Considering this question : Passing a Shapeless Extensible Record to a Function, Travis's answer shows that every function taking an extensible record as parameter must have an implicit selector as parameter. I wonder if one could factorize those declarations in case we have many functions of this kind. E.g. :
val w1 = Witness("foo1")
val w2 = Witness("foo2")
val w3 = Witness("foo3")
//Here some "magical" declarations avoiding to declara selectors in fun1, fun2, fun3 below
def fun1[L <: HList](xs: L) = ... //Access to foo1, foo2, foo3
def fun2[L <: HList](xs: L) = ... //Access to foo1, foo2, foo3
def fun3[L <: HList](xs: L) = ... //Access to foo1, foo2, foo3
Thanks
Benoit
edit on Dec 10
When trying the code of the answer, on comes with two problems :
If I call fun1 with ("foo1"->> "hello") :: ("foo2" -> 1)::("foo3" ->> 1.2)::HNiL, the result is (hello, 1, 1.2) with type (selectors.s1.Out, selectors.s2.Out, selectors.s3.Out) If I try to add 1 to the last value (1.2), Scala complains that it can't add an Int and a selectors.s3.Out ;but if I write :
val x = fun1(("foo1"->> "hello") :: ("foo2" -> 1)::("foo3" ->> 1.2)::HNil)
I can write :
x._3 == 1.2
and scala answers True!
I have tried to modify the code in this way, hopping that the types would be propagated, but it doesn't solve the problem. I can't even call fun1 with (foo1->> "hello") :: (foo2 -> 1)::(foo3 ->> 1.2)::HNil as parameter :
object foo1 extends FieldOf[String]
object foo2 extends FieldOf[Int]
object foo3 extends FieldOf[Double]
val w1 = Witness(foo1)
val w2 = Witness(foo2)
val w3 = Witness(foo3)
case class HasMyFields[L <: HList](implicit
s1: Selector[L, w1.T],
s2: Selector[L, w2.T],
s3: Selector[L, w3.T]
)
object HasMyFields {
implicit def make[L <: HList](implicit
s1: Selector[L, w1.T],
s2: Selector[L, w2.T],
s3: Selector[L, w3.T]
) = HasMyFields[L]
}
def fun1[L <: HList](xs: L)(implicit selectors: HasMyFields[L]) = {
import selectors._
(xs(foo1), xs(foo2), xs(foo3))
}
Is there a way to progress?
Benoit
You can define your own type class to gather the evidence that the record has the fields you need:
import shapeless._, ops.record.Selector, record._, syntax.singleton._
val w1 = Witness("foo1")
val w2 = Witness("foo2")
val w3 = Witness("foo3")
case class HasMyFields[L <: HList](implicit
s1: Selector[L, w1.T, String],
s2: Selector[L, w2.T, Int],
s3: Selector[L, w3.T, Double]
)
object HasMyFields {
implicit def make[L <: HList](implicit
s1: Selector[L, w1.T, String],
s2: Selector[L, w2.T, Int],
s3: Selector[L, w3.T, Double]
) = HasMyFields[L]
}
And then, for example:
def fun1[L <: HList](xs: L)(implicit selectors: HasMyFields[L]) = {
import selectors._
(xs("foo1"), xs("foo2"), xs("foo3"))
}
It's still a little verbose, especially since the import is necessary, but much less so than requiring all of the selectors individually as implicit parameters.
The out type of a given field can be specified using:
Selector[L, w1.T] { type Out = String }
Also, we can slightly simplify the syntax using a type constructor:
import shapeless._, ops.record.Selector, record._, syntax.singleton._
val w1 = Witness("foo1")
val w2 = Witness("foo2")
val w3 = Witness("foo3")
type HasFoo1[L <: HList] = Selector[L, w1.T] { type Out = String }
type HasFoo2[L <: HList] = Selector[L, w2.T]
type HasFoo3[L <: HList] = Selector[L, w3.T]
@implicitNotFound("${L} should have foo1, foo2 and foo3")
case class HasMyFields[L <: HList](implicit s1: HasFoo1[L], s2: HasFoo2[L], s3: HasFoo3[L])
object HasMyFields {
implicit def make[L <: HList : HasFoo1 : HasFoo2 : HasFoo3] = HasMyFields[L]
}
def fun1[L <: HList : HasMyFields](xs: L) = {
val selectors = implicitly[HasMyFields[L]]
import selectors._
(xs("foo1").length, xs("foo2"), xs("foo3"))
}
fun1(("foo1"->> "hello") :: ("foo2" ->> 1)::("foo3" ->> 1.2)::HNil)
// Does not compile: the value in foo1 is not a String
fun1(("foo1"->> 2) :: ("foo2" ->> 1)::("foo3" ->> 1.2)::HNil)
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