In Java, Object is the base class, which is extended by all other classes. What is the equivalent base class in Scala?
Not so fast! ;)
The question seems somewhat ill-posed, because primitive types do not extend from Object in Java. Therefore, Java simply isn't like Scala. The base class of all reference types (what corresponds to classes that subclass Object in Java) is actually AnyRef, not Any.
Indeed, Scala Tour page about unified types states:
AnyRef represents reference types. All non-value types are defined as reference types. Every user-defined type in Scala is a subtype of AnyRef. If Scala is used in the context of a Java runtime environment, AnyRef corresponds to java.lang.Object.
The base class of all types (Any) and the base class of all reference types (AnyRef) are two separate concepts.
I got the answer. It is of type Any.
http://joelabrahamsson.com/learning-scala-part-eight-scalas-type-hierarchy-and-object-equality/
Related
What is the advantage of using an abstract class instead of a trait (apart from performance)? It seems like abstract classes can be replaced by traits in most cases.
I can think of two differences
Abstract classes can have constructor parameters as well as type parameters. Traits can have only type parameters. There was some discussion that in future even traits can have constructor parameters
Abstract classes are fully interoperable with Java. You can call them from Java code without any wrappers. Traits are fully interoperable only if they do not contain any implementation code
There's a section in Programming in Scala called "To trait, or not to trait?" which addresses this question. Since the 1st ed is available online, I'm hoping it's OK to quote the whole thing here. (Any serious Scala programmer should buy the book):
Whenever you implement a reusable collection of behavior, you will
have to decide whether you want to use a trait or an abstract class.
There is no firm rule, but this section contains a few guidelines to
consider.
If the behavior will not be reused, then make it a concrete class. It
is not reusable behavior after all.
If it might be reused in multiple, unrelated classes, make it a trait.
Only traits can be mixed into different parts of the class hierarchy.
If you want to inherit from it in Java code, use an abstract class.
Since traits with code do not have a close Java analog, it tends to be
awkward to inherit from a trait in a Java class. Inheriting from a
Scala class, meanwhile, is exactly like inheriting from a Java class.
As one exception, a Scala trait with only abstract members translates
directly to a Java interface, so you should feel free to define such
traits even if you expect Java code to inherit from it. See Chapter 29
for more information on working with Java and Scala together.
If you plan to distribute it in compiled form, and you expect outside
groups to write classes inheriting from it, you might lean towards
using an abstract class. The issue is that when a trait gains or loses
a member, any classes that inherit from it must be recompiled, even if
they have not changed. If outside clients will only call into the
behavior, instead of inheriting from it, then using a trait is fine.
If efficiency is very important, lean towards using a class. Most Java
runtimes make a virtual method invocation of a class member a faster
operation than an interface method invocation. Traits get compiled to
interfaces and therefore may pay a slight performance overhead.
However, you should make this choice only if you know that the trait
in question constitutes a performance bottleneck and have evidence
that using a class instead actually solves the problem.
If you still do not know, after considering the above, then start by
making it as a trait. You can always change it later, and in general
using a trait keeps more options open.
As #Mushtaq Ahmed mentioned, a trait cannot have any parameters passed to the primary constructor of a class.
Another difference is the treatment of super.
The other difference between classes and traits is that whereas in classes, super calls are statically bound, in traits, they are dynamically bound. If you write super.toString in a class, you know exactly which method implementation will be invoked. When you write the same thing in a trait, however, the method implementation to invoke for the super call is undefined when you define the trait.
See the rest of Chapter 12 for more details.
Edit 1 (2013):
There is a subtle difference in the way abstract classes behaves compared to traits. One of the linearization rules is that it preserves the inheritance hierarchy of the classes, which tends to push abstract classes later in the chain while traits can happily be mixed in. In certain circumstances, it's actually preferable to be in latter position of the class linearization, so abstract classes could be used for that. See constraining class linearization (mixin order) in Scala.
Edit 2 (2018):
As of Scala 2.12, trait's binary compatibility behavior has changed. Prior to 2.12, adding or removing a member to the trait required recompilation of all classes that inherit the trait, even if the classes have not changed. This is due to the way traits were encoded in JVM.
As of Scala 2.12, traits compile to Java interfaces, so the requirement has relaxed a bit. If the trait does any of the following, its subclasses still require recompilation:
defining fields (val or var, but a constant is ok – final val without result type)
calling super
initializer statements in the body
extending a class
relying on linearization to find implementations in the right supertrait
But if the trait does not, you can now update it without breaking binary compatibility.
