Question

继Witness that an abstract type implements a typeclass之后我试图在下面的代码片段中并行比较这两种方法：

// We want both ParamaterizedTC and WithAbstractTC (below) to check that 
// their B parameter implements AddQuotes 
abstract class AddQuotes[A] {
  def inQuotes(self: A): String = s"${self.toString}"  
}
implicit val intAddQuotes = new AddQuotes[Int] {}

abstract class ParamaterizedTC[A, _B](implicit ev: AddQuotes[_B]) {
  type B = _B
  def getB(self: A): B 
  def add1ToB(self: A): String = ev.inQuotes(getB(self)) // TC witness does not need to be at method level
}

abstract class WithAbstractTC[A] private { 
  // at this point the compiler has not established that type B implements AddQuotes, even if we have created
  // this instance via the apply[A, _B] constructor below...
  type B 
  def getB(self: A): B
  def add1ToB(self: A)(implicit ev: AddQuotes[B]): String = 
    ev.inQuotes(getB(self)) // ... so here the typeclass witness has to occur on the method level
}
object WithAbstractTC {
  // This constructor checks that B implements AddQuotes
  def apply[A, _B: AddQuotes](getB: A => _B): WithAbstractTC[A] = new WithAbstractTC[A] { 
    type B = _B 
    def getB(self: A): B = getB(self)
  }
  // But we could also have a constructor that does not check, so the compiler can never be certain that 
  // for a given instance of WithAbstractTC, type B implements AddQuotes
  def otherConstructor[A, _B](getB: A => _B): WithAbstractTC[A] { type B = _B } = new WithAbstractTC[A] { 
    type B = _B 
    def getB(self: A): B = getB(self)
  }
}

case class Container[B: AddQuotes]( get: B )

// These are both fine
implicit def containerIsParamaterized[B: AddQuotes]: ParamaterizedTC[Container[B], B] = 
  new ParamaterizedTC[Container[B], B] { def getB(self: Container[B]): B = self.get }
implicit def containerIsWithAbstract[_B: AddQuotes]: WithAbstractTC[Container[_B]] = 
  WithAbstractTC[Container[_B], _B](self => self.get)

val contIsParamaterized: ParamaterizedTC[Container[Int], Int] = 
  implicitly[ParamaterizedTC[Container[Int], Int]]
val contIsWithAbstract: WithAbstractTC[Container[Int]] = 
  implicitly[WithAbstractTC[Container[Int]]]

implicitly[AddQuotes[contIsParamaterized.B]]
implicitly[AddQuotes[contIsWithAbstract.B]] // This is not fine

我的结论（如果我错了，请纠正我）是，如果类型类见证人存在于公共构造函数中（如下面的ParamaterizedTC所示），则编译器始终可以确定B实现了AddQuotes。而如果将这个见证人放在类型类同伴对象的构造函数中（例如WithAbstractTC），那么它就不能。与基于抽象类型的方法相比，这在某种程度上改变了基于类型参数的方法的使用。

Answer 1

差别在于：在ParametrizedTC中，该类的作用域是隐式的，而在WithAbstractTC中则没有。但是当您有抽象类型时，没有什么可以阻止您添加它：

abstract class WithAbstractTC2[A] private { 
  type B 
  implicit val ev: AddQuotes[B]
  def getB(self: A): B
  def add1ToB(self: A): String = 
    ev.inQuotes(getB(self))
}

def apply[A, _B](getB: A => _B)(implicit _ev: AddQuotes[_B]): WithAbstractTC2[A] = new WithAbstractTC2[A] { 
  type B = _B
  implicit val ev: AddQuotes[B] = _ev
  def getB(self: A): B = getB(self)
}

不幸的是，类似的东西

def apply[A, _B: AddQuotes](getB: A => _B): WithAbstractTC2[A] = new WithAbstractTC2[A] { 
  type B = _B
  implicit val ev: AddQuotes[B] = implicitly[AddQuotes[_B]]
  def getB(self: A): B = getB(self)
}

因为它将在最接近的范围中选择隐式：它正在尝试定义的隐式。

Answer 2

implicitly[AddQuotes[contIsWithAbstract.B]]拒绝编译与单个/多个构造函数/ apply方法或类型参数/类型成员差异无关。你只是失去类型的改进随处可见。编译器无法检查您是否丢失了类型优化。您有权放弃类型的优化而放弃类型。

如果您恢复类型细化，则代码会编译

object WithAbstractTC {
  def apply[A, _B: AddQuotes](getB: A => _B): WithAbstractTC[A] {type B = _B} = 
//                                                              ^^^^^^^^^^^^^
    new WithAbstractTC[A] {
      type B = _B
      def getB(self: A): B = getB(self)
    }
  ...
}

implicit def containerIsWithAbstract[_B: AddQuotes]: 
  WithAbstractTC[Container[_B]] { type B = _B } =
//                              ^^^^^^^^^^^^^^^
  WithAbstractTC[Container[_B], _B](self => self.get)

val contIsWithAbstract: WithAbstractTC[Container[Int]] { type B = Int } =
//                                                     ^^^^^^^^^^^^^^^^
  shapeless.the[WithAbstractTC[Container[Int]]]
//^^^^^^^^^^^^^

implicitly[AddQuotes[contIsWithAbstract.B]] // compiles

请注意，implicitly失去了类型限制，shapeless.the是安全版本。

当implicitly不够具体时 https://typelevel.org/blog/2014/01/18/implicitly_existential.html

如何通过抽象隐式对类成员类型类使用类级别的隐式约束，请参见 @AlexeyRomanov 的答案。

使用类型参数与抽象类型实现类型类

2 个答案: