java Constructor can not convert to scala

Question

i know java Constructor convert into Scala code. but this is my project Generic class implementation. i am using hibernate ,spring in this project and i create genericDAO trait but can't create It's implementation or i can't convert this java constructor to scala

This are 2 variables

private Class<T> entityClass;

private String entityClassName;

This is GenericDAOImpl class code here GenericDAOImpl is the constructor

@SuppressWarnings("unchecked")
        public GenericDAOImpl() {
            Type genericSuperclass;
            Class<?> parametrizedClass = getClass();
            do {
                genericSuperclass = parametrizedClass.getGenericSuperclass();
                if (genericSuperclass instanceof Class) {
                    parametrizedClass = (Class<?>) genericSuperclass;
                }
            } while (genericSuperclass != null
                    && !(genericSuperclass instanceof ParameterizedType));
            this.entityClass = (Class<T>) ((ParameterizedType) genericSuperclass)
                    .getActualTypeArguments()[0];
            if (entityClass != null) {
                entityClassName = entityClass.getSimpleName();
            }
        }

Thanks From Milano

EDIT

I tried this

@SuppressWarnings("unchecked")
 def this(T,ID){
   var genericSuperclass:Type;
   var parametrizedClass:Class[?]=getClass
   do {
            genericSuperclass = parametrizedClass.getGenericSuperclass()
            if (genericSuperclass instanceof[Class]) {
                parametrizedClass = (Class<?>) genericSuperclass
            }
        } while (genericSuperclass != null
                && !(genericSuperclass instanceof [ParameterizedType]))
        this.entityClass = (Class[T]) ((ParameterizedType) genericSuperclass)
                .getActualTypeArguments()[0]
        if (entityClass != null) {
            entityClassName = entityClass.getSimpleName()
        }

 }


 And this got compilation Error

@milano: you should study scala a little before dealing with reflection or other complicated things. Your "tried" code is very far from valid scala code. — senia
– senia, Commented Jul 8, 2013 at 9:49
@milano: s\Class[?]\Class[_]\ , s\instanceof[Class]\isInstanceOf[Class]\ , s\(Class<?>) genericSuperclass\genericSuperclass.asInstanceOf[Class[_]]\ , s\[0]\(0)\ , remove def this(T,ID){ (class body is construvtor), there is no do{...} while in scala, you should initialize even vars. — senia
– senia, Commented Jul 8, 2013 at 10:01
@milano - try and work it out for yourself. Look things up, think about it. At the moment senia is just rewriting your code and you're not learning anything. — selig
– selig, Commented Jul 8, 2013 at 10:38

Mark Butler · Accepted Answer · 2013-09-16 08:04:43Z

I can see some problems with the Java to Scala conversion. Specifically I would replace Java's instanceof with

isInstanceOf[SomeClass]

Casts change from

(Class[T])

to

.asInstanceOf[Class[T]]

Add the class bound ? changes to _ (I think)

Array indexes change from square brackets to round brackets e.g.

.getActualTypeArguments[0]

to

.getActualTypeArguments(0)

Applying these changes to the original Java I get

class GenericDAOImpl[T] {
var genericSuperclass: Type = null
var parametrizedClass: Class[_] = getClass()
do {
    genericSuperclass = parametrizedClass.getGenericSuperclass()
    if (genericSuperclass.isInstanceOf[Class])
        parametrizedClass = genericSuperclass.asInstanceOf[Class[_]]
} while (genericSuperclass != null
                && !(genericSuperclass.isInstanceOf[ParameterizedType]))
val entityClass = genericSuperclass.asInstanceOf[ParameterizedType]
                .getActualTypeArguments()(0).asInstanceOf[Class[T]]
val entityClassName =
  if (entityClass != null) 
      entityClass.getSimpleName()
  else
      null

but ... it doesn't make sense to me, I would need to see more of your original class to get it to work.

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