Scala Mutually Convertible Generic Types - scala

I'm very new to Scala programming, and I really like the degree to which code is composable. I wanted to write some traits that deal with two related objects that are convertible to each other, and build more functionality by continuing to extend that trait so that when I create objects I can specify the related types for my generics. Here is a working toy example of the type of code I'm talking about:
trait FirstConverter[First] {
def toFirst: First
}
trait SecondConverter[Second] {
def toSecond: Second
}
trait TwoWayConverter[First <: SecondConverter[Second], Second <: FirstConverter[First]] {
def firstToSecond(x: First) = x.toSecond
def secondToFirst(x: Second) = x.toFirst
}
trait RoundTripConverter[First <: SecondConverter[Second], Second <: FirstConverter[First]] extends TwoWayConverter[First, Second] {
def firstToFirst(x: First) = secondToFirst(firstToSecond(x))
def secondToSecond(x: Second) = firstToSecond(secondToFirst(x))
}
case class A(s: String) extends SecondConverter[B] {
def toSecond: B = B((s.toInt) + 1)
}
case class B(i: Int) extends FirstConverter[A] {
def toFirst: A = A((i * 2).toString)
}
object ABConverter extends RoundTripConverter[A, B]
object Main {
def main(args: Array[String]): Unit = {
println(ABConverter firstToSecond A("10")) // 11
println(ABConverter secondToFirst B(42)) // 84
println(ABConverter firstToFirst A("1")) // 4
println(ABConverter secondToSecond B(2)) // 5
}
}
While this works, I'm not sure if it's idiomatic Scala. I'm asking if there are any tricks to make the type definitions more concise and if I can somehow define the type restrictions only once and have them used by multiple traits which extend other traits.
Thanks in advance!

One way to improve your design would be to use a type class instead of inheriting from FirstConverter and SecondConverter. That way you could use multiple conversion functions for the same types and convert between classes you don't control yourself.
One way would be to create a type class which can convert an A into a B :
trait Converter[A, B] {
def convert(a: A): B
}
trait TwoWayConverter[A, B] {
def firstToSecond(a: A)(implicit conv: Converter[A, B]): B = conv.convert(a)
def secondToFirst(b: B)(implicit conv: Converter[B, A]): A = conv.convert(b)
}
trait RoundTripConverter[A, B] extends TwoWayConverter[A, B] {
def firstToFirst(a: A)(implicit convAB: Converter[A, B], convBA: Converter[B, A]) =
secondToFirst(firstToSecond(a))
def secondToSecond(b: B)(implicit convAB: Converter[A, B], convBA: Converter[B, A]) =
firstToSecond(secondToFirst(b))
}
We could create type class instances for the following two classes Foo and Bar similar to your A and B
case class Foo(s: String)
case class Bar(i: Int)
implicit val convFooBarFoor = new Converter[Foo, Bar] {
def convert(foo: Foo) = Bar((foo.s toInt) + 1)
}
implicit val convBarFoo = new Converter[Bar, Foo] {
def convert(bar: Bar) = Foo((bar.i * 2) toString)
}
We then could create a FooBarConverter :
object FooBarConverter extends RoundTripConverter[Foo, Bar]
FooBarConverter firstToSecond Foo("10") // Bar(11)
FooBarConverter secondToFirst Bar(42) // Foo(84)
FooBarConverter firstToFirst Foo("1") // Foo(4)
FooBarConverter secondToSecond Bar(2) // Bar(5)
The only problem is because we can not pass parameters to a trait, we can not limit the types to types with a Converter type class instance. So you can create the StringIntConverter below even if no Converter[String, Int] and/or Convert[Int, String] instances exist.
object StringIntConverter extends TwoWayConverter[String, Int]
You cannot call StringIntConverter.firstToSecond("a") because the firstToSecond method needs the implicit evidence of the two mentioned type class instances.

