获取抽象超类上泛型类型参数的实际类型

Question

I have a class like:

public abstract class BaseDao<T extends PersistentObject> {

  protected Class<T> getClazz() {
     return T.class;
  }

  // ...

}

But the compiler says to T.class; : Illegal class literal for the type parameter T .

How can I get the class of T ?

Answer 1

It's definitely possible to extract it from Class#getGenericSuperclass() because it's not defined during runtime, but during compiletime by FooDao extends BaseDao<Foo> .

Here's a kickoff example how you could extract the desired generic super type in the constructor of the abstract class, taking a hierarchy of subclasses into account (along with a real world use case of applying it on generic EntityManager methods without the need to explicitly supply the type):

public abstract class BaseDao<E extends BaseEntity> {

    @PersistenceContext
    private EntityManager em;

    private Class<E> type;

    @SuppressWarnings("unchecked") // For the cast on Class<E>.
    public BaseDao() {
        Type type = getClass().getGenericSuperclass();

        while (!(type instanceof ParameterizedType) || ((ParameterizedType) type).getRawType() != BaseDao.class) {
            if (type instanceof ParameterizedType) {
                type = ((Class<?>) ((ParameterizedType) type).getRawType()).getGenericSuperclass();
            } else {
                type = ((Class<?>) type).getGenericSuperclass();
            }
        }

        this.type = (Class<E>) ((ParameterizedType) type).getActualTypeArguments()[0];
    }

    public E find(Long id) {
        return em.find(type, id);
    }

    public List<E> list() {
        return em.createQuery(String.format("SELECT e FROM %s e ORDER BY id", type.getSimpleName()), type).getResultList();
    }

    // ...
}

Answer 2

Actually, this is not as easy as it seems. There is a problem when you have rich type hierarchy and want to get generic parameter in the supertype. For example, you may have the following hierarchy:

public abstract class BaseDao<T extends BaseEntity> {
...
}

public abstract class SpecialDao<X extends SomeType, E extends BaseEntity> extends BaseDao<E> {
...
}

public class MyDao extends SpecialDao<TypeImpl, EntityImpl> {
...
}

Calling getClass().getGenericSuperclass() in an instance of MyDao returns SpecialDao<TypeImpl, EntityImpl> , but when you call it inside BaseDao method, you don't know how deep the generic hierarchy is. Moreover, as far as I know, you cannot obtain generic supertype of a supertype. Thus, when you invoke getClass().getGenericSuperclass().getRawType().getGenericSuperclass() (with some typecasting omitted for readability), you'll get BaseDao<E> (notice <E> instead of <T> ). Since getRawType() strips all type-variable mapping from the type, we're starting with unmapped type variables X and E . Then getGenericSuperclass() just maps these type variables to their positions in BaseDao .

This behavior can be used so that we keep mapping from type variables to their actual values while traversing the type hierarchy. When we hit the class we want, we simply look up its type parameters in the map. Here is the code:

@SuppressWarnings("unchecked")
public static <T> Class<T> getGenericClassParameter(final Class<?> parameterizedSubClass, final Class<?> genericSuperClass, final int pos) {
    // a mapping from type variables to actual values (classes)
    Map<TypeVariable<?>, Class<?>> mapping = new HashMap<>();

    Class<?> klass = parameterizedSubClass;
    while (klass != null) {
        Type type = klass.getGenericSuperclass();
        if (type instanceof ParameterizedType) {
            ParameterizedType parType = (ParameterizedType) type;
            Type rawType = parType.getRawType();
            if (rawType == genericSuperClass) {
                // found
                Type t = parType.getActualTypeArguments()[pos];
                if (t instanceof Class<?>) {
                    return (Class<T>) t;
                } else {
                    return (Class<T>) mapping.get((TypeVariable<?>)t);
                }
            }
            // resolve
            Type[] vars = ((GenericDeclaration)(parType.getRawType())).getTypeParameters();
            Type[] args = parType.getActualTypeArguments();
            for (int i = 0; i < vars.length; i++) {
                if (args[i] instanceof Class<?>) {
                    mapping.put((TypeVariable)vars[i], (Class<?>)args[i]);
                } else {
                    mapping.put((TypeVariable)vars[i], mapping.get((TypeVariable<?>)(args[i])));
                }
            }
            klass = (Class<?>) rawType;
        } else {
            klass = klass.getSuperclass();
        }
    }
    throw new IllegalArgumentException("no generic supertype for " + parameterizedSubClass + " of type " + genericSuperClass);
}

Answer 3

If Spring framework is available, you can do like here:

import org.springframework.core.GenericTypeResolver;

public abstract class BaseDao<T extends PersistentObject> {

    protected Class<T> getClazz() {
        return (Class<T>) GenericTypeResolver.resolveTypeArgument(getClass(), BaseDao.class);
    }

}

Answer 4

If your class is abstract, you can try with this:

public class<T> getClassOfT() {
    final ParameterizedType type = (ParameterizedType) this.getClass()
            .getGenericSuperclass();
    Class<T> clazz = (Class<T>) type.getActualTypeArguments()[0];
    return clazz;
}

This only work if the instance is a direct subclass , and the type of the class you want is the first one (see the [0]).

If you have a large hierarchy of dao's, you can try fidn the BaseDao recursively and get the parametrized type

See a example here (see the output in the bottom)

Cheers and sorry for my bad english

Answer 5

Common way to sort this issue in a secure way is to add a constructor to store the class of the type. Example in your context:

public abstract class BaseDao<T extends PersistentObject> {
  private Class<T> classT;

  BaseDao(Class<T> classT){
    this.classT=classT;
  }

  protected Class<T> getClazz() {
     return classT;
  }

  // ...

}

Answer 6

您可以为此查看TypeTools ：

Class<T> t = (Class<T>)TypeResolver.resolveRawArgument(BaseDao.class, getClass());

Answer 7

You can pass the type of class through the constructor of the abstract class, like this.

public abstract class BaseDao<T extends PersistentObject> {

    private final Class<T> typeClass;

    //Here in the constructor, you are asking for the actual Type Class.
    BaseDao(Class<T> typeClass) {
        this.typeClass = typeClass;
    }

   // ...

}

And in the concrete class pass the class type in the constructor.

public class BaseDaoImpl extends BaseDao<ActualType> {
   
   BaseDaoImpl() {
       //Here in the implementation, you pass the type class
       super(ActualType.class);
   }
}

The problem with this approach is that you cannot ensure that the type you are passing through the constructor is the actual type is being set in generic, so this can lead to issues if developers do not make sure they are passing the correct type through the constructor.