How entity comparison works in JPQL WHERE clause?

Question

How is entities comparison (equality) statement evaluated in JPQL: is it by identity comparison, or by equals(), or sth else?

I've spent a couple of hours of googling and going through the specs of Hibernate and JPA but still can't find how it works. Consider the following entities:

class MyProductType{Integer id;}
class MyProduct{Integer id; MyProductType pType;}

And now the JPQL/HQL query:

SELECT t FROM MyProductType t, MyProduct p WHERE p.pType = t

(I know it's an ugly query, just focus on the where clause semantics.)

So how is p.pType = t evaluated?

JSR 317 mentions "entity_expression" comparison but it's behaviour is not clarified.

EDIT: What I dislike about Rika's suggestion below is that the .id approach includes implicit inner join which is usually not what you want in case the query employs an OUTER (LEFT) join.

Answer 1

I found this at http://www.objectdb.com/java/jpa/query/jpql/comparison, it is very interesting, good question.

Instances of user defined classes (entity classes and embeddable classes) can be compared by using the equality operators (=, <>, ==, !=). For entities, e1 = e2 if e1 and e2 have the same type and the same primary key value. For embeddable objects, e1 = e2 if e1 and e2 have exactly the same content.

So it seems it checks the primary key value of the objects and the type of the objects. So it seems with p.pType = t, it will check the (assuming the id is the primary key) the id of p.pType, with the id of t and see if they are equal. Then it will check to see if both entities are of the same type or MyProductType.