How to use implicitly with a function? - scala

Hi I read the interesting post from Debasish about the implicitly function. I have wrote this code:
def find[C <: Business](id: String) = {
collection.findOneByID(id).map(x=> implicitly[DBObject => C].apply(x))
}
but it fails to compile with this compiler message:
could not find implicit value for parameter e: (com.mongodb.casbah.commons.Imports.DBObject) => C
what is my fault? anyone can help me?
UPDATE
My idea was this:
find is declared in a trait don't know nothing about DBObject, I don't want to put this dependency.
trait BusinessRepository {
def find[C <: Business](id: String): Option[C]
}
class MongoBusinessRepository {
val collection = ..
def find[C <: Business](id: String): Option[C] = {
collection.findOneByID(id).map(x=> implicitly[DBObject => C].apply(x))
}
implicit def DBObject2Hotel(x: DBObject): Hotel = {
// ...
// returning Hotel
}
}
case class Hotel(...) extends Business(...)

implicitly is just a convenience method to look up an implicit value that you know already exists. So it fails to compile when there is no such implicit value in scope.
A possible use case is when you use shortcut syntax for context bounds:
def find[C: Numeric](a: C, b: C): C = implicitly[Numeric[C]].plus(a, b)
Of course, in this example, the explicit form is less verbose
def find[C](a: C, b: C)(implicit n: Numeric[C]): C = n.plus(a, b)
You will find more thorough explanations in this Stackoverflow thread.
What I imagine you had in mind with your method is rather
def find[C <: Business](id: String)(implicit fun: DBObject => C) =
collection.findOneByID(id).map(fun)

I think that the problem comes from the fact that Scala compiler try to find an implicit definition of a function DBObject => C and that the only implicit definition that he can find is DBObject => Hotel that could be a solution but it's not strict. With your solution, the compiler is not able to know what should be C.
Maybe you should consider defining a DBObject2Business and so implicitly define a DBObject => Business function, or change your design to define C in a concrete class.

Related

How to implement Applicative for State[S, A] in scala / cats

In this session SystemFw gives an example of implementing State[S, A] wih vanilla scala, When follow the exmple, I run into a trouble in supplying a Applicative definition for the vanilla State type (In order to get the commands.traverse work. See code here
I tried to make a implicit def to solve the Applicative instance, but didn't figure out how to deal with the 2 type parameter.
How to implement Applicative for this type:
case class State[S, A](modify: S => (A, S)) {
def runA(initial: S): A = modify(initial)._1
def flatMap[B](f: A => State[S, B]): State[S, B] =
State { s =>
val (result, nextState) = modify(s)
f(result).modify(nextState)
}
}
Wrong code:
implicit def stateApplicative[S, A]: Applicative[State[S, A]] = new Applicative[State[S, A]] {
override def pure[A](x: A): State[A???] = ??? // error
override def ap[A, B](ff: State[A => B])(fa: State[A???]): State[B] = ??? // error
}
Basically, the solution to this problem is always fixing one type parameter.
In the case of a State you want the value inside the state to change but no the type of the state itself so you fix S.
So you can create an application for a given particular state, for example Int
type IntState[A] = State[A, Int]
implicit final val intStateApplicative: Applicative[IntState] =
new Applicative[IntState] {
// Some implementation.
}
However, after you fill the implementation you will see that the fact of knowing that S was Int was meaningless. And that we could just copy and paste the whole code if S would have been String or whatever.
So, what we want is a way to say that this works for any S, we can do that with a type lambda (i.e. a function in the type level).
type StateTypeFun[S] = { type F[A] = State[A, S] }
implicit final def stateApplicative[S]: Applicative[StateTypeFun[S]#F] =
new Applicative[StateTypeFun[S]#F] {
// Some implementation.
}
And that is how we solve this problem.
Note that the type alias is unnecessary but makes the code easier to read, but you could have done Applicative[({ type F[A] = State[A, S]})#F].
BTW, because this necessity of creating type lambdas is somewhat common in Scala 2 we have kind projector, and Scala 3 has proper syntax support for it.

