I am implementing an extension of ml.Transformer in Spark; but this question is Scala specific. Here is an example object (part of a Class/Object pair):
abstract class UDFTransformer(func: UserDefinedFunction,
inputFieldNames: Seq[String],
outputFieldName: String) extends Transformer with MLWritable with Serializable {
... definitions here ...
}
object UDFTransformer extends MLReadable[UDFTransformer] {
// Since there are no parameters associted with the UDF, there is nothing to save!
class Writer(instance: UDFTransformer) extends MLWriter {
override protected def saveImpl(path: String): Unit = {}
}
abstract protected class Reader extends MLReader[UDFTransformer]
override def read: MLReader[UDFTransformer] = new Reader
override def load(path: String): UDFTransformer = super.load(path)
}
The new Reader does not compile because the class is abstract and cannot be instantiated. But; any child class will have to define it; along with its necessary members. I cannot just make read abstract as well, this gives me a warning Only classes can have declared but undefined methods.
The fundamental problem is that each child class of my UDFTransformer is going to wrap a specific UDF. Therefore, the reader needs to be able to generate a specific UDF object; this can't be declared in the superclass. But this 'factory' belongs in the companion object, not in the abstract class itself.
How can I go about building a companion object for an abstract class that can leave the definition of read undefined?
The normal way to do it is by creating an abstract class or trait for the companion objects. Something like
abstract class UDFTransformerCompanion[T <: UDFTransformer] extends MLReadable[T] {
abstract def read: MLReader[T]
override def load(path: String): T = super.load(path)
}
class SomeTransformer extends UDFTransformer { ... }
object SomeTransformer extends UDFTransformerCompanion[SomeTransformer] {
override def read: MLReader[SomeTransformer] = ...
}
Not sure why you have the load = super.load override, and it doesn't look like you can have a companion object for the UDFTransformer itself, at least not one extending this abstract class.
See GenericTraversableTemplate for a standard library example.
Related
Let's say we have an abstract class (same question for traits also):
abstract class TypeParser[C <: Changes : TypeTag] extends Serializable {
val enrichmentType: EnrichmentType
protected def parseChanges(row: Row): C
}
Where implementations look like the following:
object PlaceholderParser extends TypeParser[PlaceholderChanges] {
val enrichmentType: EnrichmentType = PlaceholderType
override protected def parseChanges(row: Row): PlaceholderChanges = ???
}
The above implementation is a singleton, however, it can't be forced to be a singleton for future implementations. So one can simply implement it as a class, for example:
class PlaceholderParser2() extends TypeParser[PlaceholderChanges2] {
val enrichmentType: EnrichmentType = PlaceholderType2
override protected def parseChanges(row: Row): PlaceholderChanges2 = ???
}
Is there any way of forcing implementations to be a singleton?
side question: is there any advantage of forcing it?
To our advantage all objects extend an interface called Singleton.
You can't extend it directly but we can use a Scala feature called self-types to enforce that all subtypes of TypeParser to also be Singletons (i.e. objects)
abstract class TypeParser[C <: Changes : TypeTag] extends Serializable {
self: Singleton =>
...
}
Assuming I have a simple abstract base class like so:
abstract class MyAbstractBaseClass {
def hello : Unit
}
and then I write a "stacking" trait like so:
trait MyTrait extends MyAbstractBaseClass {
abstract override def hello : Unit =
{
super.hello
println("How are you?");
}
}
then why won't Scala let me define a subclass as follows:
class MyClass extends MyAbstractBaseClass with MyTrait {
override def hello : Unit = println("Hello!")
}
error: overriding method hello in trait MyTrait of type => Unit;
method hello needs `abstract override' modifiers
If I try their suggestion of using 'abstract override':
class MyClass extends MyAbstractBaseClass with MyTrait {
abstract override def hello : Unit = println("Hello!")
}
error: `abstract override' modifier only allowed for members of traits
Can anyone help me understand this?
P.S. I know that the below does work:
class MyClass extends MyAbstractBaseClass {
override def hello : Unit = println("Hello!")
}
val x = new MyClass with MyTrait
x.hello
Hello!
How are you?
but am trying to understand why the former does not.
In summary: why can't I provide an implementation of the abstract base class - while also taking advantage of the trait's functionality?
The trick is that you can't have an "abstract" method in the flow of the linearization, that is called from a super call.
Try this, you will see it compiles:
abstract class MyAbstractBaseClass {
def hello : Unit
}
class SubClass extends MyAbstractBaseClass {
def hello {
println("toto")
}
}
trait MyTrait extends MyAbstractBaseClass {
abstract override def hello : Unit =
{
super.hello
println("How are you?")
}
}
class MyClass extends SubClass with MyTrait { //note the CONCRETE SubClass here
override def hello : Unit = println("Hello!")
}
new MyClass().hello
You got the error, because the compiler starts with MyTrait (at the top of the stack, so the first to be called) and that MyTrait calls through super an abstract method... (of MyAbstractBaseClass) => it crashes since your super call can't target immediately a concrete method.
In my code snippet, you will notice that MyTrait is "at the top" (during linearization) of a concrete Subclass class, that makes the trick.
