I have 3rd party class A:
class A {
def methodA = ...
}
I want to use use trait to add a new method methodT to an instance of A
trait Atrait[...] {
def methodT = {
// how to get a reference of instance of type A?
}
}
This methodT is specific to some situation, so I should use constraint in the trait. But I could not figure it out. Also, how can I invoke instance of A's method in a trait?
UPDATE
Trait doesn't work this way. See answer for alternative solution.
This is the standard pattern for adding a method to a 3rd party class:
class A
implicit class ExtendedA(val a: A) extends AnyVal {
def methodT: Unit = { println("called A.methodT") }
}
Then you can do:
val a = new A
a.methodT
Related
I want to extend a trait from Scala object and override those methods which are in trait. So my doubt is those methods will become static to that Object
or instance methods, and is this good approach to extend from trait to Scala Object. Please help on this
trait A{
def show:Unit
}
object B extends A{
override def show(): Unit = {
println("inside Object")
}
}
There are no static methods in Scala. object can indeed extend a trait. Overriden methods, like show, do not become static methods, instead they belong to a single instance of B.type. This is the singleton pattern provided by Scala's object definition facility.
Try the following in Scala REPL:
object B
B
It should output something like
res0: B.type = B$#5688722f
Note how the value B has type B.type, so B is just a value/instance, nothing to do with statics.
Hm, I think a common example/usecase of what you've just described is extending the App trait and overriding the main definition.
object test extends App
{
override def main (args: Array[String]): Unit = {
println("Hello, let's get started")
}
}
In general though, why don't you define the class itself to extend the trait?
If you are going to instantiate new instances of B using B() (instead of new B()) it makes sense to do this.
trait A{
def show:Unit
}
object B { // companion aka singleton object
def apply(){
...
}
}
class B extends A{
override def show(): Unit = {
println("inside Object")
}
}
When you have a parent:
abstract class Parent {
def something(arg: ???): Parent = ???
}
and
class Child extends Parent {}
I would like
val updatedChild = new Child().something(...)
updatedChild to be of type Child and not of type Parent, is it possible ?
One way to do it, is to parametrize the parent:
abstract class Parent[T <: Parent[T]] {
def something(arg: Foo): T
}
class Child(val foo: String) extends Parent[Child] {
def something(arg: String) = return new Child(arg)
}
Sometimes, you can also get away with using this.type:
class Parent {
def something(arg: Foo): this.type = this
}
class Child {
override def something(arg: Foo) = this
}
But the latter method only works if all you ever want to return is this (this.type is not Parent or Child, but a specific type that only has one instance - this).
Here is a proposal that actually compiles:
abstract class Parent[Repr <: Parent[Repr]] {
def something(arg: Int): Repr
}
This is something you can do, at least it's not explicitly discouraged. Standard collection library uses it a lot, see e.g. IterableLike as a typical example of such F-bounded polymorphism.
It seems that you can do :
class Parent[THIS <: Parent[THIS]] {
def something: THIS
}
And that seems to work.
I am not sure if this is something you should do tho.
Both Andrey's and Dima's answers cover one way to solve the problem using only oo-patterns.
However I would like to point out another approach called typeclasses (which is more common in functional languages), that would be helpful if you are planning to write generic functions using your interface.
First, instead of having a parent class, you have an interface that describes the operations that can be performed on instances of the typeclass.
trait Typeclass[T] {
def something(t: T)(arg: Foo): T
}
Then, you would define your types, this time they don't extend any parent class, thus they don't have to override nothing.
class Child {
...
}
Now, you have to prove that your type is an instance of the type class.
(A common place to do that is in the companion object of the class).
object Child {
implicit final val ChildTypeclass: Typeclass[Child] = new Typeclass[Child] {
override def something(child: Child)(arg: Foo): Child = ???
}
}
Finally, you define a generic method that can operate on any type T as long as there is an instance of your typeclass for that type.
def generic[T](t: T, arg: Foo)(implicit tt: Typeclass[T]): T =
tt.something(t)(arg)
Bonus, if you want to recover the "dot notation" you can add an Ops pattern to your Typeclass.
object syntax {
object typeclass {
implicit final class TypeclassOps[T](val t: T) extends AnyVal {
final def something(arg: Foo)(implicit tt: Typelcass[T]) =
tt.something(t)(arg)
}
}
}
import syntax.typeclasss._
def generic[T: Typelcass](t: T, arg: Foo): T
t.something(arg)
val newChild = generic(new Child, new Foo)
// newChild: Child = ???
Also, a common approach is to define the something method in your class and the typeclass instance forwards the call to the one defined in the class, this way you can use your method in any instance of Child without having to put all the typeclass machinery.
I must say that this is useful for very high-level abstractions to which you plan to provide instances for many types (even types outside your control like any of the standard collection types) and write very generic functions that can operate on any of these.
If not, F-bounded types seems like the more rational solution.
Given a superclass or trait, and assuming an open hierarchy, how can I enforce that all extending classes implement a particular type class?
