Scala type parameters recurrence - scala

In Scala you can define parameters that descend from other one that take the first one as parameter. For example in Lift you can find such things in Record and Mapper
MongoMetaRecord[BaseRecord <: MongoRecord[BaseRecord]]
What does it mean and how is that useful ?

This is a pattern often used to let an abstract class know about the type of the actual concrete class that extends it. It is sometimes useful to know what that final concrete type is — for instance, to use it as the return type of a method that produces a copy of the current object.
Suppose you want to do this — let an abstract class Abstract know about the concrete type implementing it. You could start by defining a type parameter, maybe like this:
trait Abstract[A] {
def copyObject: A = ...
}
But then you realize that actually, A should be a subclass of Abstract itself, as you don't want subclasses to provide a random parametrization. So you could add this:
trait Abstract[A <: Abstract]
... but you'll soon realize that Abstract has turned into a generic type as well, so you'll rather need this:
trait Abstract[A <: Abstract[A]]
As a final step, you'll probably want to make A covariant if you can, so as to allow intermediate abstract classes along the inheritance path from Abstract to the final concrete class:
trait Abstract[+A <: Abstract[A]]
class Concrete1 extends Abstract[Concrete1]
trait RefinedAbstract[+A <: RefinedAbstract[A]] extends Abstract[A]
class Concrete2 extends RefinedAbstract[Concrete2]
This means that every non-leaf (abstract) type should be parametrized, and that only the final concrete class will be able to drop the type parameter.

Related

Scala abstract class type with a Lower Type Bounds. I could not understand

The code loos like this.
I can't figure out why the planType in this brackets .And I also write some tests to conquer it, but no value.
abstract class QueryPlan[PlanType <: QueryPlan[PlanType]] extends TreeNode[PlanType] {}
The writer of this abstract class wants the child classes to be only able to use PlanType which are of the type QueryPlan[PlanType] or its subtypes.
This is also termed Type-Narrowing in scala documentation. The QueryPlan[PlanType] is the upper-type bound in this example.
So, the class QueryPlan[PlanType <: QueryPlan[PlanType]] cannot have all the PlanTypes that are inherited from its parent TreeNode[PlanType]
In the example below, the type on left is narrowest, while Any on the right will be the widest type
QueryPlan[PlanType] <: TreeNode[PlanType] <: ParentPlanType <: Any

Scala trait `this.type` in type parameter

Have a look at these two simple traits:
trait TreeNode1[S] {
def subNodes: List[S]
}
trait TreeNode2 {
def subNodes: List[this.type]
}
(Not best naming, renamed them just to be brief.)
TreeNode1 defines a tree node with its children access, pointing their type S.
TreeNode2 defines the same, but its children have the same type as the class the current trait is mixed in (another words, tree node with uniform subnodes).
In theory TreeNode2 is a particular case of TreeNode1:
trait TreeNode2 extends TreeNode1[this.type] {...}
But Scala will not compile TreeNode2 with such an extension, because this.type can not be used in such way, although there are no any inconsistencies with its working in runtime.
How can I get around this situation? Or Scala do not offer such poorly used mechanism?
The reason I need this construction is the following:
I have another trait that requires TreeNode1 to be mixed in. I also have some class that mixes TreeNode1 with quite another children type. But I also have several classes which have the same type as they are:
class SomeTreeNode extends TreeNode1[SomeTreeNode]
So it will look prettier if I use TreeNode2 for it:
class SomeTreeNode extends TreeNode2
Implementing the same logic. But for using TreeNode2 should be a case of TreeNode1, which it actually is, but Scala doesn't agree with me.
P.S. At least it is wondering as theoretical question about Scala, not for a wide practical use.
its children have the same type as the class the current trait is mixed
No. This is a common misunderstanding. this.type is the singleton type of this; i.e. the type whose only two values are this and null. All children of a TreeNode2 instance must be the same instance.
To answer the other part of your question, one option is to make S a type member instead of a type parameter:
trait TreeNode1 {
type S
def subNodes: List[S]
}
object TreeNode1 {
// use TreeNode1.Aux[S] where you had TreeNode1[S] originally
type Aux[T] = TreeNode1 { type S = T }
}
trait TreeNode2 {
type S = this.type // not really what you want
def subNodes: List[this.type]
}
(so-called Aux pattern), but whether this works for you depends on how they are used.