I am new to scala. I don't understand scala traits properly. I have read it is similar to java interfaces but the methods need not be abstract. But how can I declare a scala trait and instantiate it in the following code. BTW, the following code is working fine.
trait fooable {
def foo: Unit = {
println("This is foo")
}
}
object Main {
def main(args: Array[String]): Unit = {
println("This is morking")
val foo = new fooable{}
foo.foo
}
}
Output -
This is morking
This is foo
Thanks in advance.
Scala traits are more general than both Java interfaces and abstract classes.
You can use a trait as an interface, you can use it to store some implementation, you can use it simply to define a common super-type:
trait Message
case class Text(text: String) extends Message
case class Data(data: ByteString) extends Message
Multiple traits can be 'mixed in' a class:
class MyClass extends TraitA with TraitB with TraitC
where any conflict with identically named methods is resolved by the simple rule: the last trait takes precedence. This code:
trait TraitA { def print() { println("A") } }
trait TraitB { def print() { println("B") } }
trait TraitC { def print() { println("C") } }
new MyClass.print()
will print "C".
Scala traits can't be instantiated. You are creating an anonymous class in you example. If you add an abstract method to your trait it will not compile.
Unrelated note:
It is a good practice to write braces "()" in methods with side effects. Your method foo has a side effect: it prints something. So you should write "foo()".
When you instantiate a trait you create an instance of an anonymous class that extends that trait it works the same way as creating anonymous classes of interfaces in java. If you had any unimplemented methods in the trait fooable the compiler would've forced you to implement them on spot when you created your anonymous class.
Indeed your code should work fine and output the same result you mentioned. However the important point to note here is TRAIT CAN NEVER EVER be instantiated. Its the same concept that Java interface can never ever be instantiated.
When scala compiler notice the code foo = new fooable{}, It internally creates an anonymous class which extends your fooable trait and thus it inherits foo() method due to inheritance. See following code snippet:
scala> val foo = new fooable{}
foo: fooable = $anon$1#277c0f21
Thus when you call foo.foo, At runtime it invokes a anonymous's class(i.e. $anon$1#277c0f21) inherited method foo().
The same understanding is true in case of Java as well, Following is perfectly legal java code:
Runnable r = new Runnable() {
public void run() {
System.out.println(" Cat");
}
};
r.run()
Happy learning !
An Scala Trait is abstract and can't be instantiated. In the code above its instantiated as
val foo = new fooable{}
and NOT as
val foo = new fooable()
The curly braces created some anonymous class which is not fooable but an empty one
What are Traits?
Traits are similar to interfaces in any other language. Scala allows traits to be instantiate during definition as well as during construction. For example if we have a trait and a abstract class
trait testTrait {
}
abstract class BaseClass {}
class Derive extends BaseClass with testTrait {
}
// ---------------------- OR----------------------
// Traits can be used during instantiation
class Derive extends BaseClass {
}
val classInst = new Derive with testTrait
Traits can also be chained using multiple with trait
I would like to implement a trait with a self-type like the following:
trait HasEquipment {
this: {def loadEquipment: List[Equipment]} =>
//more methods
}
Now i can mix this trait into classes and these classes have to define a "loadEquipment" Method:
case class Course(id: Long) extends HasEquipment {
def loadEquipment: List[Equipment] = {
//implementation
}
}
However I don't want to expose to loadEquipment Method to everyone but only to the trait. So what i would like to do is make loadEquipment private in the Course class so that only the trait has access to it. Making it private prevents the trait from accessing the method.
Is there another way to make loadEquipment only accessible from within the trait?
Not sure why you want a self-type here. An abstract, protected member works just as well:
trait HasEquipment {
protected def loadEquipment: List[Equipment]
}
case class Course(id: Long) extends HasEquipment {
override protected def loadEquipment: List[Equipment] = ???
}
As a side note, structural types in Scala make use of reflection, which is the reason why they're usually avoided.
Consider the following:
trait TestTrait {
def doStuff()
}
final class TestClass {
// ...
}
I would like to instantiate an instance of TestClass that implements the method in the trait. The following does not compile:
// Illegal inheritance from final class TestClass
val t = new TestClass with TestTrait {
def doStuff() {
println("doing stuff")
}
}
This makes sense, since the anonymous class created would extend the final class. What I'm really after is an anonymous implementation of the trait mixed in to an instance of the final class.
The following works, but seems a bit roundabout. Is there a way to do this directly without the declared trait implementation?
trait TestTraitImpl extends TestTrait {
def doStuff() {
println("doing stuff")
}
}
val t = new TestClass with TestTraitImpl
As it turns out, you can't do the latter either.
final class TestClass
trait TestTrait
val t = new TestClass with TestTrait
Apparently, any Foo with Bar is creation of a new anonymous type, and thus final classes cannot have traits as mixins. This adds a great deal of theoretical significance to making a class final, as it prevents not just the concept of inheritance, but also stackable modification.
Method resolution rules are different in the two cases. In the first case an anonymous class is constructed first by whatever happens to be the methods/attributes of the type:
final class TestClass with TestTrait
and then you are trying to override a method of that, which conflicts with the final qualifier of TestClass.
In the second case you explicitly specify that you are overriding TestTrait behavior, and then the overriden behavior is mixed into TestClass.
I think it's perfectly fine to use the second method, and that it conveys more clearly what the intention is.