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> {
}
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>
}
}
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();
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!
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
Child extends Generic<Integer>
)OR
new Generic<Integer>() {};
) 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];
}
}
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.
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>());
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.
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.
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]);
}
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 ";
}
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(i);
}
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 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;
}
}
I'm using workaround for this:
class MyClass extends Foo<T> {
....
}
MyClass myClassInstance = MyClass.class.newInstance();
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() );
}
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
}
}
You can't do it because of type erasure. See also Stack Overflow question Java generics - type erasure - when and what happens.
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();
}
}
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 );
}
}
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);
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 have an (ugly but effective) solution for this problem, which I used recently:
import java.lang.reflect.TypeVariable;
public static <T> Class<T> getGenericClass()
{
__<T> ins = new __<T>();
TypeVariable<?>[] cls = ins.getClass().getTypeParameters();
return (Class<T>)cls[0].getClass();
}
private final class __<T> // generic helper class which does only provide type information
{
private __()
{
}
}
Source: Stackoverflow.com