I return the type of T? (your question)
return realType;
}
}
No matter how many level does your class hierarchy has,
this solution still works, for example:
public class FirstLevelChild<T> extends GenericClass<T> {
}
public class SecondLevelChild extends FirstLevelChild<String> {
}
In this case, getMyType() = java.lang.String

Here is my trick:
public class Main {
public static void main(String[] args) throws Exception {
System.out.println(Main.<String> getClazz());
}
static <T> Class getClazz(T... param) {
return param.getClass().getComponentType();
}
}

Just in case you use store a variable using the generic type you can easily solve this problem adding a getClassType method as follows:
public class Constant<T> {
private T value;
#SuppressWarnings("unchecked")
public Class<T> getClassType () {
return ((Class<T>) value.getClass());
}
}
I use the provided class object later to check if it is an instance of a given class, as follows:
Constant<?> constant = ...;
if (constant.getClassType().equals(Integer.class)) {
Constant<Integer> integerConstant = (Constant<Integer>)constant;
Integer value = integerConstant.getValue();
// ...
}

Here is my solution. The examples should explain it. The only requirement is that a subclass must set the generic type, not an object.
import java.lang.reflect.AccessibleObject;
import java.lang.reflect.Field;
import java.lang.reflect.Method;
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;
import java.lang.reflect.TypeVariable;
import java.util.HashMap;
import java.util.Map;
public class TypeUtils {
/*** EXAMPLES ***/
public static class Class1<A, B, C> {
public A someA;
public B someB;
public C someC;
public Class<?> getAType() {
return getTypeParameterType(this.getClass(), Class1.class, 0);
}
public Class<?> getCType() {
return getTypeParameterType(this.getClass(), Class1.class, 2);
}
}
public static class Class2<D, A, B, E, C> extends Class1<A, B, C> {
public B someB;
public D someD;
public E someE;
}
public static class Class3<E, C> extends Class2<String, Integer, Double, E, C> {
public E someE;
}
public static class Class4 extends Class3<Boolean, Long> {
}
public static void test() throws NoSuchFieldException {
Class4 class4 = new Class4();
Class<?> typeA = class4.getAType(); // typeA = Integer
Class<?> typeC = class4.getCType(); // typeC = Long
Field fieldSomeA = class4.getClass().getField("someA");
Class<?> typeSomeA = TypeUtils.getFieldType(class4.getClass(), fieldSomeA); // typeSomeA = Integer
Field fieldSomeE = class4.getClass().getField("someE");
Class<?> typeSomeE = TypeUtils.getFieldType(class4.getClass(), fieldSomeE); // typeSomeE = Boolean
}
/*** UTILS ***/
public static Class<?> getTypeVariableType(Class<?> subClass, TypeVariable<?> typeVariable) {
Map<TypeVariable<?>, Type> subMap = new HashMap<>();
Class<?> superClass;
while ((superClass = subClass.getSuperclass()) != null) {
Map<TypeVariable<?>, Type> superMap = new HashMap<>();
Type superGeneric = subClass.getGenericSuperclass();
if (superGeneric instanceof ParameterizedType) {
TypeVariable<?>[] typeParams = superClass.getTypeParameters();
Type[] actualTypeArgs = ((ParameterizedType) superGeneric).getActualTypeArguments();
for (int i = 0; i < typeParams.length; i++) {
Type actualType = actualTypeArgs[i];
if (actualType instanceof TypeVariable) {
actualType = subMap.get(actualType);
}
if (typeVariable == typeParams[i]) return (Class<?>) actualType;
superMap.put(typeParams[i], actualType);
}
}
subClass = superClass;
subMap = superMap;
}
return null;
}
public static Class<?> getTypeParameterType(Class<?> subClass, Class<?> superClass, int typeParameterIndex) {
return TypeUtils.getTypeVariableType(subClass, superClass.getTypeParameters()[typeParameterIndex]);
}
public static Class<?> getFieldType(Class<?> clazz, AccessibleObject element) {
Class<?> type = null;
Type genericType = null;
if (element instanceof Field) {
type = ((Field) element).getType();
genericType = ((Field) element).getGenericType();
} else if (element instanceof Method) {
type = ((Method) element).getReturnType();
genericType = ((Method) element).getGenericReturnType();
}
if (genericType instanceof TypeVariable) {
Class<?> typeVariableType = TypeUtils.getTypeVariableType(clazz, (TypeVariable) genericType);
if (typeVariableType != null) {
type = typeVariableType;
}
}
return type;
}
}

If you have a class like:
public class GenericClass<T> {
private T data;
}
with T variable, then you can print T name:
System.out.println(data.getClass().getSimpleName()); // "String", "Integer", etc.

