首先,我不知道如何正确标记我的问题。这也可能是我找不到有用资源的原因。任何提示都受到高度赞赏。
trait Context[T]
{
self =>
trait Rule
{
def apply( value: T ): Boolean
}
implicit class RichRule[A <: Rule]( a: A )
{
def and[B <: Rule]( b: B ): and[A, B] = self.and( a, b )
def or[B <: Rule]( b: B ): or[A, B] = self.or( a, b )
}
sealed trait Group[A <: Rule, B <: Rule] extends Rule
{
def a: A
def b: B
override def apply( value: T ) = ???
}
case class and[A <: Rule, B <: Rule]( a: A, b: B ) extends Group[A, B]
case class or[A <: Rule, B <: Rule]( a: A, b: B ) extends Group[A, B]
}
鉴于上述代码,我现在可以用这种方式定义和链接Rules
:
new Context[String]
{
class MyRule extends Rule
{
override def apply( value: String ) = true
}
case class A() extends MyRule
case class B() extends MyRule
val x1: A and B or A = A() and B() or A()
}
这按照我的意图运作,但现在是棘手的部分。我想介绍一个类型类Combination
,它解释了如何加入两个规则。
trait Combination[-A <: Rule, -B <: Rule]
{
type Result <: Rule
def combine( a: A, b: B ): Result
}
trait AndCombination[-A <: Rule, -B <: Rule] extends Combination[A, B]
trait OrCombination[-A <: Rule, -B <: Rule] extends Combination[A, B]
此类型类现在应该与运算符一起传递。
implicit class RichRule[A <: Rule]( a: A )
{
def and[B <: Rule]( b: B )( implicit c: AndCombination[A, B] ): and[A, B] = ???
def or[B <: Rule]( b: B )( implicit c: OrCombination[A, B] ): or[A, B] = self.or( a, b )
}
经过一些调整后仍然有效。
implicit val c1 = new Combination[MyRule, MyRule]
{
type Result = MyRule
def combine( a: A, b: B ): MyRule = a
}
val x: A and B = A() and B()
但如果它变得更复杂,事情就会崩溃。
A() and B() and A()
将引发隐式遗漏错误:Combination[and[A, B], A]
丢失。但我希望它使用and[A, B]
(type Result = MyRule
)的隐式组合的结果,它已经知道如何处理(Combination[and[A, B]#Result, A]
)。
对我来说,保留组合规则val x: A and B or A
的类型信息非常重要,将它们折叠到最终结果类型很容易,但不是我想要的。
尽管我尽可能接近,但它编译失败了。
trait Context[T]
{
self =>
trait Rule
trait Value extends Rule
trait Group[A <: Rule, B <: Rule] extends Rule
{
def a: A
def b: B
implicit val resolver: Resolver[_ <: Group[A, B]]
}
case class and[A <: Rule, B <: Rule]( a: A, b: B )( implicit val resolver: Resolver[and[A, B]] ) extends Group[A, B]
implicit class RichRule[A <: Rule]( a: A )
{
def and[B <: Rule]( b: B )( implicit resolver: Resolver[and[A, B]] ) = self.and[A, B]( a, b )
}
trait Resolver[-A <: Rule]
{
type R <: Value
def resolve( a: A ): R
}
}
object O extends Context[String]
{
implicit val c1 = new Resolver[A and A]
{
override type R = A
override def resolve( a: O.and[A, A] ) = ???
}
implicit def c2[A <: Value, B <: Value, C <: Value]( implicit r1: Resolver[A and B] ) = new Resolver[A and B and C]
{
override type R = C
override def resolve( a: A and B and C ): C =
{
val x: r1.R = r1.resolve( a.a )
new c2( x )
???
}
}
class c2[A <: Value, B <: Value]( val a: A )( implicit r2: Resolver[A and B] ) extends Resolver[A and B]
{
override type R = B
override def resolve( a: O.and[A, B] ) = a.b
}
case class A() extends Value
val x: A and A and A = A() and A() and A()
}
答案 0 :(得分:9)
您的代码无法编译的原因是指令
new c2( x )
编译器需要从implicit r2: Resolver[A and B]
解析x
,并且唯一可用的类型信息是x
的类型,即r1.R
。
这类问题需要为编译器提供更多类型信息并添加一些隐式参数。如果您需要Resolver[A and B]
,则无法使用其R
类型来解析另一个Resolver[r1.R and C]
。
type ResolverAux[-A<:Rule,B] = Resolver[A] { type R = B }
如果可以,您可以重写c2的签名
implicit def c2[A <: Value, B <: Value, C <: Value,D<:Value]( implicit r1: ResolverAux[A and B,D], r2:Resolver[D and C] ):Resolver[A and B and C] = new Resolver[A and B and C]
{
override type R = C
override def resolve( a: A and B and C ): C =
{
val x: r1.R = r1.resolve( a.a )
new c2[r1.R,C]( x )
???
}
}
请注意,通过使用类型别名并引入其他通用参数,我可以表达关系r1.R1 = D
,然后将其用于解析第二个隐式r2