Poly fold return type not inferred correctly
I am trying to create a poly function that folds over a tuple of Foos:
case class Foo[A](a: A)
object extractFold extends Poly2 {
implicit def default[A, As <: HList]: Case.Aux[Foo[A], Foo[As], Foo[A :: As]] = {
???
}
}
def extract[In, A <: HList, B <: HList](keys: In)
(implicit
gen: Generic.Aux[In, A],
folder: RightFolder.Aux[A, Foo[HNil], extractFold.type, Foo[B]],
tupler: Tupler[B])
: Foo[tupler.Out] = {
???
}
val result = extract((Foo(1), Foo("a")))
The function works at runtime, but the compiler inferred result type is always Foo[Unit] which is not right - in this example it should be Foo[(Int, String)]
scala shapeless
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
add a comment |
I am trying to create a poly function that folds over a tuple of Foos:
case class Foo[A](a: A)
object extractFold extends Poly2 {
implicit def default[A, As <: HList]: Case.Aux[Foo[A], Foo[As], Foo[A :: As]] = {
???
}
}
def extract[In, A <: HList, B <: HList](keys: In)
(implicit
gen: Generic.Aux[In, A],
folder: RightFolder.Aux[A, Foo[HNil], extractFold.type, Foo[B]],
tupler: Tupler[B])
: Foo[tupler.Out] = {
???
}
val result = extract((Foo(1), Foo("a")))
The function works at runtime, but the compiler inferred result type is always Foo[Unit] which is not right - in this example it should be Foo[(Int, String)]
scala shapeless
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
add a comment |
I am trying to create a poly function that folds over a tuple of Foos:
case class Foo[A](a: A)
object extractFold extends Poly2 {
implicit def default[A, As <: HList]: Case.Aux[Foo[A], Foo[As], Foo[A :: As]] = {
???
}
}
def extract[In, A <: HList, B <: HList](keys: In)
(implicit
gen: Generic.Aux[In, A],
folder: RightFolder.Aux[A, Foo[HNil], extractFold.type, Foo[B]],
tupler: Tupler[B])
: Foo[tupler.Out] = {
???
}
val result = extract((Foo(1), Foo("a")))
The function works at runtime, but the compiler inferred result type is always Foo[Unit] which is not right - in this example it should be Foo[(Int, String)]
scala shapeless
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
I am trying to create a poly function that folds over a tuple of Foos:
case class Foo[A](a: A)
object extractFold extends Poly2 {
implicit def default[A, As <: HList]: Case.Aux[Foo[A], Foo[As], Foo[A :: As]] = {
???
}
}
def extract[In, A <: HList, B <: HList](keys: In)
(implicit
gen: Generic.Aux[In, A],
folder: RightFolder.Aux[A, Foo[HNil], extractFold.type, Foo[B]],
tupler: Tupler[B])
: Foo[tupler.Out] = {
???
}
val result = extract((Foo(1), Foo("a")))
The function works at runtime, but the compiler inferred result type is always Foo[Unit] which is not right - in this example it should be Foo[(Int, String)]
scala shapeless
scala shapeless
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
asked Jan 18 at 23:27
AlexFoxGillAlexFoxGill
5,27013250
5,27013250
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Maybe someone with a better understanding of shapeless can provide you a better answer. According to my understanding the problem lies at the type inference step. If you specify all the types explicitly as in
val result: Foo[(Int, String)] = extract[(Foo[Int], Foo[String]),
Foo[Int] :: Foo[String] :: HNil,
Int :: String :: HNil]((Foo(1), Foo("a")))
the code correctly typechecks. Obviously you don't want to specify those types explicitly though.
