如何使用 HList 现有成员创建类型类
How to create typeclass with HList existential member
我正在尝试创建模仿无形类型类的自定义类型类。它看起来像这样:
trait View[Record] {
type Result <: HList
def apply(r: Record): Result
}
object View extends LowPriorityLiftFunction1{
type Aux[Record, L <: HList] = View[Record] {type Result = L}
implicit def atView[Record: View] = at[Record](implicitly[View[Record]].apply)
}
假设我提供这样的功能:
object toHView extends ->( (_:Int) + 1)
implicit def provideView[Record, L <: HList]
(implicit generic: Generic.Aux[Record, L],
mapper: Mapper[toHView.type, L])
: View.Aux[Record, mapper.Out] =
new View[Record] {
type Result = mapper.Out
def apply(r: Record) = mapper(generic.to(r))
}
所以如果我们定义:
case class Viewable(x: Int, y: Int, z : Int)
case class NotViewable(x: Int, y: Long, z : Int)
然后
val view = View(Viewable(1, 2, 3)) // is 2 :: 3 :: 4 :: HNil
val noView = View(NotViewable(1, 2, 3)) // is HNil
如果我尝试获取这里的麻烦
view.head
我有
Error:could not find implicit value for parameter
c: IsHCons[View[Viewable]#Result]
我如何定义这个类型类以便以后有效地使用它的所有类型成员?
当然我可以去掉类型成员:
trait View[Record, Result <: HList] {
def apply(r: Record): Result
}
object View extends LowPriorityLiftFunction1{
implicit def atView[Record, Result]
(implicit view: View[Record, Result]) = at[Record](view.apply)
}
object toHView extends ->((_: Int) + 1)
implicit def provideView[Record, L <: HList]
(implicit generic: Generic.Aux[Record, L],
mapper: Mapper[toHView.type, L])
: View[Record, mapper.Out] =
new View[Record, mapper.Out] {
type Result = mapper.Out
def apply(r: Record) = mapper(generic.to(r))
}
但从这一点开始
val view = View(Viewable(1, 2, 3))
我遇到“不明确的隐式值”问题
好的,这是:更改
implicit def atView[Record: View] = at[Record](implicitly[View[Record]].apply)
到
implicit def atView[Record](implicit v: View[Record]) = at[Record](v.apply(_))
原因是 implicitly
在处理精化类型成员时会失去精度,因此您的 HList
(在本例中为 Int :: Int :: Int :: HNil
)不是预期的精化类型,而是编译器吐出一个相当无用的 View#Result
.
使用隐式参数而不是上下文绑定似乎可以保留改进后的类型。
此外,shapeless' the
是 implicitly
的替代方案,它保留了类型改进,although it doesn't seem to work in this case。
这是 implicitly
丢失精度的示例,取自 the
implementation in shapeless:
scala> trait Foo { type T ; val t: T }
defined trait Foo
scala> implicit val intFoo: Foo { type T = Int } = new Foo { type T = Int ; val t = 23 }
intFoo: Foo{type T = Int} = $anon$1@6067b682
scala> implicitly[Foo].t // implicitly loses precision
res0: Foo#T = 23
scala> implicitly[Foo].t+13
<console>:13: error: type mismatch;
found : Int(13)
required: String
implicitly[Foo].t+13
^
scala> the[Foo].t // the retains it
res1: Int = 23
scala> the[Foo].t+13
res2: Int = 36
我正在尝试创建模仿无形类型类的自定义类型类。它看起来像这样:
trait View[Record] {
type Result <: HList
def apply(r: Record): Result
}
object View extends LowPriorityLiftFunction1{
type Aux[Record, L <: HList] = View[Record] {type Result = L}
implicit def atView[Record: View] = at[Record](implicitly[View[Record]].apply)
}
假设我提供这样的功能:
object toHView extends ->( (_:Int) + 1)
implicit def provideView[Record, L <: HList]
(implicit generic: Generic.Aux[Record, L],
mapper: Mapper[toHView.type, L])
: View.Aux[Record, mapper.Out] =
new View[Record] {
type Result = mapper.Out
def apply(r: Record) = mapper(generic.to(r))
}
所以如果我们定义:
case class Viewable(x: Int, y: Int, z : Int)
case class NotViewable(x: Int, y: Long, z : Int)
然后
val view = View(Viewable(1, 2, 3)) // is 2 :: 3 :: 4 :: HNil
val noView = View(NotViewable(1, 2, 3)) // is HNil
如果我尝试获取这里的麻烦
view.head
我有
Error:could not find implicit value for parameter
c: IsHCons[View[Viewable]#Result]
我如何定义这个类型类以便以后有效地使用它的所有类型成员?
当然我可以去掉类型成员:
trait View[Record, Result <: HList] {
def apply(r: Record): Result
}
object View extends LowPriorityLiftFunction1{
implicit def atView[Record, Result]
(implicit view: View[Record, Result]) = at[Record](view.apply)
}
object toHView extends ->((_: Int) + 1)
implicit def provideView[Record, L <: HList]
(implicit generic: Generic.Aux[Record, L],
mapper: Mapper[toHView.type, L])
: View[Record, mapper.Out] =
new View[Record, mapper.Out] {
type Result = mapper.Out
def apply(r: Record) = mapper(generic.to(r))
}
但从这一点开始
val view = View(Viewable(1, 2, 3))
我遇到“不明确的隐式值”问题
好的,这是:更改
implicit def atView[Record: View] = at[Record](implicitly[View[Record]].apply)
到
implicit def atView[Record](implicit v: View[Record]) = at[Record](v.apply(_))
原因是 implicitly
在处理精化类型成员时会失去精度,因此您的 HList
(在本例中为 Int :: Int :: Int :: HNil
)不是预期的精化类型,而是编译器吐出一个相当无用的 View#Result
.
使用隐式参数而不是上下文绑定似乎可以保留改进后的类型。
此外,shapeless' the
是 implicitly
的替代方案,它保留了类型改进,although it doesn't seem to work in this case。
这是 implicitly
丢失精度的示例,取自 the
implementation in shapeless:
scala> trait Foo { type T ; val t: T }
defined trait Foo
scala> implicit val intFoo: Foo { type T = Int } = new Foo { type T = Int ; val t = 23 }
intFoo: Foo{type T = Int} = $anon$1@6067b682
scala> implicitly[Foo].t // implicitly loses precision
res0: Foo#T = 23
scala> implicitly[Foo].t+13
<console>:13: error: type mismatch;
found : Int(13)
required: String
implicitly[Foo].t+13
^
scala> the[Foo].t // the retains it
res1: Int = 23
scala> the[Foo].t+13
res2: Int = 36