Related

Type Class for Related Types

Let's say we have the following traits:
trait MyValue
object MyValue {
case class MyBoolean(record: Boolean) extends MyValue
case class MyLong(record: Long) extends MyValue
}
trait MyValueExtractor[T] {
def apply(record: T): Option[MyValue]
}
trait MyThing[T] {
def name: String
def myValueExtractor: MyValueExtractor[T]
def myValue(record: T): Option[MyValue] = myValueExtractor(record)
}
What I want is something like this but without the second type parameter.
Note: I can't actually update the MyThing trait; I'm just using this as an illustration of the intended functionality.
trait MyThing[T, U] {
def name: String
def myValueExtractor: MyValueExtractor[T]
def myValue(record: T): Option[MyValue] = myValueExtractor(record)
def myRelatedValue(record: T): Option[U]
}
I'm wondering if I could use the type class pattern to help solve this (i.e., import some rich class that implicitly gives me a myRelatedValue method)?
Here's the rub. Every time T (above) is MyValue.MyBoolean, U must be a String. Every time T is MyValue.MyLong, U must be a Double. In other words, there's a sort of underlying mapping between T and U.
Is there a good way to do this using type class?
Sure. You just need to define some Mapping typeclass with implementations for your desired pairs of types. Then MyThing can have a method that takes an implicit typeclass instance and simply invokes its method.
Here's the code (I removed the unneeded details)
// types
case class MyBoolean(record: Boolean)
case class MyLong(record: Long)
// trait which uses the Mapping typeclass
trait MyThing[T] {
def myRelatedValue[U](record: T)(implicit ev: Mapping[T, U]): Option[U] = ev.relatedValue(record)
}
// typeclass itself
trait Mapping[T, U] {
def relatedValue(record: T): Option[U]
}
object Mapping {
implicit val boolStringMapping = new Mapping[MyBoolean, String] {
def relatedValue(record: MyBoolean) = Some(record.record.toString)
}
implicit val longDoubleMapping = new Mapping[MyLong, Double] {
def relatedValue(record: MyLong) = Some(record.record)
}
}
// usage
val myBoolThing = new MyThing[MyBoolean] {}
val myLongThing = new MyThing[MyLong] {}
val myStringThing = new MyThing[String] {}
myBoolThing.myRelatedValue(MyBoolean(true)) // Some(true)
myLongThing.myRelatedValue(MyLong(42L)) // Some(42.0)
myStringThing.myRelatedValue("someString") // error: could not find implicit value
Note that e.g. myBoolThing.myRelatedValue(MyBoolean(true)) will yield a type Option[U]. However, since myRelatedValue is parameterized, you can help the compiler and invoke it as myBoolThing.myRelatedValue[String](MyBoolean(true)), in which case you will obtain an Option[String]. If you try something other than String for MyBoolean, you will get an error.

List of elements implementing typeclass A and B in Scala

I got the following issue when trying to use typeclasses throughout my project.
trait FooAble[T] { def fa(t: T): List[T] }
object Foo { def apply[T](t: T) = implicitly[FooAble[T]].fa(t) }
trait BarAble[T] { def fb(t: T): Double }
object Bar { def apply[T](t: T) = implicitly[BarAble[T]].fb(t) }
And would like to be able to do the following:
// xs contains elements of type A and B which are subclasses of the trait Something
def f(xs: List[Something]) = {
val something = xs.map(Foo)
val somethingElse = xs.map(Bar)
}
However, this would not work as we don't know if Something implements A[]and B[], no implicit implementation found. What do I need to do so that the elements of the list xs implement the typeclasses FooAble and BarAble?
I think this question: What are type classes in Scala useful for? will help you to understand the proper use (& usefulness) of type classes.
Am just extending the answer by Kevin Wright in the above link for your use case (if I understand your need correctly!):
trait Addable[T] {
def zero: T
def append(a: T, b: T): T
}
trait Productable[T] {
def zero: T
def product(a: T, b: T): T
}
implicit object IntIsAddable extends Addable[Int] {
def zero = 0
def append(a: Int, b: Int) = a + b
}
implicit object IntIsProductable extends Productable[Int] {
def zero = 1
def product(a: Int, b: Int) = a*b
}
def sumAndProduct[T](xs: List[T])(implicit addable: Addable[T], productable: Productable[T]) =
(xs.foldLeft(addable.zero)(addable.append), xs.foldLeft(productable.zero)(productable.product))
So akin to above, in your use case, you need to provide implicit objects which implement your type classes FooAble & BarAble and your method signature for function f becomes:
def f[Something](xs: List[Something])(implicit fooable: FooAble[Something], barable: BarAble[Something])

Scala factory for generic types using the apply method?