Scala GADT instance Functor

In Haskell I got:
data Foo a where
Bar :: a -> Foo a
Map :: (a -> b) -> Foo a -> Foo b
instance Functor Foo where
fmap = Map
In Scala I came up with:
import cats.Functor
trait Foo[A]
case class Bar[A](t: A) extends Foo[A]
case class Map[A,B](fun: A => B,foo: Foo[A]) extends Foo[B]
implicit val fooFunctor: Functor[Foo] = new Functor[Foo] {
def map[A,B](fa: Foo[A])(f: A => B) = Map(f,fa)
}
But Bar(40).map(_+2) gives me:
error: value map is not a member of Bar[Int]
I fairly new to Scala and don't know the way of inheritance that well.
What am I missing?
You need to upcast Bar(40) to Foo explicitly, and import the syntax implicits:
import cats.syntax.functor._
(Bar(40): Foo[Int]).map(_+2)
You need the upcast because Scala will infer the type Bar[Int] for the expression Bar(40) and this then interferes with finding the appropriate implicit that adds the map method. For this reason you sometimes see helper methods in the companion object to do the upcast for you:
object Foo {
def bar[A](a: A): Foo[A] = Bar(a)
// etc.
}
I thought a little bit about how one could solve this problem. Basically, we want to provide the functor operations even if only a super type is a functor. It seems to be impossible to express, that some higher-kinded type is the super type of an other, e.g. something like this is not possible: def bla[F[_], R[_] <: F]
We can however provide a conversion from R[] to F[] implicitly together with the functor for B:
abstract class ProvidesFor[R,F[_],FF[_[_]],A](val value: FF[F]) {
def convert(r: R): F[A]
}
implicit def providesFunctorForBar[A] =
new ProvidesFor[Bar[A],Foo,Functor,A](Functor[Foo]) {
override def convert(r: Bar[A]): Foo[A] = r
}
So the implict val will provide us with a functor for Foo and a convertion from Bar to Foo.
Now we can provide an implicit converstion to Functor.Ops like this:
implicit class FunctorOps[R[_],F[_],A](target: R[A])(implicit tc: ProvidesFor[R[A],F,Functor,A])
extends Functor.Ops[F,A] {
override val self = tc.convert(target)
override val typeClassInstance = tc.value
override def map[B](f: A => B): F[B] = typeClassInstance.map(self)(f)
override def imap[B](f: A => B)(g: B => A): F[B] = map(f)
}
Now this works as expected:
Bar(1).imap(_+2)(_+5)
We can also do the same for Map:
implicit def providesFunctorForMap[A,B] =
new ProvidesFor[Map[A,B],Foo,Functor,B](Functor[Foo]) {
override def convert(r: Map[A,B]): Foo[B] = r
}
Map((_:Int) + 1,Bar(5)).map(_+2)
For some strange reason I had to implement map and imap when extending Functor.Ops even though these methods are not abstract. In fact it compiles fine when I do not implement them but it fails at runtime with an AbstractMethodError. So somehow the compiler thinks the implementations are there but they are not. I suspect they are using some kind of byte code optimization tool that removes those implementations.

Provide type-class when implementing/override method

I would prefer to work more with type-classes but having some issues:
Given the following interface
trait Processor[A] {
def process[B](f: A => B): Processor[B]
}
I have an implementation that needs an Ordering[A] for some other reasons. Hence the method process needs an Ordering[B] to construct a Processor[B].The following is what I would like to do, but it does obviously not work:
class Plant[A, OA <: Ordering[A]] extends Processor[A] {
def process[B:Ordering](f: A => B): Processor[B] = null // Plant[B, OB <: Ordering[B]]
}
How can I provide the Ordering[B] for the implementation of process?
I know the reason for this is, that Ordering[A] is passed as an implicit second argument. I don't know but shouldn't there be special support for type-classes in Scala similar to Haskell to recognize what I want (only allow Bs that have an Ordering) in the implementation above without this "workaround"?
No, and this shouldn't work at all. Given your defnition of Processor, this code will compile:
val processor: Processor[Int] = foo() // foo() is some function that returns a Processor
processor.process[Object](x => new Object())
Now if foo is actually implemented as
def foo() = new Plant[Int]()
then its process method won't work with B = Object.
I'm not sure I fully understand what you're trying to achieve, but perhaps you could make Processor generic on the context bound?
trait Processor[A, Bound[_]] {
def process[B: Bound](f: A => B): Processor[B, Any]
}
class Plant[A: Ordering] extends Processor[A, Ordering] {
def process[B: Ordering](f: A => B) = ???
}

What are the limitations on inference of higher-kinded types in Scala?

