Getting type from class name with Java generics

Question

I have following class, I need to get type in constructor, how can I do that?

public abstract class MyClass<T> {
    public MyClass()
    {
        // I need T type here ...
    }
}

EDIT:

Here is concrete example what I want to achieve:

public abstract class Dao<T> {
    public void save(GoogleAppEngineEntity entity)
    {
        // save entity to datastore here
    }

    public GoogleAppEngineEntity getEntityById(Long id)
    {
        // return entity of class T, how can I do that ??
    }
}

What I want to do is to have this class extended to all other DAOs, because other DAOs have some queries that are specific to those daos and cannot be general, but these simple queries should be generally available to all DAO interfaces/implementations...

Depends on what you want to do with it. Do you want to instantiate a T? Do you want to call another function with a T? — dplass
– dplass, Commented Jun 3, 2010 at 21:21
I have dao that all other daos extend, so this extended dao will have some basic operation that can be same for all daos, so I need to know class where resulted objects should be mapped — newbie
– newbie, Commented Jun 3, 2010 at 21:30
I'm not sure I understand, but since you have T, you can "map" it in the various other methods in your DAO, e.g., T doSomething(T obj) — dplass
– dplass, Commented Jun 4, 2010 at 13:06
You need to expand and give concrete examples of what you're trying to accomplish. What you want to do isn't generally possible due to erasure, and has a very bad design smell. — Mark Peters
– Mark Peters, Commented Jun 4, 2010 at 14:32

dplass · Accepted Answer · 2010-06-03 21:27:14Z

7

You can get it, to some degree... not sure if this is useful:

import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;

abstract class MyClass<T> {
  public MyClass() {        
    Type genericSuperclass = this.getClass().getGenericSuperclass();
    if (genericSuperclass instanceof ParameterizedType) {
      ParameterizedType pt = (ParameterizedType) genericSuperclass;
      Type type = pt.getActualTypeArguments()[0];
      System.out.println(type); // prints class java.lang.String for FooClass
    }
  }
}

public class FooClass extends MyClass<String> { 
  public FooClass() {
    super();
  }
  public static void main(String[] args) {
    new FooClass();
  }
}

answered Jun 3, 2010 at 21:27

dplass

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1 Comment

Mark Peters Over a year ago

Note that if you did new ConcreteMyClass<String>(), where public class ConcreteMyClass<T> extends MyClass<T> you would just get "T", since the type isn't reified.

taer · Accepted Answer · 2010-06-04 17:29:00Z

4

We've done this

public abstract BaseClass<T>{
protected Class<? extends T> clazz;

    public BaseClass(Class<? extends T> theClass)
    {
        this.clazz = theClass;
    }
...
}

And in the subclasses,

public class SubClass extends BaseClass<Foo>{
    public SubClass(){
       super(Foo.class);
    }
}

answered Jun 4, 2010 at 17:29

taer

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Comments

user268396 · Accepted Answer · 2010-06-03 21:53:04Z

0

And you cannot simply add a constructor parameter?

public abstract class MyClass<T> { 
  public MyClass(Class<T> type) {
    // do something with type?
  }
}

answered Jun 3, 2010 at 21:53

user268396

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Comments

Powerlord · Accepted Answer · 2010-06-09 21:01:12Z

0

If I'm not reading this wrong, wouldn't you just want

public <T> void save(T entity)

and

public <T> T getEntityById(Long id)

for your method signatures?

answered Jun 9, 2010 at 21:01

Powerlord

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