In my code I have something like this:
trait MyObj
trait Companion {
type C <: MyObj
}
Then I have several pairs of classes and companion objects, where the companions extend Companion and the classes extend MyObj and define a constructor with a String parameter. I can't just make case classes inheriting from MyObj, cause then I can't make the companions extend Companion, but I end up repeating the same code in every companion:
class Foo(name: String) extends MyObj
object Foo extends Companion {
type C = Foo
def apply(name: String) = new Foo(name)
// new C(name) works too
}
I'd like to move the implementation of apply to the Companion trait:
trait Companion {
type C <: MyObj { def this(name: String) }
def apply(name: String) = new C(name)
}
and now the inheriting objects just need to specify the type of C. But this doesn't compile. Is there any way of telling the compiler that C must have a specific constructor, so that I can call new C, without having to manually resort to reflection?
What about adding ClassTag or TypeTag? It can be some sort of intermediate solution towards reflection:
abstract class Companion[C <: MyObj](implicit ct:ClassTag[C]) {
def apply(name: String):C =
ct.getConstructor(classOf[String]).invoke(Array[Any](name)).asInstanceOf[C]
}
(not sure in the exact syntax)
Related
I'm writing a type-safe code and want to replace apply() generated for case classes with my own implementation. Here it is:
import shapeless._
sealed trait Data
case object Remote extends Data
case object Local extends Data
case class SomeClass(){
type T <: Data
}
object SomeClass {
type Aux[TT] = SomeClass { type T = TT }
def apply[TT <: Data](implicit ev: TT =:!= Data): SomeClass.Aux[TT] = new SomeClass() {type T = TT}
}
val t: SomeClass = SomeClass() // <------------------ still compiles, bad
val tt: SomeClass.Aux[Remote.type] = SomeClass.apply[Remote.type] //compiles, good
val ttt: SomeClass.Aux[Data] = SomeClass.apply[Data] //does not compile, good
I want to prohibit val t: SomeClass = SomeClass() from compiling. Is it possible to do somehow except do not SomeClass to be case class?
There is a solution that is usually used if you want to provide some smart constructor and the default one would break your invariants. To make sure that only you can create the instance you should:
prevent using apply
prevent using new
prevent using .copy
prevent extending class where a child could call the constructor
This is achieved by this interesing patten:
sealed abstract case class MyCaseClass private (value: String)
object MyCaseClass {
def apply(value: String) = {
// checking invariants and stuff
new MyCaseClass(value) {}
}
}
Here:
abstract prevents generation of .copy and apply
sealed prevents extending this class (final wouldn't allow abstract)
private constructor prevents using new
While it doesn't look pretty it's pretty much bullet proof.
As #LuisMiguelMejíaSuárez pointed out this is not necessary in your exact case, but in general that could be used to deal with edge cases of case class with a smart constructor.
UPDATE:
In Scala 3 you only need to do
case class MyCaseClass private (value: String)
and it will prevent usage of: apply, new and copy from outside of this class and its companion.
This behavior was ported to Scala 2.13 with option -Xsource:3 enabled. You have to use at least 2.13.2 as in 2.13.1 this flag doesn't fix the issue.
So you can make the constructor private and ensure that T is also something different to Nothing.
I believe the best way to ensure the constructor is private (as well as many other things as #MateuszKubuszok show) is to use a (sealed) trait instead of a class:
(if you can not use a trait for whatever reasons, please refer to Mateusz's answer)
import shapeless._
sealed trait Data
final case object Remote extends Data
final case object Local extends Data
sealed trait SomeClass {
type T <: Data
}
object SomeClass {
type Aux[TT] = SomeClass { type T = TT }
def apply[TT <: Data](implicit ev1: TT =:!= Data, ev2: TT =:!= Nothing): Aux[TT] =
new SomeClass { override final type T = TT }
}
Which works like this:
SomeClass() // Does not compile.
SomeClass.apply[Remote.type] // Compiles.
SomeClass.apply[Data] // Does not compile.
You can see it running here.
If you want to prohibit using some of auto-generated methods of a case class you can define the methods (with proper signature) manually (then they will not be generated) and make them private (or private[this]).
Try
object SomeClass {
type Aux[TT] = SomeClass { type T = TT }
def apply[TT <: Data](implicit ev: TT =:!= Data): SomeClass.Aux[TT] = new SomeClass() {type T = TT}
private def apply(): SomeClass = ??? // added
}
val t: SomeClass = SomeClass() // doesn't compile
val tt: SomeClass.Aux[Remote.type] = SomeClass.apply[Remote.type] //compiles
val ttt: SomeClass.Aux[Data] = SomeClass.apply[Data] //doesn't compile
In principle, the methods (apply, unapply, copy, hashCode, toString) can be generated not by compiler itself but with macro annotations. Then you can choose any subset of them and modify their generation as you want.
Generate apply methods creating a class
how to efficiently/cleanly override a copy method
Also the methods can be generated using Shapeless case classes a la carte. Then you can switch on/off the methods as desired too.
https://github.com/milessabin/shapeless/blob/master/examples/src/main/scala/shapeless/examples/alacarte.scala
https://github.com/milessabin/shapeless/blob/master/core/src/test/scala/shapeless/alacarte.scala
I want to restrict creation of a Scala class to use the companion object constructor:
case class Foo private (bar: String, baz: Int)
Now I can't use the new keyword; I can only create a foo like Foo("catch", 22). But let's say I want to hook into the apply method on object creation to provide some additional functionality whenever an instance is created. Can I do that in the trait?
object Foo extends SpecialApply[Foo]
trait SpecialApply[C] {
override def apply(args: ???) = ???