Scala: force type parameter to be a case class

I have an abstract class Model from which I create case classes:
abstract class Model
case class User(.) extends Model
an abstract class Table taking such a Model as type parameter, used in one of its default concrete methods:
abstract class Table[M <: Model] {
def parser = SomeExternalBuilder[M]
}
The meaning is rather simple: "Give every instance of Table a default parser based on its own class".
The problem is that SomeExternalBuilder will only accept a case class as argument ("case class expected: M"), so it does not compile.
Can I make Table take only case classes as type parameter?
I have seen a few answers providing a missing copy method (ref1, ref2), so I tried this:
trait Model[T] {
def copy: T
}
abstract class Table[M <: Model[M]]
but now case class User extends Model[User] and must overwrite copy too, every function creating a Model takes a type parameter, and honestly the code quickly starts being atrocious, all that for that single line in Table.
Is there no better way than copying that def parser line in every child's body?
Edit: N.B. The real function is def parser: anorm.Macro.namedParser[M] from the "anorm" library for Play.
Edit: Source of the type check by this macro: https://github.com/playframework/anorm/blob/0a1b19055ba3e3749044ad8a54a6b2326235f7c8/core/src/main/scala/anorm/Macro.scala#L117
The problem is that SomeExternalBuilder will only accept a case class as argument ("case class expected: M"), so it does not compile.
I don't think you can ever get such a message from Scala compiler itself, which means that SomeExternalBuilder.apply is a macro. It requires a specific case class in order to know its fields, so that it doesn't matter if you could limit M to be a case class (which you can't): it still wouldn't accept a type parameter.
What you can do is create a macro annotation, so that when you write e.g.
#HasModel
class SomeTable extends Table[SomeModel] {
...
}
the val parser = namedParser[SomeModel] is generated automatically.
Alternately, write #HasModel[SomeModel] class SomeTable { ... } and generate extends Table[SomeModel] as well.
It wouldn't be hard (as macros go), but you still need to annotate each class extending Table.
Not fool proof solution but worth a try
case classes extend Product and Serialisable. Constraint Product with Serialisable will help you get some type safety. M can be any class which extends Product with Serialisable. But Product is extended by case class mostly
abstract class Table[M <: (Product with Serializable)] {
def parser = SomeExternalBuilder[M]
}

What is the difference between mixins and inheritance?