Use an abstract method that returns the class type then use it in that class and wherever you extend generic class you will have to implement that abstract method to return the required class type
public class AbsractService<T>{
public abstract Class<T> getClassType ();
.......
}
at runtime
class AnimalService extends AbstractService<Animal>{
#Override
public Class<Animal> getClassType (){
return Animal.class;
}
.....
}

public static final Class<?> getGenericArgument(final Class<?> clazz)
{
return (Class<?>) ((ParameterizedType) clazz.getGenericSuperclass()).getActualTypeArguments()[0];
}

If you are working with spring:
public static Class<?>[] resolveTypeArguments(Class<?> parentClass, Class<?> subClass) {
if (subClass.isSynthetic()) {
return null;
}
return GenericTypeResolver.resolveTypeArguments(subClass, parentClass);
}
By the way, GenericTypeResolver will still get null for the non-subclasses class like the question mentioned, because the generic info of such class was completely erased after compilation.
The only way to solve this question may be:
public class GenericClass<T>
{
private final Class<T> clazz;
public Foo(Class<T> clazz) {
this.clazz= clazz;
}
public Type getMyType()
{
return clazz;
}
}

If you cannot change the generic class and use one of the method already explained on this page, then simple approach would be to get the type class based on the runtime instance class name.
Class getType(GenericType runtimeClassMember){
if (ClassA.class.equals(runtimeClassMember.getClass()){
return TypeForClassA.class;
} else if (ClassB.class.equals(runtimeClassMember.getClass()){
return TypeForClassB.class;
}
//throw an expectation or do whatever you want for the cases not described in the if section.
}

I did the same as #Moesio Above but in Kotlin it could be done this way:
class A<T : SomeClass>() {
var someClassType : T
init(){
this.someClassType = (javaClass.genericSuperclass as ParameterizedType).actualTypeArguments[0] as Class<T>
}
}

This was inspired by Pablo's and CoolMind's answers.
Occasionally I have also used the technique from kayz1's answer (expressed in many other answers as well), and I believe it is a decent and reliable way to do what the OP asked.
I chose to define this as an interface (similar to PJWeisberg) first because I have existing types that would benefit from this functionality, particularly a heterogeneous generic union type:
public interface IGenericType<T>
{
Class<T> getGenericTypeParameterType();
}
Where my simple implementation in a generic anonymous interface implementation looks like the following:
//Passed into the generic value generator function: toStore
//This value name is a field in the enclosing class.
//IUnionTypeValue<T> is a generic interface that extends IGenericType<T>
value = new IUnionTypeValue<T>() {
...
private T storedValue = toStore;
...
#SuppressWarnings("unchecked")
#Override
public Class<T> getGenericTypeParameterType()
{
return (Class<T>) storedValue.getClass();
}
}
I imagine this could be also implemented by being built with a class definition object as the source, that's just a separate use-case.
I think the key is as many other answers have stated, in one way or another, you need to get the type information at runtime to have it available at runtime; the objects themselves maintain their type, but erasure (also as others have said, with appropriate references) causes any enclosing/container types to lose that type information.