For whatever it is worth, Odersky et al's Programming in Scala recommends that, when you doubt, you use traits. You can always change them into abstract classes later on if needed.
Other than the fact that you cannot directly extend multiple abstract classes, but you can mixin multiple traits into a class, it's worth mentioning that traits are stackable, since super calls in a trait are dynamically bound (it is referring a class or trait mixed before current one).
From Thomas's answer in Difference between Abstract Class and Trait:
trait A{
def a = 1
}
trait X extends A{
override def a = {
println("X")
super.a
}
}
trait Y extends A{
override def a = {
println("Y")
super.a
}
}
scala> val xy = new AnyRef with X with Y
xy: java.lang.Object with X with Y = $anon$1#6e9b6a
scala> xy.a
Y
X
res0: Int = 1
scala> val yx = new AnyRef with Y with X
yx: java.lang.Object with Y with X = $anon$1#188c838
scala> yx.a
X
Y
res1: Int = 1
When extending an abstract class, this shows that the subclass is of a similar kind. This is not neccessarily the case when using traits, I think.
In Programming Scala the authors say that abstract classes make a classical object oriented "is-a" relationship while traits are a scala-way of composition.
Abstract classes can contain behaviour - They can parameterized with constructor args (which traits can't) and represent a working entity. Traits instead just represent a single feature, an interface of one functionality.
A class can inherit from multiple traits but only one abstract class.
Abstract classes can have constructor parameters as well as type parameters. Traits can have only type parameters. For example, you can’t say trait t(i: Int) { }; the i parameter is illegal.
Abstract classes are fully interoperable with Java. You can call them from Java code without any wrappers. Traits are fully interoperable only if they do not contain any implementation code.
I have a Java method which takes in a java.lang.Object.
Now I'm trying to pass a Scala Int to it, but the compiler complains about type mismatch of Int and Object.
This feels quite strange as I suppose a Scala Int inherits from AnyRef and should be the same as java.lang.Object?
Scala Int is not extending AnyRef, though - it extends AnyVal, which is not cast-compatible with java.lang.Object, and is more like Java int. Scala also have Integer class, which is just an alias for java.lang.Integer - use that one if you need to pass it as an object.
There're two braches of Scala classes inherited from the Scala root class - Any:
AnyVal: All value based types, e.g. Int, Byte, Double... etc. This is the counterpart of Java raw value classes, i.e. int, float... etc.
AnyRef: All reference based types, e.g. String, List, including boxed Java types, such as java.lang.Integer... etc.
(See diagram in https://docs.scala-lang.org/tour/unified-types.html)
The AnyRef is equivalent to java.lang.Object. Therefore, when calling Java methods taking Object, a AnyVal cannot be directly transferred before boxing. This can be done through explicit type declaration:
JavaClass.methodWithObject(10: java.lang.Integer)
More discussion refer to:
Result type of an implicit conversion must be more specific than AnyRef
Is that possible to create a type class based on type parameter by inheriting type parameter like in the code below?
I do suspect that this is fundamentally not possible as this is a scala limitation as functional language...
But if possible what I have to change in the code below??
class MyInt[T : Numeric] extends T {
}
It's not possible. You can extend a class (traits included), but you cannot extend a type, and T is a type.
Even if it were possible, what would it then mean to have MyInt[Int] (which is final), or MyInt[List[String]] (which is not a class) ? It would get quite complicated for the compiler very quickly.
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.
Java does not only have types that extend java.lang.Object (aka scala.AnyRef), but primitive types, e.g. 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.
What is the advantage of using an abstract class instead of a trait (apart from performance)? It seems like abstract classes can be replaced by traits in most cases.
I can think of two differences
Abstract classes can have constructor parameters as well as type parameters. Traits can have only type parameters. There was some discussion that in future even traits can have constructor parameters
Abstract classes are fully interoperable with Java. You can call them from Java code without any wrappers. Traits are fully interoperable only if they do not contain any implementation code
There's a section in Programming in Scala called "To trait, or not to trait?" which addresses this question. Since the 1st ed is available online, I'm hoping it's OK to quote the whole thing here. (Any serious Scala programmer should buy the book):
Whenever you implement a reusable collection of behavior, you will
have to decide whether you want to use a trait or an abstract class.
There is no firm rule, but this section contains a few guidelines to
consider.
If the behavior will not be reused, then make it a concrete class. It
is not reusable behavior after all.