I am trying to understand the Mixins in the context of scala. In particular I wanted to know difference between concepts of inheritance and Mixins.
The definition of Mixin in wiki says :
A mixin class acts as the parent class, containing the desired functionality. A subclass can then inherit or simply reuse this functionality, but not as a means of specialization. Typically, the mixin will export the desired functionality to a child class, without creating a rigid, single "is a" relationship. Here lies the important difference between the concepts of mixins and inheritance, in that the child class can still inherit all the features of the parent class, but, the semantics about the child "being a kind of" the parent need not be necessarily applied.
In the above definition, I am not able to understand the statements marked in bold. what does it mean that
A subclass can inherit functionality in mixin but not as a means of specialization
In mixins, the child inherits all features of parent class but semantics about the child "being a kind" the parent need not be necessarily applied. - How can a child extend a parent and not necessarily a kind of Parent ? Is there an example like that.
I'm not sure I understood your question properly, but if I did, you're asking how something can inherit without really meaning the same thing as inheriting.
Mixins, however, aren't inheritance – it's actually more similar to dynamically adding a set of methods into an object. Whereas inheritance says "This thing is a kind of another thing", mixins say, "This object has some traits of this other thing." You can see this in the keyword used to declare mixins: trait.
To blatantly steal an example from the Scala homepage:
abstract class Spacecraft {
def engage(): Unit
}
trait CommandoBridge extends Spacecraft {
def engage(): Unit = {
for (_ <- 1 to 3)
speedUp()
}
def speedUp(): Unit
}
trait PulseEngine extends Spacecraft {
val maxPulse: Int
var currentPulse: Int = 0
def speedUp(): Unit = {
if (currentPulse < maxPulse)
currentPulse += 1
}
}
class StarCruiser extends Spacecraft
with CommandoBridge
with PulseEngine {
val maxPulse = 200
}
In this case, the StarCruiser isn't a CommandoBridge or PulseEngine; it has them, though, and gains the methods defined in those traits. It is a Spacecraft, as you can see because it inherits from that class.
It's worth mentioning that when a trait extends a class, if you want to make something with that trait, it has to extend that class. For example, if I had a class Dog, I couldn't have a Dog with PulseEngine unless Dog extended Spacecraft. In that way, it's not quite like adding methods; however, it's still similar.
A trait (which is called mixin when mixed with a class) is like an interface in Java (though there are many differences) where you can add additional features to a class without necessarily having "is a" relationship. Or you can say that generally traits bundle up features which can be used by multiple independent classes.
To give you an example from Scala library, Ordered[A] is a trait which provides implementation for some basic comparison operations (like <, <=, >, >=) to classes that can have data with natural ordering.
For example, let's say you have your own class Number and subclasses EvenNumber and OddNumber as shown below.
class Number(val num : Int) extends Ordered[Number] {
override def compare(that : Number) = this.num - that.num
}
trait Half extends Number {
def half() = num / 2
}
trait Increment extends Number {
def increment() = num + 1
}
class EvenNumber(val evenNum : Int) extends Number(evenNum) with Half
class OddNumber(val oddNum : Int) extends Number(oddNum) with Increment
In the example above, classes EvenNumber and OddNumber share is a relationship with Number but EvenNumber does not have "is a" relation with Half neither OddNumber share "is a" relation with Increment.
Another important point is even though class Number uses extends Ordered syntax, it means that Number has an implicit is a relationship with superclass of Ordered ie Any.
I think its very usage dependent. Scala being a multi-paradigm language makes it powerful as well as a bit confusing at times.
I think Mixins are very powerful when used the right way.
Mixins should be used to introduce behavior and reduce bolierplate.
A trait in Scala can have implementations and it is tempting to extend them and use them.
Traits could be used for inheritance. It can also be called mixins however that in my opinion is not the best way to use mixin behavior. In this case you could think of traits as Java Abstract Classes. Wherein you get subclasses that are "type of" the super class (the trait).
However Traits could be used as proper mixins as well. Now using a trait as a mixin depends on the implementation that is "how you mix it in". Mostly its a simple question to ask yourself . It is "Is the subclass of the trait truly a kind of the trait or are the behaviors in the trait behaviors that reduce boilerplate".
Typically it is best implemented by mixing in traits to objects rather than extending the trait to create new classes.
For example consider the following example:
//All future versions of DAO will extend this
trait AbstractDAO{
def getRecords:String
def updateRecords(records:String):Unit
}
//One concrete version
trait concreteDAO extends AbstractDAO{
override def getRecords={"Here are records"}
override def updateRecords(records:String){
println("Updated "+records)
}
}
//One concrete version
trait concreteDAO1 extends AbstractDAO{
override def getRecords={"Records returned from DAO2"}
override def updateRecords(records:String){
println("Updated via DAO2"+records)
}
}
//This trait just defines dependencies (in this case an instance of AbstractDAO) and defines operations based over that
trait service{
this:AbstractDAO =>
def updateRecordsViaDAO(record:String)={
updateRecords(record)
}
def getRecordsViaDAO={
getRecords
}
}
object DI extends App{
val wiredObject = new service with concreteDAO //injecting concrete DAO to the service and calling methods
wiredObject.updateRecords("RECORD1")
println(wiredObject.getRecords)
val wiredObject1 = new service with concreteDAO1
wiredObject1.updateRecords("RECORD2")
println(wiredObject1.getRecords)
}
concreteDAO is a trait which extends AbstractDAO - This is inheritance
val wiredObject = new service with concreteDAO -
This is proper mixin behavior
Since the service trait mandates the mixin of a AbstractDAO. It would be just wrong for Service to extend ConcreteDAO anyways because the service required AbstractDAO it is not a type of AbstractDAO.
Instead you create instances of service with different mixins.
The difference between mixin and inheritance is at semantic level. At syntax level they all are the same.
To mix in a trait, or to inherit from a trait, they all use extends or with which is the same syntax.
At semantic level, a trait that is intended to be mixed in usually doesn't have a is a relationship with the class mixining it which differs to a trait that is intended to be inherited.
To me, whether a trait is a mixin or parent is very subjective, which often time is a source of confusion.
I think it is talking about the actual class hierarchy. For example, a Dog is a type of Animal if it extends from the class (inheritance). It can be used wherever an Animal parameter is applicable.

