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
class A implements Comparable<String>
, the actual type parameter is String
, but it CANNOT tell that in Set<String> a = new TreeSet<String>()
, the actual type parameter is String
. In fact, the type parameter information is "erased" after compilation, as explained in other answers.
java.lang.Class cannot be cast to java.lang.reflect.ParameterizedType
for this answer.
Class-Mate
from the Jackson folks. I wrote a gist here gist.github.com/yunspace/930d4d40a787a1f6a7d1
(Class<T>) ((ParameterizedType)getClass().getSuperclass().getGenericSuperclass()).getActualTypeArguments()
to get to actual type arguments.
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.lang.IllegalArgumentException: class com.google.common.reflect.TypeToken isn't parameterized
. So I changed the line new TypeToken(getClass()) { }
to new TypeToken<T>(getClass()) { }
. Now, code runs fine, but Type is still 'T'. See this: gist.github.com/m-manu/9cda9d8f9d53bead2035
GenericClass
, you should make that class abstract
so wrong usage doesn’t compile.
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;
private final Class<T> type;
Type t = //String[]
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:
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.
actualTypeArguments
via my IDE's debugger but I cannot figure out how to actually access it.
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;
}
}
Type[] actualTypeArguments = parameterizedType.getActualTypeArguments();
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
Exception in thread "main" java.lang.NullPointerException at Main$ClassA.findTypeArguments(Main.java:54) at Main$ClassA.findTypeArguments(Main.java:54) at Main$ClassA.findTypeArguments(Main.java:54) at Main$ClassA.<init>(Main.java:43) at Main.main(Main.java:61)
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();
}
}
T
is a type variable. In the case that T
is a type variable, the varargs creates an array of erasure of T
. See e.g. http://ideone.com/DIPNwd.
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();
// ...
}
value
is null
? Second of all, what if value
is a subclass of T
? Constant<Number> c = new Constant<Number>(new Integer(0)); Class<Number> n = c.getClassType();
returns Integer.class
when it should return Number.class
. It would be more correct to return Class<? extends T>
. Integer.class
is a Class<? extends Number>
but not a Class<Number>
.
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;
}
}
public static final Class<?> getGenericArgument(final Class<?> clazz)
{
return (Class<?>) ((ParameterizedType) clazz.getGenericSuperclass()).getActualTypeArguments()[0];
}
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.
data == null
. In this case we cannot get a type.
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.
}
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<MyType>{
public Class<MyType> getClassType (){
return MyType.class;
}
.....
}
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();
}
}
WeakReference
? Please provide some explanation with your answer, not just some code.
SomeClass<MyClass>
you can instantiate SomeClass
and call getType
on that instance and have the runtime to be MyClass
.
WeakReference
? What you said is not different from most of the other answers.
AtomicReference
, List
, Set
).
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);
}
}
(T...t)
. (That's why this code doesn't work.)
Success story sharing
Foo foo1 = GetDao<Foo>(Foo.class).get(Foo.class, 1)
public static <T> GenericClass<T> of(Class<T> type) {...}
and then call it as such:GenericClass<String> var = GenericClass.of(String.class)
. A bit nicer.