If it might be reused in multiple, unrelated classes, make it a trait.
Only traits can be mixed into different parts of the class hierarchy.
If you want to inherit from it in Java code, use an abstract class.
Since traits with code do not have a close Java analog, it tends to be
awkward to inherit from a trait in a Java class. Inheriting from a
Scala class, meanwhile, is exactly like inheriting from a Java class.
As one exception, a Scala trait with only abstract members translates
directly to a Java interface, so you should feel free to define such
traits even if you expect Java code to inherit from it. See Chapter 29
for more information on working with Java and Scala together.
If you plan to distribute it in compiled form, and you expect outside
groups to write classes inheriting from it, you might lean towards
using an abstract class. The issue is that when a trait gains or loses
a member, any classes that inherit from it must be recompiled, even if
they have not changed. If outside clients will only call into the
behavior, instead of inheriting from it, then using a trait is fine.
If efficiency is very important, lean towards using a class. Most Java
runtimes make a virtual method invocation of a class member a faster
operation than an interface method invocation. Traits get compiled to
interfaces and therefore may pay a slight performance overhead.
However, you should make this choice only if you know that the trait
in question constitutes a performance bottleneck and have evidence
that using a class instead actually solves the problem.
If you still do not know, after considering the above, then start by
making it as a trait. You can always change it later, and in general
using a trait keeps more options open.
As #Mushtaq Ahmed mentioned, a trait cannot have any parameters passed to the primary constructor of a class.
Another difference is the treatment of super.
The other difference between classes and traits is that whereas in classes, super calls are statically bound, in traits, they are dynamically bound. If you write super.toString in a class, you know exactly which method implementation will be invoked. When you write the same thing in a trait, however, the method implementation to invoke for the super call is undefined when you define the trait.
See the rest of Chapter 12 for more details.
Edit 1 (2013):
There is a subtle difference in the way abstract classes behaves compared to traits. One of the linearization rules is that it preserves the inheritance hierarchy of the classes, which tends to push abstract classes later in the chain while traits can happily be mixed in. In certain circumstances, it's actually preferable to be in latter position of the class linearization, so abstract classes could be used for that. See constraining class linearization (mixin order) in Scala.
Edit 2 (2018):
As of Scala 2.12, trait's binary compatibility behavior has changed. Prior to 2.12, adding or removing a member to the trait required recompilation of all classes that inherit the trait, even if the classes have not changed. This is due to the way traits were encoded in JVM.
As of Scala 2.12, traits compile to Java interfaces, so the requirement has relaxed a bit. If the trait does any of the following, its subclasses still require recompilation:
defining fields (val or var, but a constant is ok – final val without result type)
calling super
initializer statements in the body
extending a class
relying on linearization to find implementations in the right supertrait
But if the trait does not, you can now update it without breaking binary compatibility.
For whatever it is worth, Odersky et al's Programming in Scala recommends that, when you doubt, you use traits. You can always change them into abstract classes later on if needed.
Other than the fact that you cannot directly extend multiple abstract classes, but you can mixin multiple traits into a class, it's worth mentioning that traits are stackable, since super calls in a trait are dynamically bound (it is referring a class or trait mixed before current one).
From Thomas's answer in Difference between Abstract Class and Trait:
trait A{
def a = 1
}
trait X extends A{
override def a = {
println("X")
super.a
}
}
trait Y extends A{
override def a = {
println("Y")
super.a
}
}
scala> val xy = new AnyRef with X with Y
xy: java.lang.Object with X with Y = $anon$1#6e9b6a
scala> xy.a
Y
X
res0: Int = 1
scala> val yx = new AnyRef with Y with X
yx: java.lang.Object with Y with X = $anon$1#188c838
scala> yx.a
X
Y
res1: Int = 1
When extending an abstract class, this shows that the subclass is of a similar kind. This is not neccessarily the case when using traits, I think.
In Programming Scala the authors say that abstract classes make a classical object oriented "is-a" relationship while traits are a scala-way of composition.
Abstract classes can contain behaviour - They can parameterized with constructor args (which traits can't) and represent a working entity. Traits instead just represent a single feature, an interface of one functionality.
A class can inherit from multiple traits but only one abstract class.
Abstract classes can have constructor parameters as well as type parameters. Traits can have only type parameters. For example, you can’t say trait t(i: Int) { }; the i parameter is illegal.
Abstract classes are fully interoperable with Java. You can call them from Java code without any wrappers. Traits are fully interoperable only if they do not contain any implementation code.