In the following simplified sample code:
case class One[A](a: A) // An identity functor
case class Twice[F[_], A](a: F[A], b: F[A]) // A functor transformer
type Twice1[F[_]] = ({type L[α] = Twice[F, α]}) // We'll use Twice1[F]#L when we'd like to write Twice[F]
trait Applicative[F[_]] // Members omitted
val applicativeOne: Applicative[One] = null // Implementation omitted
def applicativeTwice[F[_]](implicit inner: Applicative[F]): Applicative[({type L[α] = Twice[F, α]})#L] = null
I can call applicativeTwice on applicativeOne, and type inference works, as soon as I try to call it on applicativeTwice(applicativeOne), inference fails:
val aOK = applicativeTwice(applicativeOne)
val bOK = applicativeTwice[Twice1[One]#L](applicativeTwice(applicativeOne))
val cFAILS = applicativeTwice(applicativeTwice(applicativeOne))
The errors in scala 2.10.0 are
- type mismatch;
found : tools.Two.Applicative[[α]tools.Two.Twice[tools.Two.One,α]]
required: tools.Two.Applicative[F]
- no type parameters for method applicativeTwice:
(implicit inner: tools.Two.Applicative[F])tools.Two.Applicative[[α]tools.Two.Twice[F,α]]
exist so that it can be applied to arguments
(tools.Two.Applicative[[α]tools.Two.Twice[tools.Two.One,α]])
--- because ---
argument expression's type is not compatible with formal parameter type;
found : tools.Two.Applicative[[α]tools.Two.Twice[tools.Two.One,α]]
required: tools.Two.Applicative[?F]
Why wouldn't "?F" match with anything (of the right kind) ?
Ultimately I'd like applicativeTwice to be an implicit function, but I'd have to get the type inference working first.
I have seen similar questions, and the answers pointed to limitations in the type inference algorithms. But this case seems pretty limitative, and must be quite an annoyance in monad transformers, so I suspect I'm missing some trick to work around this.
You've hit a common annoyance: SI-2712. For clarity, I'm going to minimize your code a bit:
import language.higherKinds
object Test {
case class Base[A](a: A)
case class Recursive[F[_], A](fa: F[A])
def main(args: Array[String]): Unit = {
val one = Base(1)
val two = Recursive(one)
val three = Recursive(two) // doesn't compile
println(three)
}
}
This demonstrates the same type error as yours:
argument expression's type is not compatible with formal parameter type;
found : Test.Recursive[Test.Base,Int]
required: ?F
val three = Recursive(two) // doesn't compile
^
First a bit of syntax and terminology you probably already know:
In Scala we say that a plain, unparameterized data type (such as Int) has kind _. It's monomorphic.
Base, on the other hand, is parameterized. we can't use it as the type of a value without providing the type it contains, so we say has kind _[_]. It's rank-1 polymorphic: a type constructor that takes a type.
Recursive goes further still: it has two parameters, F[_] and A. The number of type parameters don't matter here, but their kinds do. F[_] is rank-1 polymorphic, so Recursive is rank-2 polymorphic: it's a type constructor that takes a type constructor.
We call anything rank two or above higher-kinded, and this is where the fun starts.
Scala in general doesn't have trouble with higher-kinded types. This is one of several key features that distinguishes its type system from, say, Java's. But it does have trouble with partial application of type parameters when dealing with higher-kinded types.
Here's the problem: Recursive[F[_], A] has two type parameters. In your sample code, you did the "type lambda" trick to partially apply the first parameter, something like:
val one = Base(1)
val two = Recursive(one)
val three = {
type λ[α] = Recursive[Base, α]
Recursive(two : λ[Int])
}
This convinces the compiler that you're providing something of the correct kind (_[_]) to the Recursive constructor. If Scala had curried type parameter lists, I'd definitely have used that here:
case class Base[A](a: A)
case class Recursive[F[_]][A](fa: F[A]) // curried!
def main(args: Array[String]): Unit = {
val one = Base(1) // Base[Int]
val two = Recursive(one) // Recursive[Base][Int]
val three = Recursive(two) // Recursive[Recursive[Base]][Int]
println(three)
}
Alas, it does not (see SI-4719). So, to the best of my knowledge, the most common way of dealing with this problem is the "unapply trick," due to Miles Sabin. Here is a greatly simplified version of what appears in scalaz:
import language.higherKinds
trait Unapply[FA] {
type F[_]
type A
def apply(fa: FA): F[A]
}
object Unapply {
implicit def unapply[F0[_[_], _], G0[_], A0] = new Unapply[F0[G0, A0]] {
type F[α] = F0[G0, α]
type A = A0
def apply(fa: F0[G0, A0]): F[A] = fa
}
}
In somewhat hand-wavey terms, this Unapply construct is like a "first-class type lambda." We define a trait representing the assertion that some type FA can be decomposed into a type constructor F[_] and a type A. Then in its companion object, we can define implicits to provide specific decompositions for types of various kinds. I've only defined here the specific one that we need to make Recursive fit, but you could write others.
With this extra bit of plumbing, we can now do what we need:
import language.higherKinds
object Test {
case class Base[A](a: A)
case class Recursive[F[_], A](fa: F[A])
object Recursive {
def apply[FA](fa: FA)(implicit u: Unapply[FA]) = new Recursive(u(fa))
}
def main(args: Array[String]): Unit = {
val one = Base(1)
val two = Recursive(one)
val three = Recursive(two)
println(three)
}
}
Ta-da! Now type inference works, and this compiles. As an exercise, I'd suggest you create an additional class:
case class RecursiveFlipped[A, F[_]](fa: F[A])
... which isn't really different from Recursive in any meaningful way, of course, but will again break type inference. Then define the additional plumbing needed to fix it. Good luck!
Edit
You asked for a less simplified version, something aware of type-classes. Some modification is required, but hopefully you can see the similarity. First, here's our upgraded Unapply:
import language.higherKinds
trait Unapply[TC[_[_]], FA] {
type F[_]
type A
def TC: TC[F]
def apply(fa: FA): F[A]
}
object Unapply {
implicit def unapply[TC[_[_]], F0[_[_], _], G0[_], A0](implicit TC0: TC[({ type λ[α] = F0[G0, α] })#λ]) =
new Unapply[TC, F0[G0, A0]] {
type F[α] = F0[G0, α]
type A = A0
def TC = TC0
def apply(fa: F0[G0, A0]): F[A] = fa
}
}
Again, this is completely ripped off from scalaz. Now some sample code using it:
import language.{ implicitConversions, higherKinds }
object Test {
// functor type class
trait Functor[F[_]] {
def map[A, B](fa: F[A])(f: A => B): F[B]
}
// functor extension methods
object Functor {
implicit class FunctorOps[F[_], A](fa: F[A])(implicit F: Functor[F]) {
def map[B](f: A => B) = F.map(fa)(f)
}
implicit def unapply[FA](fa: FA)(implicit u: Unapply[Functor, FA]) =
new FunctorOps(u(fa))(u.TC)
}
// identity functor
case class Id[A](value: A)
object Id {
implicit val idFunctor = new Functor[Id] {
def map[A, B](fa: Id[A])(f: A => B) = Id(f(fa.value))
}
}
// pair functor
case class Pair[F[_], A](lhs: F[A], rhs: F[A])
object Pair {
implicit def pairFunctor[F[_]](implicit F: Functor[F]) = new Functor[({ type λ[α] = Pair[F, α] })#λ] {
def map[A, B](fa: Pair[F, A])(f: A => B) = Pair(F.map(fa.lhs)(f), F.map(fa.rhs)(f))
}
}
def main(args: Array[String]): Unit = {
import Functor._
val one = Id(1)
val two = Pair(one, one) map { _ + 1 }
val three = Pair(two, two) map { _ + 1 }
println(three)
}
}
Note (3 years later, July 2016), scala v2.12.0-M5 is starting to implement SI-2172 (support for higher order unification)
See commit 892a6d6 from Miles Sabin
-Xexperimental mode now only includes -Ypartial-unification
It follows Paul Chiusano's simple algorithm:
// Treat the type constructor as curried and partially applied, we treat a prefix
// as constants and solve for the suffix. For the example in the ticket, unifying
// M[A] with Int => Int this unifies as,
//
// M[t] = [t][Int => t] --> abstract on the right to match the expected arity
// A = Int --> capture the remainder on the left
The test/files/neg/t2712-1.scala includes:
package test
trait Two[A, B]
object Test {
def foo[M[_], A](m: M[A]) = ()
def test(ma: Two[Int, String]) = foo(ma) // should fail with -Ypartial-unification *disabled*
}
And (test/files/neg/t2712-2.scala):
package test
class X1
class X2
class X3
trait One[A]
trait Two[A, B]
class Foo extends Two[X1, X2] with One[X3]
object Test {
def test1[M[_], A](x: M[A]): M[A] = x
val foo = new Foo
test1(foo): One[X3] // fails with -Ypartial-unification enabled
test1(foo): Two[X1, X2] // fails without -Ypartial-unification
}