According to my understanding the compiler can't infer good B and tupler.Out because they are not coupled tight enough to In and A. One way you can work this around is by introducing an intermediate trait like this:
trait Extractor[L <: HList, HF] {
type FR <: HList
type TR
val folder: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR]]
val tupler: Tupler.Aux[FR, TR]
}
object Extractor {
type Aux[L <: HList, HF, FR0 <: HList, TR0] = Extractor[L, HF] {type FR = FR0; type TR = TR0}
implicit def wrap[L <: HList, In, HF, FR0 <: HList, TR0](implicit folder0: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR0]],
tupler0: Tupler.Aux[FR0, TR0]) = new Extractor[L, HF] {
type FR = FR0
type TR = TR0
override val folder = folder0
override val tupler = tupler0
}
}
and then using it like this:
def extract[In, A <: HList, B <: HList, C](keys: In)
(implicit gen: Generic.Aux[In, A],
extractor: Extractor.Aux[A, extractFold.type, B, C])
: Foo[C] = {
val hli = gen.to(keys)
val fr = extractor.folder(hli, Foo(HNil))
Foo(extractor.tupler(fr.a))
}
This is a hacky solution but at least it seem to work (see also online demo).
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
thank you for the reply. this approach obeys inference if i type it likeval result: Foo[(Int, String)] = extract(...), without the type hint it resolves toFoo[Nothing]
– AlexFoxGill
Jan 21 at 11:17
@AlexFoxGill, I probably miss something but this is not what I see. If I writeval r2 = extract((Foo(1), Foo("a")))and then try to assign it toval r3a:Foo[(Int, String)] = r2, it works, butval r3b:Foo[Nothing] = r2fails. See the modified demo. Could you tell me how you reproduce the issue?
– SergGr
Jan 21 at 13:00
perhaps i missed something, i'll try again and confirm
– AlexFoxGill
Jan 21 at 13:03
add a comment |
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1 Answer
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1 Answer
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Maybe someone with a better understanding of shapeless can provide you a better answer. According to my understanding the problem lies at the type inference step. If you specify all the types explicitly as in
val result: Foo[(Int, String)] = extract[(Foo[Int], Foo[String]),
Foo[Int] :: Foo[String] :: HNil,
Int :: String :: HNil]((Foo(1), Foo("a")))
the code correctly typechecks. Obviously you don't want to specify those types explicitly though.
According to my understanding the compiler can't infer good B and tupler.Out because they are not coupled tight enough to In and A. One way you can work this around is by introducing an intermediate trait like this:
trait Extractor[L <: HList, HF] {
type FR <: HList
type TR
val folder: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR]]
val tupler: Tupler.Aux[FR, TR]
}
object Extractor {
type Aux[L <: HList, HF, FR0 <: HList, TR0] = Extractor[L, HF] {type FR = FR0; type TR = TR0}
implicit def wrap[L <: HList, In, HF, FR0 <: HList, TR0](implicit folder0: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR0]],
tupler0: Tupler.Aux[FR0, TR0]) = new Extractor[L, HF] {
type FR = FR0
type TR = TR0
override val folder = folder0
override val tupler = tupler0
}
}
and then using it like this:
def extract[In, A <: HList, B <: HList, C](keys: In)
(implicit gen: Generic.Aux[In, A],
extractor: Extractor.Aux[A, extractFold.type, B, C])
: Foo[C] = {
val hli = gen.to(keys)
val fr = extractor.folder(hli, Foo(HNil))
Foo(extractor.tupler(fr.a))
}
This is a hacky solution but at least it seem to work (see also online demo).
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
thank you for the reply. this approach obeys inference if i type it likeval result: Foo[(Int, String)] = extract(...), without the type hint it resolves toFoo[Nothing]
– AlexFoxGill
Jan 21 at 11:17
@AlexFoxGill, I probably miss something but this is not what I see. If I writeval r2 = extract((Foo(1), Foo("a")))and then try to assign it toval r3a:Foo[(Int, String)] = r2, it works, butval r3b:Foo[Nothing] = r2fails. See the modified demo. Could you tell me how you reproduce the issue?
– SergGr
Jan 21 at 13:00
perhaps i missed something, i'll try again and confirm
– AlexFoxGill
Jan 21 at 13:03
add a comment |
Maybe someone with a better understanding of shapeless can provide you a better answer. According to my understanding the problem lies at the type inference step. If you specify all the types explicitly as in
val result: Foo[(Int, String)] = extract[(Foo[Int], Foo[String]),
Foo[Int] :: Foo[String] :: HNil,
Int :: String :: HNil]((Foo(1), Foo("a")))
the code correctly typechecks. Obviously you don't want to specify those types explicitly though.