It might be useful to someone. You can Use java.lang.ref.WeakReference;
this way:
class SomeClass<N>{
WeakReference<N> variableToGetTypeFrom;
N getType(){
return variableToGetTypeFrom.get();
}
}

I found this to be a simple understandable and easily explainable solution
public class GenericClass<T> {
private Class classForT(T...t) {
return t.getClass().getComponentType();
}
public static void main(String[] args) {
GenericClass<String> g = new GenericClass<String>();
System.out.println(g.classForT());
System.out.println(String.class);
}
}

Return class object of generic in java [duplicate]

I'm creating a generic class and in one of the methods I need to know the Class of the generic type currently in use. The reason is that one of the method's I call expects this as an argument.
Example:
public class MyGenericClass<T> {
public void doSomething() {
// Snip...
// Call to a 3rd party lib
T bean = (T)someObject.create(T.class);
// Snip...
}
}
Clearly the example above doesn't work and results in the following error: Illegal class literal for the type parameter T.
My question is: does someone know a good alternative or workaround for this?

Still the same problems : Generic informations are erased at runtime, it cannot be recovered. A workaround is to pass the class T in parameter of a static method :
public class MyGenericClass<T> {
private final Class<T> clazz;
public static <U> MyGenericClass<U> createMyGeneric(Class<U> clazz) {
return new MyGenericClass<U>(clazz);
}
protected MyGenericClass(Class<T> clazz) {
this.clazz = clazz;
}
public void doSomething() {
T instance = clazz.newInstance();
}
}
It's ugly, but it works.

I was just pointed to this solution:
import java.lang.reflect.ParameterizedType;
public abstract class A<B> {
public Class<B> g() throws Exception {
ParameterizedType superclass =
(ParameterizedType) getClass().getGenericSuperclass();
return (Class<B>) superclass.getActualTypeArguments()[0];
}
}
This works if A is given a concrete type by a subclass:
new A<String>() {}.g() // this will work
class B extends A<String> {}
new B().g() // this will work
class C<T> extends A<T> {}
new C<String>().g() // this will NOT work

Unfortunately Christoph's solution as written only works in very limited circumstances. [EDIT: as commented below I no longer remember my reasoning for this sentence and it is likely wrong: "Note that this will only work in abstract classes, first of all."] The next difficulty is that g() only works from DIRECT subclasses of A. We can fix that, though:
private Class<?> extractClassFromType(Type t) throws ClassCastException {
if (t instanceof Class<?>) {
return (Class<?>)t;
}
return (Class<?>)((ParameterizedType)t).getRawType();
}
public Class<B> g() throws ClassCastException {
Class<?> superClass = getClass(); // initial value
Type superType;
do {
superType = superClass.getGenericSuperclass();
superClass = extractClassFromType(superType);
} while (! (superClass.equals(A.class)));
Type actualArg = ((ParameterizedType)superType).getActualTypeArguments()[0];
return (Class<B>)extractClassFromType(actualArg);
}
This will work in many situations in practice, but not ALL the time. Consider:
public class Foo<U,T extends Collection<?>> extends A<T> {}
(new Foo<String,List<Object>>() {}).g();
This will throw a ClassCastException, because the type argument here isn't a Class or a ParameterizedType at all; it's the TypeVariable T. So now you would be stuck trying to figure out what type T was supposed to stand for, and so on down the rabbit hole.
I think the only reasonable, general answer is something akin to Nicolas's initial answer -- in general, if your class needs to instantiate objects of some other class that is unknown at compile-time, users of your class need to pass that class literal (or, perhaps, a Factory) to your class explicitly and not rely solely on generics.

i find another way to obtain the Class of the generic object
public Class<?> getGenericClass(){
Class<?> result =null;
Type type =this.getClass().getGenericSuperclass();
if(type instanceofParameterizedType){
ParameterizedType pt =(ParameterizedType) type;
Type[] fieldArgTypes = pt.getActualTypeArguments();
result =(Class<?>) fieldArgTypes[0];
}
return result;
}