Suppose that I have the following trait that defines an interface and takes a couple of type parameters...
trait Foo[A, B] {
// implementation details not important
}
I want to use the companion object as a factory for concrete implementations of the trait. I also want to force users to use the Foo interface instead of sub-classing So I hide the concrete implementations in the companion object like so:
object Foo {
def apply[A, B](thing: Thing): Foo[A, B] = {
???
}
private case class FooImpl[A1, B1](thing: Thing) extends Foo[A1, B1]
private case class AnotherFooImpl[A2, B1](thing: Thing) extends Foo[A2, B1]
}
I want to be able to use the factory as follows:
val foo = Foo[A1, B1](thing) // should be an instance of FooImpl
val anotherFoo = Foo[A2, B1](thing) // should be an instance of AnotherFooImpl
How do I implement the apply method to make this happen? This SO post seems close to the mark.
How about:
trait Foo[A, B]
trait Factory[A, B] {
def make(thing: Thing): Foo[A, B]
}
class Thing
object Foo {
def apply[A, B](thing: Thing)(implicit ev: Factory[A, B]) = ev.make(thing)
private case class FooImpl[A, B](thing: Thing) extends Foo[A, B]
private case class AnotherFooImpl[A, B](thing: Thing) extends Foo[A, B]
implicit val fooImplFactory: Factory[Int, String] = new Factory[Int, String] {
override def make(thing: Thing): Foo[Int, String] = new FooImpl[Int, String](thing)
}
implicit val anotherFooImplFactory: Factory[String, String] = new Factory[String, String] {
override def make(thing: Thing): Foo[String, String] = new AnotherFooImpl[String, String](thing)
}
And now:
def main(args: Array[String]): Unit = {
import Foo._
val fooImpl = Foo[Int, String](new Thing)
val anotherFooImpl = Foo[String, String](new Thing)
println(fooImpl)
println(anotherFooImpl)
}
Yields:
FooImpl(testing.X$Thing#4678c730)
AnotherFooImpl(testing.X$Thing#c038203)
Using TypeTags (to overcome erasure of type parameters), we can call the respective hidden implementations based on the type parameters passed in to the apply method like below. It correctly instantiates the respective implementations but the type information for Foo is lost, in fact its coming some garbage like _202 etc? I don't know why that is happening and how to retain the correct types for Foo. Maybe someone can throw light on this.
trait Foo[A,B]
object Foo {
def apply[A: TypeTag, B: TypeTag](thing: Thing) =
if(typeTag[A] == typeTag[Int])
FooImpl(thing)
else if(typeTag[A] == typeTag[String])
AnotherFooImpl(thing)
else
new Foo[Double,Double] {}
private case class FooImpl(thing: Thing) extends Foo[Int, String]
private case class AnotherFooImpl(thing: Thing) extends Foo[String, String]
}
Foo[Int,String](new Thing) // Foo[_202, _203] = FooImpl($sess.cmd123$Thing#50350b75)
The actual types for _203 and _203 are: ???
// type _203 >: String with _201, type _202 >: Int with _200
Foo[String,String](new Thing) //Foo[_202, _203] = AnotherFooImpl($sess.cmd123$Thing#51d80d6)