According to my understanding the compiler can't infer good B and tupler.Out because they are not coupled tight enough to In and A. One way you can work this around is by introducing an intermediate trait like this:
trait Extractor[L <: HList, HF] {
type FR <: HList
type TR
val folder: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR]]
val tupler: Tupler.Aux[FR, TR]
}
object Extractor {
type Aux[L <: HList, HF, FR0 <: HList, TR0] = Extractor[L, HF] {type FR = FR0; type TR = TR0}
implicit def wrap[L <: HList, In, HF, FR0 <: HList, TR0](implicit folder0: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR0]],
tupler0: Tupler.Aux[FR0, TR0]) = new Extractor[L, HF] {
type FR = FR0
type TR = TR0
override val folder = folder0
override val tupler = tupler0
}
}
and then using it like this:
def extract[In, A <: HList, B <: HList, C](keys: In)
(implicit gen: Generic.Aux[In, A],
extractor: Extractor.Aux[A, extractFold.type, B, C])
: Foo[C] = {
val hli = gen.to(keys)
val fr = extractor.folder(hli, Foo(HNil))
Foo(extractor.tupler(fr.a))
}
This is a hacky solution but at least it seem to work (see also online demo).
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
thank you for the reply. this approach obeys inference if i type it likeval result: Foo[(Int, String)] = extract(...), without the type hint it resolves toFoo[Nothing]
– AlexFoxGill
Jan 21 at 11:17
@AlexFoxGill, I probably miss something but this is not what I see. If I writeval r2 = extract((Foo(1), Foo("a")))and then try to assign it toval r3a:Foo[(Int, String)] = r2, it works, butval r3b:Foo[Nothing] = r2fails. See the modified demo. Could you tell me how you reproduce the issue?
– SergGr
Jan 21 at 13:00
perhaps i missed something, i'll try again and confirm
– AlexFoxGill
Jan 21 at 13:03
add a comment |
Maybe someone with a better understanding of shapeless can provide you a better answer. According to my understanding the problem lies at the type inference step. If you specify all the types explicitly as in
val result: Foo[(Int, String)] = extract[(Foo[Int], Foo[String]),
Foo[Int] :: Foo[String] :: HNil,
Int :: String :: HNil]((Foo(1), Foo("a")))
the code correctly typechecks. Obviously you don't want to specify those types explicitly though.
According to my understanding the compiler can't infer good B and tupler.Out because they are not coupled tight enough to In and A. One way you can work this around is by introducing an intermediate trait like this:
trait Extractor[L <: HList, HF] {
type FR <: HList
type TR
val folder: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR]]
val tupler: Tupler.Aux[FR, TR]
}
object Extractor {
type Aux[L <: HList, HF, FR0 <: HList, TR0] = Extractor[L, HF] {type FR = FR0; type TR = TR0}
implicit def wrap[L <: HList, In, HF, FR0 <: HList, TR0](implicit folder0: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR0]],
tupler0: Tupler.Aux[FR0, TR0]) = new Extractor[L, HF] {
type FR = FR0
type TR = TR0
override val folder = folder0
override val tupler = tupler0
}
}
and then using it like this:
def extract[In, A <: HList, B <: HList, C](keys: In)
(implicit gen: Generic.Aux[In, A],
extractor: Extractor.Aux[A, extractFold.type, B, C])
: Foo[C] = {
val hli = gen.to(keys)
val fr = extractor.folder(hli, Foo(HNil))
Foo(extractor.tupler(fr.a))
}
This is a hacky solution but at least it seem to work (see also online demo).
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
Maybe someone with a better understanding of shapeless can provide you a better answer. According to my understanding the problem lies at the type inference step. If you specify all the types explicitly as in
val result: Foo[(Int, String)] = extract[(Foo[Int], Foo[String]),
Foo[Int] :: Foo[String] :: HNil,
Int :: String :: HNil]((Foo(1), Foo("a")))
the code correctly typechecks. Obviously you don't want to specify those types explicitly though.