I will elaborate on Christoph's solution.
Here is the ClassGetter abstract class:
private abstract class ClassGetter<T> {
public final Class<T> get() {
final ParameterizedType superclass = (ParameterizedType)
getClass().getGenericSuperclass();
return (Class<T>)superclass.getActualTypeArguments()[0];
}
}
Here is a static method which uses the above class to find a generic class' type:
public static <T> Class<T> getGenericClass() {
return new ClassGetter<T>() {}.get();
}
As an example of it's usage, you could make this method:
public static final <T> T instantiate() {
final Class<T> clazz = getGenericClass();
try {
return clazz.getConstructor((Class[])null).newInstance(null);
} catch (Exception e) {
return null;
}
}
And then use it like this:
T var = instantiate();

public class DatabaseAccessUtil {
EntityManagerFactory entitymanagerfactory;
EntityManager entitymanager;
public DatabaseAccessUtil() {
entitymanagerfactory=Persistence.createEntityManagerFactory("bookmyshow");
entitymanager=entitymanagerfactory.createEntityManager();
}
public void save (T t) {
entitymanager.getTransaction().begin();
entitymanager.persist(t);
entitymanager.getTransaction().commit();
}
public void update(T t) {
entitymanager.getTransaction().begin();
entitymanager.persist(t);
entitymanager.getTransaction().commit();
}
public void delete(T t) {
entitymanager.getTransaction().begin();
entitymanager.remove(t);
entitymanager.getTransaction().commit();
}
public Object retrieve(Query query) {
return query.getSingleResult();
}
//call the method - retrieve(object,requiredclass.class)
public Object retrieve(Object primaryKey,class clazz) throws Exception {
return entitymanager.find(clazz,primaryKey);
}
}

Get generic type of class at runtime

How can I achieve this?
public class GenericClass<T>
{
public Type getMyType()
{
//How do I return the type of T?
}
}
Everything I have tried so far always returns type Object rather than the specific type used.

As others mentioned, it's only possible via reflection in certain circumstances.
If you really need the type, this is the usual (type-safe) workaround pattern:
public class GenericClass<T> {
private final Class<T> type;
public GenericClass(Class<T> type) {
this.type = type;
}
public Class<T> getMyType() {
return this.type;
}
}

I have seen something like this
private Class<T> persistentClass;
public Constructor() {
this.persistentClass = (Class<T>) ((ParameterizedType) getClass()
.getGenericSuperclass()).getActualTypeArguments()[0];
}
in the hibernate GenericDataAccessObjects Example

Generics are not reified at run-time. This means the information is not present at run-time.
Adding generics to Java while mantaining backward compatibility was a tour-de-force (you can see the seminal paper about it: Making the future safe for the past: adding genericity to the Java programming language).
There is a rich literature on the subject, and some people are dissatisfied with the current state, some says that actually it's a lure and there is no real need for it. You can read both links, I found them quite interesting.

Use Guava.
import com.google.common.reflect.TypeToken;
import java.lang.reflect.Type;
public abstract class GenericClass<T> {
private final TypeToken<T> typeToken = new TypeToken<T>(getClass()) { };
private final Type type = typeToken.getType(); // or getRawType() to return Class<? super T>
public Type getType() {
return type;
}
public static void main(String[] args) {
GenericClass<String> example = new GenericClass<String>() { };
System.out.println(example.getType()); // => class java.lang.String
}
}
A while back, I posted some full-fledge examples including abstract classes and subclasses here.
Note: this requires that you instantiate a subclass of GenericClass so it can bind the type parameter correctly. Otherwise it'll just return the type as T.

Java generics are mostly compile time, this means that the type information is lost at runtime.
class GenericCls<T>
{
T t;
}
will be compiled to something like
class GenericCls
{
Object o;
}
To get the type information at runtime you have to add it as an argument of the ctor.
class GenericCls<T>
{
private Class<T> type;
public GenericCls(Class<T> cls)
{
type= cls;
}
Class<T> getType(){return type;}
}
Example:
GenericCls<?> instance = new GenericCls<String>(String.class);
assert instance.getType() == String.class;

Sure, you can.
Java does not use the information at run time, for backwards compatibility reasons. But the information is actually present as metadata and can be accessed via reflection (but it is still not used for type-checking).
From the official API:
http://download.oracle.com/javase/6/docs/api/java/lang/reflect/ParameterizedType.html#getActualTypeArguments%28%29
However, for your scenario I would not use reflection. I'm personally more inclined to use that for framework code. In your case I would just add the type as a constructor param.