Implicit parameters won't work on unapply. How to hide ubiquitous parameters from extractors?

Apparently unapply/unapplySeq in extractor objects do not support implicit parameters. Assuming here an interesting parameter a, and a disturbingly ubiquitous parameter b that would be nice to hide away, when extracting c.
[EDIT]: It appears something was broken in my intellij/scala-plugin installation that caused this. I cannot explain. I was having numerous strange problems with my intellij lately. After reinstalling, I can no longer reprodce my problem. Confirmed that unapply/unapplySeq do allow for implicit parameters! Thanks for your help.
This does not work (**EDIT:yes, it does):**
trait A; trait C; trait B { def getC(a: A): C }
def unapply(a:A)(implicit b:B):Option[C] = Option(b.getC(a))
In my understanding of what an ideal extractor should be like, in which the intention is intuitively clear also to Java folks, this limitation basically forbids extractor objects which depend on additional parameter(s).
How do you typically handle this limitation?
So far I've got those four possible solutions:
1) The simplest solution that I want to improve on: don't hide b, provide parameter b along with a, as normal parameter of unapply in form of a tuple:
object A1{
def unapply(a:(A,B)):Option[C] = Option(a._2.getC(a._1)) }
in client code:
val c1 = (a,b) match { case A1(c) => c1 }
I don't like it because there is more noise deviating that deconstruction of a into c is important here. Also since java folks, that have to be convinced to actually use this scala code, are confronted with one additional synthactic novelty (the tuple braces). They might get anti-scala aggressions "What's all this? ... Why then not use a normal method in the first place and check with if?".
2) define extractors within a class encapsulating the dependence on a particular B, import extractors of that instance. At import site a bit unusual for java folks, but at pattern match site b is hidden nicely and it is intuitively evident what happens. My favorite. Some disadvantage I missed?
class BDependent(b:B){
object A2{
def unapply(a:A):Option[C] = Option(b.getC(a))
} }
usage in client code:
val bDeps = new BDependent(someB)
import bDeps.A2
val a:A = ...
val c2 = a match { case A2(c) => c }
}
3) declare extractor objects in scope of client code. b is hidden, since it can use a "b" in local scope. Hampers code reuse, heavily pollutes client code (additionally, it has to be stated before code using it).
4) have unapply return Option of function B => C. This allows import and usage of an ubitious-parameter-dependent extractor, without providing b directly to the extractor, but instead to the result when used. Java folks maybe confused by usage of function values, b not hidden:
object A4{
def unapply[A,C](a:A):Option[B => C] = Option((_:B).getC(a))
}
then in client code:
val b:B = ...
val soonAC: B => C = a match { case A4(x) => x }
val d = soonAC(b).getD ...
Further remarks:
As suggested in this answer, "view bounds" may help to get extractors work with implicit conversions, but this doesn't help with implicit parameters. For some reason I prefer not to workaround with implicit conversions.
looked into "context bounds", but they seem to have the same limitation, don't they?
In what sense does your first line of code not work? There's certainly no arbitrary prohibition on implicit parameter lists for extractor methods.
Consider the following setup (I'm using plain old classes instead of case classes to show that there's no extra magic happening here):
class A(val i: Int)
class C(val x: String)
class B(pre: String) { def getC(a: A) = new C(pre + a.i.toString) }
Now we define an implicit B value and create an extractor object with your unapply method:
implicit val b = new B("prefix: ")
object D {
def unapply(a: A)(implicit b: B): Option[C] = Option(b getC a)
}
Which we can use like this:
scala> val D(c) = new A(42)
c: C = C#52394fb3
scala> c.x
res0: String = prefix: 42
Exactly as we'd expect. I don't see why you need a workaround here.
The problem you have is that implicit parameters are compile time (static) constraints, whereas pattern matching is a runtime (dynamic) approach.
trait A; trait C; trait B { def getC(a: A): C }
object Extractor {
def unapply(a: A)(implicit b: B): Option[C] = Some(b.getC(a))
}
// compiles (implicit is statically provided)
def withImplicit(a: A)(implicit b: B) : Option[C] = a match {
case Extractor(c) => Some(c)
case _ => None
}
// does not compile
def withoutImplicit(a: A) : Option[C] = a match {
case Extractor(c) => Some(c)
case _ => None
}
So this is a conceptual problem, and the solution depends on what you actually want to achieve. If you want something along the lines of an optional implicit, you might use the following:
sealed trait FallbackNone {
implicit object None extends Optional[Nothing] {
def toOption = scala.None
}
}
object Optional extends FallbackNone {
implicit def some[A](implicit a: A) = Some(a)
final case class Some[A](a: A) extends Optional[A] {
def toOption = scala.Some(a)
}
}
sealed trait Optional[+A] { def toOption: Option[A]}
Then where you had implicit b: B you will have implicit b: Optional[B]:
object Extractor {
def unapply(a:A)(implicit b: Optional[B]):Option[C] =
b.toOption.map(_.getC(a))
}
def test(a: A)(implicit b: Optional[B]) : Option[C] = a match {
case Extractor(c) => Some(c)
case _ => None
}
And the following both compile:
test(new A {}) // None
{
implicit object BImpl extends B { def getC(a: A) = new C {} }
test(new A {}) // Some(...)
}