How to determine the class of a generic type?

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;
  }