Java program convert to scala the interoperability problems

Question

Here java list convertion errors are occured

Scala Code

 @SuppressWarnings("unchecked") 
  @Override
  def getAllStudents():List[Student] = {
    return getSession().createQuery("from Student where isDelete =  'false' ")
    .list()  **here error occured and that shows below **
     }

here i import this statement but no change

import scala.collection.JavaConverters._

Error type mismatch; found : java.util.List[?0] where type ?0 required: scala.collection.immutable.List[com.model.domain.entity.Student]

The Java Code

@SuppressWarnings("unchecked")
    @Override
    public List<Student> getAllStudents() {
        return getSession().createQuery(
                "from Student where isDelete =  'false' ").list();
    }

Where you have placed import scala.collection.JavaConverters._? — om-nom-nom
– om-nom-nom, Commented Jul 4, 2013 at 7:07

Alexey Romanov · Accepted Answer · 2013-07-04 07:50:40Z

6

Your own answer is incorrect. Instead, if you are implementing an interface (or extending a class) which needs to return a Java list, you should do this:

def getAllStudents() : java.util.List[Student] = {
  getSession().createQuery("from Student where isDelete =  'false' ")
    .list()
 }

If you don't need Java list here, then you should instead do

import scala.collection.JavaConverters._

def getAllStudents() : Seq[Student] = {
  getSession().createQuery("from Student where isDelete =  'false' ")
    .list().asScala
 }

answered Jul 4, 2013 at 7:50

Alexey Romanov

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1 Comment

Petr Over a year ago

Worth mentioning that if you use JavaConverters to convert let's say a java.util.List to a Seq and later back to a java.util.List, it remembers the original list and yields it at the later conversion. See Time complexity of JavaConverters asScala method.

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