Where would "abstract override" in subtrait with no implementation be useful, if any?

Given the following example of two traits with one extending another with no implementation of def a in each:
scala> trait A { def a: String }
defined trait A
scala> trait B extends A { abstract override def a: String }
defined trait B
Is the construct useful at all? What are the use cases?
I think the answer is essentially the same as the one linked in your comment. In Scala, the abstract keyword for methods isn't required, since the compiler can figure out whether it's abstract or not based on whether or not it has an implementation. So it's usage here is superfluous.
The override keyword is also not required for methods that are implementing an abstract method (or I guess not doing anything at all, in this case). So really, B is equivalent to:
trait B extends A { def a: String }
Or really just (since B will be assumed to be abstract):
trait B extends A
Similarly to the linked answer, I can imagine once scenario where using override might be useful for readability. If I were making the return type of a in B more specific than A, I could use override as a hint that I'm modifying the behavior in some way:
trait A {
def a: Any
}
trait B extends A {
override def a: String
}
In this case, I'm hinting that a in B might be slightly different than the inherited signature from A. Of course, this is only useful if it's known to the reader and used in a consistent manner. But I could still do the same thing without the override.
Short answer: abstract override is not useful in this case. It's basically like giving a type annotation where none would be needed.
The value added use of abstract override is for decorating an implementation that will be mixed in later, sometimes known as the "stackable trait pattern". See Why is "abstract override" required not "override" alone in subtrait?.
Abstract override indicates that you wish to override an 'abstract' method. Others address why it's useless here, so I'll focus on an example. Abstract override is best used for mixins. A simple example would be a Pollable trait:
trait Pollable{def poll:Double}
Lets say we want to weight this pollable. This trait will be a mixin for our trait. Our weighted pollable will have a weight field, which it will multiply a poll by to get a result. For example:
class OnePollable extends Pollable{
def poll:Double=1
}
val myWeightedOne=new OnePollable with WeightedPollable;
Lets try and write this trait:
//Does not compile
trait WeightedPollable extends Pollable{
var weight=1
def poll:Double=super.poll*weight
}
If you look, you'll see clearly why this doesn't work. Our trait tries to call a super type method that isn't implemented! One solution is to add a default to the super trait, Pollable:
//Don't do this!
trait Pollable{def poll:Double=1}
This sorta works here, but is sorta dumb in a lot of real world applications. The better way is this:
trait WeightedPollable extends Pollable{
var weight=1
abstract override def poll:Double=super.poll*weight
}
It's our friend the abstract override modifier! This tells the compiler that we are overriding an abstract method, but we want to use super to refer to an object we are being mixed into. This also disallows the trait being used as an interface.