使用 apply 方法的泛型类型的 Scala 工厂？

假设我有以下特征，它定义了一个接口并采用几个类型参数......

trait Foo[A, B] {

    // implementation details not important

}

我想使用伴随对象作为该特征的具体实现的工厂。我还想强制用户使用Foo接口而不是子类所以我隐藏了伴随对象中的具体实现，如下所示：

object Foo {

  def apply[A, B](thing: Thing): Foo[A, B] = {
    ???
  }

  private case class FooImpl[A1, B1](thing: Thing) extends Foo[A1, B1]

  private case class AnotherFooImpl[A2, B1](thing: Thing) extends Foo[A2, B1]

}

我希望能够按如下方式使用工厂：

val foo = Foo[A1, B1](thing)  // should be an instance of FooImpl

val anotherFoo = Foo[A2, B1](thing)  // should be an instance of AnotherFooImpl

我如何实施apply方法来实现这一点？这SO post https://stackoverflow.com/questions/30926958/scala-factory-method-with-generics似乎很接近目标。

怎么样：

trait Foo[A, B]
trait Factory[A, B] {
  def make(thing: Thing): Foo[A, B]
}

class Thing

object Foo {
def apply[A, B](thing: Thing)(implicit ev: Factory[A, B]) = ev.make(thing)

private case class FooImpl[A, B](thing: Thing) extends Foo[A, B]
private case class AnotherFooImpl[A, B](thing: Thing) extends Foo[A, B]

implicit val fooImplFactory: Factory[Int, String] = new Factory[Int, String] {
  override def make(thing: Thing): Foo[Int, String] = new FooImpl[Int, String](thing)
}

implicit val anotherFooImplFactory: Factory[String, String] = new Factory[String, String] {
  override def make(thing: Thing): Foo[String, String] = new AnotherFooImpl[String, String](thing)
}

And now:

def main(args: Array[String]): Unit = {
  import Foo._

  val fooImpl = Foo[Int, String](new Thing)
  val anotherFooImpl = Foo[String, String](new Thing)

  println(fooImpl)
  println(anotherFooImpl)
}

Yields:

FooImpl(testing.X$Thing@4678c730)
AnotherFooImpl(testing.X$Thing@c038203)