public abstract class AbstractDao<T>
{
private final Class<T> persistentClass;
public AbstractDao()
{
this.persistentClass = (Class<T>) ((ParameterizedType) this.getClass().getGenericSuperclass())
.getActualTypeArguments()[0];
}
}

I used follow approach:
public class A<T> {
protected Class<T> clazz;
public A() {
this.clazz = (Class<T>) ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0];
}
public Class<T> getClazz() {
return clazz;
}
}
public class B extends A<C> {
/* ... */
public void anything() {
// here I may use getClazz();
}
}

I dont think you can, Java uses type erasure when compiling so your code is compatible with applications and libraries that were created pre-generics.
From the Oracle Docs:
Type Erasure
Generics were introduced to the Java language to provide tighter type
checks at compile time and to support generic programming. To
implement generics, the Java compiler applies type erasure to:
Replace all type parameters in generic types with their bounds or
Object if the type parameters are unbounded. The produced bytecode,
therefore, contains only ordinary classes, interfaces, and methods.
Insert type casts if necessary to preserve type safety. Generate
bridge methods to preserve polymorphism in extended generic types.
Type erasure ensures that no new classes are created for parameterized
types; consequently, generics incur no runtime overhead.
http://docs.oracle.com/javase/tutorial/java/generics/erasure.html

Technique described in this article by Ian Robertson works for me.
In short quick and dirty example:
public abstract class AbstractDAO<T extends EntityInterface, U extends QueryCriteria, V>
{
/**
* Method returns class implementing EntityInterface which was used in class
* extending AbstractDAO
*
* #return Class<T extends EntityInterface>
*/
public Class<T> returnedClass()
{
return (Class<T>) getTypeArguments(AbstractDAO.class, getClass()).get(0);
}
/**
* Get the underlying class for a type, or null if the type is a variable
* type.
*
* #param type the type
* #return the underlying class
*/
public static Class<?> getClass(Type type)
{
if (type instanceof Class) {
return (Class) type;
} else if (type instanceof ParameterizedType) {
return getClass(((ParameterizedType) type).getRawType());
} else if (type instanceof GenericArrayType) {
Type componentType = ((GenericArrayType) type).getGenericComponentType();
Class<?> componentClass = getClass(componentType);
if (componentClass != null) {
return Array.newInstance(componentClass, 0).getClass();
} else {
return null;
}
} else {
return null;
}
}
/**
* Get the actual type arguments a child class has used to extend a generic
* base class.
*
* #param baseClass the base class
* #param childClass the child class
* #return a list of the raw classes for the actual type arguments.
*/
public static <T> List<Class<?>> getTypeArguments(
Class<T> baseClass, Class<? extends T> childClass)
{
Map<Type, Type> resolvedTypes = new HashMap<Type, Type>();
Type type = childClass;
// start walking up the inheritance hierarchy until we hit baseClass
while (!getClass(type).equals(baseClass)) {
if (type instanceof Class) {
// there is no useful information for us in raw types, so just keep going.
type = ((Class) type).getGenericSuperclass();
} else {
ParameterizedType parameterizedType = (ParameterizedType) type;
Class<?> rawType = (Class) parameterizedType.getRawType();
Type[] actualTypeArguments = parameterizedType.getActualTypeArguments();
TypeVariable<?>[] typeParameters = rawType.getTypeParameters();
for (int i = 0; i < actualTypeArguments.length; i++) {
resolvedTypes.put(typeParameters[i], actualTypeArguments[i]);
}
if (!rawType.equals(baseClass)) {
type = rawType.getGenericSuperclass();
}
}
}
// finally, for each actual type argument provided to baseClass, determine (if possible)
// the raw class for that type argument.
Type[] actualTypeArguments;
if (type instanceof Class) {
actualTypeArguments = ((Class) type).getTypeParameters();
} else {
actualTypeArguments = ((ParameterizedType) type).getActualTypeArguments();
}
List<Class<?>> typeArgumentsAsClasses = new ArrayList<Class<?>>();
// resolve types by chasing down type variables.
for (Type baseType : actualTypeArguments) {
while (resolvedTypes.containsKey(baseType)) {
baseType = resolvedTypes.get(baseType);
}
typeArgumentsAsClasses.add(getClass(baseType));
}
return typeArgumentsAsClasses;
}
}