How to remove repetitive case statements for implementations of a sealed trait

I often find that I need to extract the type of a sealed trait before doing the same thing to each implementation:
sealed trait Trait
case class Foo() extends Trait
case class Bar() extends Trait
// ... lots of other implementations
// *must* take a `Trait`, not a `T <: Trait`
def thing(t: Trait): ??? = t match {
case f: Foo => // something with the instance and specific type
case b: Bar => // something with the instance and specific type
// ... same thing again for other implementations
}
for example
// typically provided by somebody else...
trait Thing[T] { def thingy: String }
implicit def thing[T]: Thing[T] = new Thing[T] { def thingy = "stuff" }
def thing(t: Trait): String = t match {
case Foo() => implicitly[Thing[Foo]].thingy
case Bar() => implicitly[Thing[Bar]].thingy
// ...
}
I'd like to reduce the boilerplate involved in doing this.
UPDATE: nowadays we'd use typeclass derivation via shapeless. e.g. https://github.com/fommil/shapeless-for-mortals
It turns out that you can use shapeless' polymorphic functions and co-product to do this:
object thing extends Poly1 {
implicit def action[T <: Trait: Thing] = at[T](
a => implicitly[Thing[T]].thingy
)
// magic that makes it work at the sealed trait level
def apply(t: Trait): String =
Generic[Trait].to(t).map(thing).unify
}
which can then be used like
println(thing(Foo(): Trait))
I'd like to make this easier to write via an abstract class (let's forget about passing on implicit parameters to action for now), e.g.
abstract class MatchSealed[In, Out] extends Poly1 {
implicit def poly[T <: In] = at[T](action)
def action[T <: In](t: T): Out
import ops.coproduct.Mapper
def apply[R <: HList](in: In)(
implicit
g: Generic.Aux[In, R],
m: Mapper[this.type, R]
): Out = {
val self: this.type = this
g.to(in).map(self).unify
}
}
but this is failing with a missing Mapper[self.type, g.Repr] on the final line. I don't know which implicit is missing, but I suspect it is the self.type. I really want to capture realisedSelf.type but I don't know how to do that.
UPDATE: it turns out that it is not possible to obtain the Mapper because it needs access to the realised object Unable to map on HList

Overriding higher-kinded abstract types in Scala

The following code shows a shallow hierarchy where a type representing a generic binary operation is used to substantiate a parameterized abstract type in another shallow container hierarchy:
trait BinaryOp[A] extends ((A,A) => A)
trait Plus[A] extends BinaryOp[A]
trait Minus[A] extends BinaryOp[A]
trait BaseOps {
type T[A] <: BinaryOp[A]
def apply[B](one: B, two: B)(op: T[B]) = op(one, two)
}
case object PlusOp extends BaseOps {
override type T[A] = Plus[A]
}
case object MinusOp extends BaseOps {
override type T[A] = Minus[A]
}
object App {
val plus = new Plus[Int] {
def apply(i: Int, i2: Int) = i + i2
}
def main(a: Array[String]) {
val exp = Expr(PlusOp)
exp.bo(1,2)(plus)
}
}
The idea is to be able to state an operation that may be valid for many different types up front, without being tied to a type-specific operation. If I define an expression class generically, all is well
case class Expr[T <: BaseOps](bo: T = PlusOp)
However for my use case it is undesirable for Expr to to be paremeterized:
case class Expr(bo: BaseOps = PlusOp)
The following code fails without a generic Expr:
object App {
val plus = new Plus[Int] {
def apply(i: Int, i2: Int) = i + i2
}
def main(a: Array[String]) {
val exp = Expr(PlusOp)
exp.bo(1,2)(plus)
}
}
The error:
found : App.plus.type (with underlying type java.lang.Object with Plus[Int])
required: exp.bo.T[Int]
exp.bo(1,2)(plus)
This makes it seem as if the type information from the abstract type T[A] <: BinaryOp[A] is not being substantiated with the information in the subtype PlusOp, which overrides the abstract type as T[A] = Plus[A]. Is there any way to work around this without making Expr generic?
With "-Ydependent-method-types",
def Expr(_bo: BaseOps = PlusOp) = new BaseOps {
override type T[A] = _bo.T[A]
val bo: _bo.type = _bo
}
But, I don't know what this precisely means...