According to my understanding the compiler can't infer good B and tupler.Out because they are not coupled tight enough to In and A. One way you can work this around is by introducing an intermediate trait like this:
trait Extractor[L <: HList, HF] {
type FR <: HList
type TR
val folder: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR]]
val tupler: Tupler.Aux[FR, TR]
}
object Extractor {
type Aux[L <: HList, HF, FR0 <: HList, TR0] = Extractor[L, HF] {type FR = FR0; type TR = TR0}
implicit def wrap[L <: HList, In, HF, FR0 <: HList, TR0](implicit folder0: RightFolder.Aux[L, Foo[HNil], HF, Foo[FR0]],
tupler0: Tupler.Aux[FR0, TR0]) = new Extractor[L, HF] {
type FR = FR0
type TR = TR0
override val folder = folder0
override val tupler = tupler0
}
}
and then using it like this:
def extract[In, A <: HList, B <: HList, C](keys: In)
(implicit gen: Generic.Aux[In, A],
extractor: Extractor.Aux[A, extractFold.type, B, C])
: Foo[C] = {
val hli = gen.to(keys)
val fr = extractor.folder(hli, Foo(HNil))
Foo(extractor.tupler(fr.a))
}
This is a hacky solution but at least it seem to work (see also online demo).
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
edited Jan 19 at 2:28
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
answered Jan 19 at 2:04
SergGrSergGr
20.9k22243
20.9k22243
thank you for the reply. this approach obeys inference if i type it likeval result: Foo[(Int, String)] = extract(...), without the type hint it resolves toFoo[Nothing]
– AlexFoxGill
Jan 21 at 11:17
@AlexFoxGill, I probably miss something but this is not what I see. If I writeval r2 = extract((Foo(1), Foo("a")))and then try to assign it toval r3a:Foo[(Int, String)] = r2, it works, butval r3b:Foo[Nothing] = r2fails. See the modified demo. Could you tell me how you reproduce the issue?
– SergGr
Jan 21 at 13:00
perhaps i missed something, i'll try again and confirm
– AlexFoxGill
Jan 21 at 13:03
add a comment |
thank you for the reply. this approach obeys inference if i type it likeval result: Foo[(Int, String)] = extract(...), without the type hint it resolves toFoo[Nothing]
– AlexFoxGill
Jan 21 at 11:17
@AlexFoxGill, I probably miss something but this is not what I see. If I writeval r2 = extract((Foo(1), Foo("a")))and then try to assign it toval r3a:Foo[(Int, String)] = r2, it works, butval r3b:Foo[Nothing] = r2fails. See the modified demo. Could you tell me how you reproduce the issue?
– SergGr
Jan 21 at 13:00
perhaps i missed something, i'll try again and confirm
– AlexFoxGill
Jan 21 at 13:03
thank you for the reply. this approach obeys inference if i type it like
val result: Foo[(Int, String)] = extract(...), without the type hint it resolves to Foo[Nothing]– AlexFoxGill
Jan 21 at 11:17
thank you for the reply. this approach obeys inference if i type it like
val result: Foo[(Int, String)] = extract(...), without the type hint it resolves to Foo[Nothing]– AlexFoxGill
Jan 21 at 11:17
@AlexFoxGill, I probably miss something but this is not what I see. If I write
val r2 = extract((Foo(1), Foo("a"))) and then try to assign it to val r3a:Foo[(Int, String)] = r2, it works, but val r3b:Foo[Nothing] = r2 fails. See the modified demo. Could you tell me how you reproduce the issue?– SergGr
Jan 21 at 13:00
@AlexFoxGill, I probably miss something but this is not what I see. If I write
val r2 = extract((Foo(1), Foo("a"))) and then try to assign it to val r3a:Foo[(Int, String)] = r2, it works, but val r3b:Foo[Nothing] = r2 fails. See the modified demo. Could you tell me how you reproduce the issue?– SergGr
Jan 21 at 13:00
perhaps i missed something, i'll try again and confirm
– AlexFoxGill
Jan 21 at 13:03
perhaps i missed something, i'll try again and confirm
– AlexFoxGill
Jan 21 at 13:03
add a comment |
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