}
The central problem is that it seems like I don't know the constructor argument types (String and Int) in the trait. I could pass them in as a tuple, say, like this:
object Foo extends SpecialApply[(String, Int), Foo]
trait SpecialApply[A, C] {
override def apply(args: ???) = <companion object>
}
Or I could get them by reflection. Either way, I'm not sure how to define the apply method over the tuple.
Is there some combination of reflection, .tupled functions, macro, and/or shapeless magic that will make this work?
I've always used companion objects just as they should be: having some class or trait, I've defined the object with the same class name in the same file.
But I'm trying at the moment to factor out some common functionality that few companion objects share and wonder if it is safe to have something like this:
trait Label {
def label: String
}
trait InstancesMap[T <: Label] {
private var instances = Map.empty[String, T]
def init(instance: T): T = {
instances += (instance.label -> instance)
instance
}
def byLabel(label: String): T = instances(label)
}
case class EventStatus(label: String) extends Label
object EventStatus extends InstancesMap[EventStatus] {
val DRAFT = init(EventStatus("draft"))
val PUBLISHED = init(EventStatus("published"))
}
I am not sure if it is safe for case class companion object to extend some other trait. It compiles and works fine, but would be great to hear some opinions.
Of course it can, just like non-companion objects. That's actually one of the major advantages of representing "statics" as companion objects. Extending a class instead of a trait works too.
You can see it in the standard library where collection companion objects extend GenericCompanion and its various subtypes.
Given a concrete class Animal, how do I define a function that only takes a subclass of Animal?
In typical examples like this Animal is a trait so defining [A <: Animal] implies that you already pass in a subclass of Animal. However, in a scenario like below where Animal is concrete, can I exclude that as being an allowed type?
I'm working with existing generated code, and this is just a generalized example of the problem. Therefore the implication is that I can't make Animal (or the equivalent) into a trait.
See below for an example:
class Animal {
def name: String = "General Animal"
}
class Dog extends Animal {
override def name: String = "Dog"
}
// How do I limit A to be a subtype of Animal (excluding Animal itself)?
class SpecificAnimalContainer[A <: Animal](a: A) {
def specificAnimal: A = a
}
val dogContainer = new SpecificAnimalContainer[Dog](new Dog)
// I do not want this to be able to compile.
val animalContainer = new SpecificAnimalContainer[Animal](new Animal)
Using shapeless you can write:
import shapeless._
class SpecificAnimalContainer[A <: Animal](a: A)(implicit ev: A =:!= Animal) {
def specificAnimal: A = a
}
// val animalContainer = new SpecificAnimalContainer[Animal](new Animal)// doesn't compile
Otherwise you can implement similar type for implicit yourself.
Type constraint for type inequality in scala
Enforce type difference
How can I have a negation type in Scala?
It's a bit unclear what you're trying to achieve, but your problem looks exactly like a book example from Scala documentation at
https://docs.scala-lang.org/tour/upper-type-bounds.html
abstract class Pet extends Animal {}
class PetContainer[P <: Pet](p: P) {
def pet: P = p
}
class Lion extends Animal {
override def name: String = "Lion"
}
// val lionContainer = new PetContainer[Lion](new Lion)
// ^this would not compile
Hope this helps
In Scala I want to return a instance of a class for a method defined in a trait which uses generics, the code example I have is this:
File 1
package packOne
import packTwo.A
trait MyTrait[T <: MyTrait[T <: A]] {
def otherFunct(): String
def funct[T <: A](): T
}
File 2
package packTwo
import packOne.MyTrait
abstract class A(someParameter: String) {}
class B(someParameter: String) extends A(someParameter) {}
object B extends MyTrait[B] { // <--- the B inside MyTrait here is the class not the object, or at least that is what I want
def otherFunct(): String = "Hello"
def funct[B](): C = new B("hi") // <--- I think here is the key
}
basically what I want is an interface that have method to return a concrete implementation of class A, in an implementing object (which happen to be a companion object for a class extending A).
Why do I want that to be on an object?, is because I want to call that method without the need of an instance (like an static method in java), so that I can call B.funct() and have an instance of B class kind of like a factory method, for other classes extending A for example a call to X.funct will return an instance of class X.
I have tried to remove the generic type from the function definition except on the return type of the function and just leave it in the trait definition (like def funct(): T) but that does not work either.
I am quite new to Scala so if you could explain it for dummies and avoid complex scala unique concepts I would appreciate
How about simply:
trait A
class B(someParameter: String) extends A
trait MyTrait[T <: A] {
def otherFunct: String //Parentheses on parameterless methods with no side effects and no serious computation are generally unidiomatic in Scala
def funct: T //Note, no generic parameter on this method
}
object B extends MyTrait[B] {
def otherFunct = "Hello"
def funct = new B("hi")
}
And then:
B.funct //returns a new `B`
The apply method is often used in this factory style (e.g. Seq.apply() which is equivalent to Seq())