Using a generic type of a subclass within it's abstract superclass?

Question

Hi there!

Within my code a have the following abstract superclass

public abstract class AbstractClass<Type extends A> {...}

and some child classes like

public class ChildClassA extends AbstractClass<GenericTypeA> {...}

public class ChildClassB extends AbstractClass<GenericTypeB> {...}

I'm searching for an elegant way how I can use the generic type of the child classes (GenericTypeA, GenericTypeB, ...) inside the abstract class in a generic way.

To solve this problem I currently defined the method

protected abstract Class<Type> getGenericTypeClass();

in my abstract class and implemented the method

@Override
protected Class<GenericType> getGenericTypeClass() {
    return GenericType.class;
}

in every child class.

Is it possible to get the generic type of the child classes in my abstract class without implementing this helper method?

BR,

Markus

Answer 1

No, it is not possible, because the generic type information is erased during compilation, so it is not available anymore during execution.

Apart from implementing the method in each subclass as you did, the only option would be to pass the concrete class token to the base class (e.g. in the constructor), so that getGenericTypeClass can return it.

Update: on type erasure:

When a generic type is instantiated, the compiler translates those types by a technique called type erasure — a process where the compiler removes all information related to type parameters and type arguments within a class or method. Type erasure enables Java applications that use generics to maintain binary compatibility with Java libraries and applications that were created before generics.

For instance, Box<String> is translated to type Box, which is called the raw type — a raw type is a generic class or interface name without any type arguments. This means that you can't find out what type of Object a generic class is using at runtime. The following operations are not possible:

public class MyClass<E> {
    public static void myMethod(Object item) {
        if (item instanceof E) {  //Compiler error
            ...
        }
        E item2 = new E();   //Compiler error
        E[] iArray = new E[10]; //Compiler error
        E obj = (E)new Object(); //Unchecked cast warning
    }
}

Answer 2

I'm not sure I fully understand your question - <Type> is the generic type of the subclass, even when it's being expressed in the abstract class. For example, if your abstract superclass defines a method:

public void augment(Type entity) {
   ...
}

and you instantiate a ChildClassA, you'll only be able to call augment with an instance of GenericTypeA.

Now if you want a class literal, then you'll need to provide the method as you indicated. But if you just want the generic parameter, you don't need to do anything special.

Answer 3

I think its possible. I saw this was being used in the DAO patterns along with generics. e.g. Consider classes:

public class A {}
public class B extends A {}

And your generic class:

  import java.lang.reflect.ParameterizedType;
  public abstract class Test<T extends A> {

     private Class<T> theType;

     public Test()  {
        theType = (Class<T>) (
               (ParameterizedType) getClass().getGenericSuperclass())
              .getActualTypeArguments()[0];
     }

     // this method will always return the type that extends class "A"
     public Class<T> getTheType()   {
        return theType;
     }

     public void printType() {
        Class<T> clazz = getTheType();
        System.out.println(clazz);
     }
  }

You can have a class Test1 that extends Test with class B (it extends A)

  public class Test1 extends Test<B>  {

     public static void main(String[] args) {
        Test1 t = new Test1();

        Class<B> clazz = t.getTheType();

        System.out.println(clazz); // will print 'class B'
        System.out.println(printType()); // will print 'class B'
     }
  }