如何映射扩展共同特征的对象列表?
How to map over Hlist of objects that extend common trait?
我有一个特征和扩展它的对象。
trait Common[K] {
def name: String
def encode(k: K): String = name + k.toString
}
object A extends Common[Int] {
override def name: String = "a"
}
object B extends Common[Int] {
override def name: String = "b"
}
object C extends Common[Int] {
override def name: String = "c"
}
我想创建这个对象的 hlist 并在其上映射:
val hl = A :: B :: C :: HNil
val result: List[Int => String] = hl.map(EncodePoly).toList
以及实现 Poly 函数的不同尝试:
object EncodePoly extends Poly1 {
implicit def indCase[K]: Case.Aux[Common[K], K => String] = at[Common[K]] {
common => k =>
common.encode(k)
}
}
object EncodePoly extends Poly1 {
implicit def indCase[K, C <: Common[K]]: Case.Aux[C, K => String] = at[C] {
common => k =>
common.encode(k)
}
}
编译器告诉我:
Error:(45, 43) could not find implicit value for parameter mapper:
shapeless.ops.hlist.Mapper[com.test.EncodePoly.type,com.test.A.type ::
com.test.B.type :: com.test.C.type :: shapeless.HNil] val result:
List[Int => String] = hl.map(EncodePoly).toList
我也尝试过使用依赖类型作为共同特征而不是类型参数。似乎没有任何效果。我应该如何使用 hlist of objects?
你的第二个 EncodePoly
很接近,但是编译器不够聪明,无法推断 K
应该是 Int
然后 C
应该是单例类型.您可以通过使用 <:<
而不是 <:
:
对子类型关系进行编码来帮助进行类型推断
trait Common[K] {
def name: String
def encode(k: K): String = name + k.toString
}
object A extends Common[Int] {
override def name: String = "a"
}
object B extends Common[Int] {
override def name: String = "b"
}
object C extends Common[Int] {
override def name: String = "c"
}
import shapeless.{ ::, HNil, Poly1 }
object EncodePoly extends Poly1 {
implicit def indCase[K, C](implicit ev: C <:< Common[K]): Case.Aux[C, K => String] = at[C] {
common => k => common.encode(k)
}
}
然后:
scala> val hl = A :: B :: C :: HNil
hl: A.type :: B.type :: C.type :: shapeless.HNil = A$@3f044518 :: B$@282b7aad :: C$@7c130749 :: HNil
scala> val result: List[Int => String] = hl.map(EncodePoly).toList
result: List[Int => String] = List(EncodePoly$$$Lambda55/1493211716@7c987ea3, EncodePoly$$$Lambda55/1493211716@10be689, EncodePoly$$$Lambda55/1493211716@5dd3c2f2)
如果您可以在 EncodePoly
定义中将 K
修改为 Int
,那也行。
我有一个特征和扩展它的对象。
trait Common[K] {
def name: String
def encode(k: K): String = name + k.toString
}
object A extends Common[Int] {
override def name: String = "a"
}
object B extends Common[Int] {
override def name: String = "b"
}
object C extends Common[Int] {
override def name: String = "c"
}
我想创建这个对象的 hlist 并在其上映射:
val hl = A :: B :: C :: HNil
val result: List[Int => String] = hl.map(EncodePoly).toList
以及实现 Poly 函数的不同尝试:
object EncodePoly extends Poly1 {
implicit def indCase[K]: Case.Aux[Common[K], K => String] = at[Common[K]] {
common => k =>
common.encode(k)
}
}
object EncodePoly extends Poly1 {
implicit def indCase[K, C <: Common[K]]: Case.Aux[C, K => String] = at[C] {
common => k =>
common.encode(k)
}
}
编译器告诉我:
Error:(45, 43) could not find implicit value for parameter mapper: shapeless.ops.hlist.Mapper[com.test.EncodePoly.type,com.test.A.type :: com.test.B.type :: com.test.C.type :: shapeless.HNil] val result: List[Int => String] = hl.map(EncodePoly).toList
我也尝试过使用依赖类型作为共同特征而不是类型参数。似乎没有任何效果。我应该如何使用 hlist of objects?
你的第二个 EncodePoly
很接近,但是编译器不够聪明,无法推断 K
应该是 Int
然后 C
应该是单例类型.您可以通过使用 <:<
而不是 <:
:
trait Common[K] {
def name: String
def encode(k: K): String = name + k.toString
}
object A extends Common[Int] {
override def name: String = "a"
}
object B extends Common[Int] {
override def name: String = "b"
}
object C extends Common[Int] {
override def name: String = "c"
}
import shapeless.{ ::, HNil, Poly1 }
object EncodePoly extends Poly1 {
implicit def indCase[K, C](implicit ev: C <:< Common[K]): Case.Aux[C, K => String] = at[C] {
common => k => common.encode(k)
}
}
然后:
scala> val hl = A :: B :: C :: HNil
hl: A.type :: B.type :: C.type :: shapeless.HNil = A$@3f044518 :: B$@282b7aad :: C$@7c130749 :: HNil
scala> val result: List[Int => String] = hl.map(EncodePoly).toList
result: List[Int => String] = List(EncodePoly$$$Lambda55/1493211716@7c987ea3, EncodePoly$$$Lambda55/1493211716@10be689, EncodePoly$$$Lambda55/1493211716@5dd3c2f2)
如果您可以在 EncodePoly
定义中将 K
修改为 Int
,那也行。