For instance, assuming the type class Default
trait Default[T] { def default: T }
and some trait Super:
trait Super { }
I would like to enforce that the following (by itself) is not allowed:
class A(val i: Int) extends Super
...while the following is:
class B(val i: Int) extends Super
implicit val bHasDef = new Default[B] { def default = B(42) }
Assuming the above is possible, can I then access the type class evidence for the subtypes from a method within Super? I.e, something like:
trait Super {
def magic: Default[this.type] = ???
}
I hardly think you can enforce that, at least in a simple enough way, maybe it's possible with something more complex like shapeless.
What I would do is add some modifications to the super trait and make it take a self reference to Default
trait Default[T] { def default: T }
trait Super[T] {
self: Default[T] =>
}
class B(val i: Int) extends Super[Int] with Default[Int] {
override def default: Int = ???
}
class A(val i: Int) extends Super[Int] // doesn't compile, needs a Default
This should also solve the second part of your question, the disadvantage is that now one trait is bundled to the other.
I'm looking for a way to define a method that returns a type T where T = the type of the subclass.
I know I could possibly do this using abstract types, but dislike the overhead of having to redefine T for each subclass.
Some sample code:
object Helper {
def help[A <: MyClass](cls: A): Option[A] = { cls.foo() map { _.asInstanceOf[A] } }
}
class MyClass {
type T <: MyClass
def foo(): Option[T] = Some(this.asInstanceOf[T])
}
class ChildClass extends MyClass {
type T = ChildClass
}
Possibly a new language feature has made this easier? Or can I use this.type in some way? It's important to me that I be able to define a helper class that can call into foo in this way.
If you are always returning this, then you can indeed have as return type this.type. Or have you tried it already?
this.type is especially useful e.g. when you want to chain calls to the same object, or provide a static guarantee that you will be returning the same object (and not a copy). For instance, Buffers in Scala have the append operation :+, which returns a Buffer[A], and +=, which returns this.type. The former duplicates the mutable sequence; the latter guarantees that you update the original object.
To follow up on Jean-Phillippe's answer, who wrote his exactly when I'm writing mine, here's the code:
trait SomeTrait {
def foo: this.type = this
}
class UsesTrait extends SomeTrait
object Main {
def main(args: Array[String]) {
println((new UsesTrait).foo) // prints UsesTrait#<hash value>
}
}
I found the following idiom useful:
class MyClass[T] {
self: T =>
def foo(): Option[T] = Some(this)
}
class ChildClass extends MyClass[ChildClass]
new ChildClass().foo()
//--> Option[ChildClass] = Some(ChildClass#2487b1)
I'm trying to implement a Scala trait that handles the details of interfacing with a Java library that requires us to create
What I want to do is something like:
trait SomeTrait[A] extends JavaAPI {
def foo = {
callApi(classOf[A])
}
override def bar = {
foo
}
}
Note that bar is actually overriding a method from a base class, so I can't change it's signature.
I've tried several variations with Manifests, etc., but can't quite get this to work. Is there a way to get the runtime class of a parameterized type?
This flavour should do the trick:
trait SomeTrait[A] {
def foo(implicit ev: Manifest[A]) = {
callApi(ev.erasure)
}
}
update At some point, the manifest must be injected via a method parameter. A constructor would be a good choice, if traits could have them.
Actually, they can! The trait has the constructor of whatever it's mixed-in to, so if you specify an abstract manifest that deriving classes must define...
trait SomeTrait {
def ev: Manifest[_] //abstract
def foo = println(ev.erasure)
}
//this `ev` provides the implementation, note that it MUST be a val, or var
class Concrete[T](implicit val ev: Manifest[T]) extends SomeTrait
And all is good again.
You have to get the manifest in there somehow, and traits have no constructor parameters. Only you can say what tradeoff you want to make. Here's another one.
trait SomeTrait[A] {
implicit def manifesto: Manifest[A]
def foo = println(manifest[A].erasure)
}
object SomeTrait {
def apply[A: Manifest] : SomeTrait[A] = new SomeTrait[A] { def manifesto = manifest[A] }
}
Due to type erasure, the compiler has no way to figure out what the type should be within the trait. Thus what you want can't be done. However, you could make it a class. That way the compiler can pass an evidence parameter when an instance is created.
class SomeTrait[A](implicit ev: Manifest[A]) extends JavaApi {
def foo = {
callApi(ev.erasure)
}
override def bar = {
foo
}
}
It might be a little inconvenient to do so in your code, but you can do this
trait SomeTrait[A] extends JavaAPI {
def objType: Class[A]
def foo = {
callApi(objType)
}
override def bar = {
foo
}
}
object SomeImplementation with SomeTrait[SomeObject] {
val objType: Class[SomeObject] = classOf[SomeObject]
}
I know it is a little wordy, but that's the way I solved this problem. I hope to find a better solution in the future, but this is what I'm using now. Let me know if that helps you.