Scala's hierarchy mapped to Java's

Question

As "Programming in Scala: A comprehensive step-by-step Guide" states, in Scala there are not basic types values, just objects: Integers are Int instances and doubles are Double instances. I assume that these classes map to Java's Integer , Double ... classes and, therefore, are mapped as Object subclasses.

In the book, the following type hierarchy (classes as types) is presented:

Few pages after this graph is presented, you can read:

What somehow troubles me is: If Scala´s Double maps to Java's Double which is an specification of java.lang.Object and AnyRef is an alias for java.lang.Object too, should't AnyVal be a subclass of AnyRef ?

EDIT

Few pages after that I read that primitive types are not mapped to Java's primitive types wrapper classes unless their "boxed" versions are required; but I am still confused since it seems to me that not all Scala's objects are java.lang.Object sublcasses instances. That is: There are classes in Scala which could be not translated in the JVM as Object subclasses.

Answer 1

Java does not only have types that extend java.lang.Object (aka scala.AnyRef ), but primitive types, eg int , double , boolean , ... In Scala you find them under scala.Any . So a scala.Int corresponds to a Java int . Not java.lang.Integer ; not until boxing occurs, a mechanism on the JVM to be able to pass primitives to generic methods. Both Java and Scala do auto-boxing, that is construct a reference around a primitive type when a reference is needed.

The difference in Scala is, it doesn't treat scala.Int any different from say String , it doesn't matter whether the type corresponds to a JVM primitive or not. You can call methods on scala.Int as if it was any regular object. In the byte-code you will still have primitive types.

This is why Scala is sometimes called a true or more pure object-oriented language than Java.