I think there is another elegant solution.
What you want to do is (safely) "pass" the type of the generic type parameter up from the concerete class to the superclass.
If you allow yourself to think of the class type as "metadata" on the class, that suggests the Java method for encoding metadata in at runtime: annotations.
First define a custom annotation along these lines:
import java.lang.annotation.*;
#Target(ElementType.TYPE)
#Retention(RetentionPolicy.RUNTIME)
public #interface EntityAnnotation {
Class entityClass();
}
You can then have to add the annotation to your subclass.
#EntityAnnotation(entityClass = PassedGenericType.class)
public class Subclass<PassedGenericType> {...}
Then you can use this code to get the class type in your base class:
import org.springframework.core.annotation.AnnotationUtils;
.
.
.
private Class getGenericParameterType() {
final Class aClass = this.getClass();
EntityAnnotation ne =
AnnotationUtils.findAnnotation(aClass, EntityAnnotation.class);
return ne.entityClass();
}
Some limitations of this approach are:
You specify the generic type (PassedGenericType) in TWO places rather than one which is non-DRY.
This is only possible if you can modify the concrete subclasses.

Here's one way, which I've had to use once or twice:
public abstract class GenericClass<T>{
public abstract Class<T> getMyType();
}
Along with
public class SpecificClass extends GenericClass<String>{
#Override
public Class<String> getMyType(){
return String.class;
}
}

This is my solution:
import java.lang.reflect.Type;
import java.lang.reflect.TypeVariable;
public class GenericClass<T extends String> {
public static void main(String[] args) {
for (TypeVariable typeParam : GenericClass.class.getTypeParameters()) {
System.out.println(typeParam.getName());
for (Type bound : typeParam.getBounds()) {
System.out.println(bound);
}
}
}
}

Here is working solution!!!
#SuppressWarnings("unchecked")
private Class<T> getGenericTypeClass() {
    try {
        String className = ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0].getTypeName();
        Class<?> clazz = Class.forName(className);
        return (Class<T>) clazz;
    } catch (Exception e) {
        throw new IllegalStateException("Class is not parametrized with generic type!!! Please use extends <> ");
    }
}
NOTES:
Can be used only as superclass
1. Has to be extended with typed class (Child extends Generic<Integer>)
OR
2. Has to be created as anonymous implementation (new Generic<Integer>() {};)

You can't. If you add a member variable of type T to the class (you don't even have to initialise it), you could use that to recover the type.

One simple solution for this cab be like below
public class GenericDemo<T>{
private T type;
GenericDemo(T t)
{
this.type = t;
}
public String getType()
{
return this.type.getClass().getName();
}
public static void main(String[] args)
{
GenericDemo<Integer> obj = new GenericDemo<Integer>(5);
System.out.println("Type: "+ obj.getType());
}
}

To complete some of the answers here, I had to get the ParametrizedType of MyGenericClass, no matter how high is the hierarchy, with the help of recursion:
private Class<T> getGenericTypeClass() {
return (Class<T>) (getParametrizedType(getClass())).getActualTypeArguments()[0];
}
private static ParameterizedType getParametrizedType(Class clazz){
if(clazz.getSuperclass().equals(MyGenericClass.class)){ // check that we are at the top of the hierarchy
return (ParameterizedType) clazz.getGenericSuperclass();
} else {
return getParametrizedType(clazz.getSuperclass());
}
}

Here is my solution
public class GenericClass<T>
{
private Class<T> realType;
public GenericClass() {
findTypeArguments(getClass());
}
private void findTypeArguments(Type t) {
if (t instanceof ParameterizedType) {
Type[] typeArgs = ((ParameterizedType) t).getActualTypeArguments();
realType = (Class<T>) typeArgs[0];
} else {
Class c = (Class) t;
findTypeArguments(c.getGenericSuperclass());
}
}
public Type getMyType()
{
// How do I return the type of T? (your question)
return realType;
}
}
No matter how many level does your class hierarchy has,
this solution still works, for example:
public class FirstLevelChild<T> extends GenericClass<T> {
}
public class SecondLevelChild extends FirstLevelChild<String> {
}
In this case, getMyType() = java.lang.String

