使用类型参数与抽象类型实现类型类

Question

继续Witness that an abstract type implements a typeclass 我尝试在下面的代码 sn-p 中并排比较这两种方法：

// 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，如果类型类见证存在于公共构造函数中（如下面的ParamaterizedTC），那么编译器总是可以确定B 实现了AddQuotes。然而，如果这个见证被放在类型类伴随对象的构造函数中（比如WithAbstractTC），那么它就不能。这在一定程度上改变了基于类型参数的方法与基于抽象类型的方法的用法。

Answer 1

不同之处在于：在ParametrizedTC 中，您拥有类的隐含范围，而在WithAbstractTC 中，您没有。但是当你有一个抽象类型时，没有什么能阻止你添加它：

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，不起作用的是

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)
}

因为它会选择最近范围内的隐含：它试图定义的那个。

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。您只是到处丢失了类型改进。编译器无法检查您是否丢失了类型优化。您有权向上转换类型以放弃其改进。

如果你恢复类型细化代码编译

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 失去了类型改进，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型成员类型类使用类级隐式约束，请参阅@AlexeyRomanov的答案。