Here is my trick:
public class Main {
public static void main(String[] args) throws Exception {
System.out.println(Main.<String> getClazz());
}
static <T> Class getClazz(T... param) {
return param.getClass().getComponentType();
}
}

Just in case you use store a variable using the generic type you can easily solve this problem adding a getClassType method as follows:
public class Constant<T> {
private T value;
#SuppressWarnings("unchecked")
public Class<T> getClassType () {
return ((Class<T>) value.getClass());
}
}
I use the provided class object later to check if it is an instance of a given class, as follows:
Constant<?> constant = ...;
if (constant.getClassType().equals(Integer.class)) {
Constant<Integer> integerConstant = (Constant<Integer>)constant;
Integer value = integerConstant.getValue();
// ...
}

Here is my solution. The examples should explain it. The only requirement is that a subclass must set the generic type, not an object.
import java.lang.reflect.AccessibleObject;
import java.lang.reflect.Field;
import java.lang.reflect.Method;
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;
import java.lang.reflect.TypeVariable;
import java.util.HashMap;
import java.util.Map;
public class TypeUtils {
/*** EXAMPLES ***/
public static class Class1<A, B, C> {
public A someA;
public B someB;
public C someC;
public Class<?> getAType() {
return getTypeParameterType(this.getClass(), Class1.class, 0);
}
public Class<?> getCType() {
return getTypeParameterType(this.getClass(), Class1.class, 2);
}
}
public static class Class2<D, A, B, E, C> extends Class1<A, B, C> {
public B someB;
public D someD;
public E someE;
}
public static class Class3<E, C> extends Class2<String, Integer, Double, E, C> {
public E someE;
}
public static class Class4 extends Class3<Boolean, Long> {
}
public static void test() throws NoSuchFieldException {
Class4 class4 = new Class4();
Class<?> typeA = class4.getAType(); // typeA = Integer
Class<?> typeC = class4.getCType(); // typeC = Long
Field fieldSomeA = class4.getClass().getField("someA");
Class<?> typeSomeA = TypeUtils.getFieldType(class4.getClass(), fieldSomeA); // typeSomeA = Integer
Field fieldSomeE = class4.getClass().getField("someE");
Class<?> typeSomeE = TypeUtils.getFieldType(class4.getClass(), fieldSomeE); // typeSomeE = Boolean
}
/*** UTILS ***/
public static Class<?> getTypeVariableType(Class<?> subClass, TypeVariable<?> typeVariable) {
Map<TypeVariable<?>, Type> subMap = new HashMap<>();
Class<?> superClass;
while ((superClass = subClass.getSuperclass()) != null) {
Map<TypeVariable<?>, Type> superMap = new HashMap<>();
Type superGeneric = subClass.getGenericSuperclass();
if (superGeneric instanceof ParameterizedType) {
TypeVariable<?>[] typeParams = superClass.getTypeParameters();
Type[] actualTypeArgs = ((ParameterizedType) superGeneric).getActualTypeArguments();
for (int i = 0; i < typeParams.length; i++) {
Type actualType = actualTypeArgs[i];
if (actualType instanceof TypeVariable) {
actualType = subMap.get(actualType);
}
if (typeVariable == typeParams[i]) return (Class<?>) actualType;
superMap.put(typeParams[i], actualType);
}
}
subClass = superClass;
subMap = superMap;
}
return null;
}
public static Class<?> getTypeParameterType(Class<?> subClass, Class<?> superClass, int typeParameterIndex) {
return TypeUtils.getTypeVariableType(subClass, superClass.getTypeParameters()[typeParameterIndex]);
}
public static Class<?> getFieldType(Class<?> clazz, AccessibleObject element) {
Class<?> type = null;
Type genericType = null;
if (element instanceof Field) {
type = ((Field) element).getType();
genericType = ((Field) element).getGenericType();
} else if (element instanceof Method) {
type = ((Method) element).getReturnType();
genericType = ((Method) element).getGenericReturnType();
}
if (genericType instanceof TypeVariable) {
Class<?> typeVariableType = TypeUtils.getTypeVariableType(clazz, (TypeVariable) genericType);
if (typeVariableType != null) {
type = typeVariableType;
}
}
return type;
}
}

If you have a class like:
public class GenericClass<T> {
private T data;
}
with T variable, then you can print T name:
System.out.println(data.getClass().getSimpleName()); // "String", "Integer", etc.

Use an abstract method that returns the class type then use it in that class and wherever you extend generic class you will have to implement that abstract method to return the required class type
public class AbsractService<T>{
public abstract Class<T> getClassType ();
.......
}
at runtime
class AnimalService extends AbstractService<Animal>{
#Override
public Class<Animal> getClassType (){
return Animal.class;
}
.....
}

public static final Class<?> getGenericArgument(final Class<?> clazz)
{
return (Class<?>) ((ParameterizedType) clazz.getGenericSuperclass()).getActualTypeArguments()[0];
}

If you are working with spring:
public static Class<?>[] resolveTypeArguments(Class<?> parentClass, Class<?> subClass) {
if (subClass.isSynthetic()) {
return null;
}
return GenericTypeResolver.resolveTypeArguments(subClass, parentClass);
}
By the way, GenericTypeResolver will still get null for the non-subclasses class like the question mentioned, because the generic info of such class was completely erased after compilation.
The only way to solve this question may be:
public class GenericClass<T>
{
private final Class<T> clazz;
public Foo(Class<T> clazz) {
this.clazz= clazz;
}
public Type getMyType()
{
return clazz;
}
}

If you cannot change the generic class and use one of the method already explained on this page, then simple approach would be to get the type class based on the runtime instance class name.
Class getType(GenericType runtimeClassMember){
if (ClassA.class.equals(runtimeClassMember.getClass()){
return TypeForClassA.class;
} else if (ClassB.class.equals(runtimeClassMember.getClass()){
return TypeForClassB.class;
}
//throw an expectation or do whatever you want for the cases not described in the if section.
}

I did the same as #Moesio Above but in Kotlin it could be done this way:
class A<T : SomeClass>() {
var someClassType : T
init(){
this.someClassType = (javaClass.genericSuperclass as ParameterizedType).actualTypeArguments[0] as Class<T>
}
}

This was inspired by Pablo's and CoolMind's answers.
Occasionally I have also used the technique from kayz1's answer (expressed in many other answers as well), and I believe it is a decent and reliable way to do what the OP asked.
I chose to define this as an interface (similar to PJWeisberg) first because I have existing types that would benefit from this functionality, particularly a heterogeneous generic union type:
public interface IGenericType<T>
{
Class<T> getGenericTypeParameterType();
}
Where my simple implementation in a generic anonymous interface implementation looks like the following:
//Passed into the generic value generator function: toStore
//This value name is a field in the enclosing class.
//IUnionTypeValue<T> is a generic interface that extends IGenericType<T>
value = new IUnionTypeValue<T>() {
...
private T storedValue = toStore;
...
#SuppressWarnings("unchecked")
#Override
public Class<T> getGenericTypeParameterType()
{
return (Class<T>) storedValue.getClass();
}
}
I imagine this could be also implemented by being built with a class definition object as the source, that's just a separate use-case.
I think the key is as many other answers have stated, in one way or another, you need to get the type information at runtime to have it available at runtime; the objects themselves maintain their type, but erasure (also as others have said, with appropriate references) causes any enclosing/container types to lose that type information.

It might be useful to someone. You can Use java.lang.ref.WeakReference;
this way:
class SomeClass<N>{
WeakReference<N> variableToGetTypeFrom;
N getType(){
return variableToGetTypeFrom.get();
}
}

I found this to be a simple understandable and easily explainable solution
public class GenericClass<T> {
private Class classForT(T...t) {
return t.getClass().getComponentType();
}
public static void main(String[] args) {
GenericClass<String> g = new GenericClass<String>();
System.out.println(g.classForT());
System.out.println(String.class);
}
}