scala - 评估抽象类型成员

Question

为什么这段代码无法编译？在 Scala 2.13.6 上测试。有没有办法让 scalac 知道 SizeOfType[Unit]#Size 实际上是 Short？

trait SizeOfType[T] {
  type Size
  def getSize(): Size
}

object SizeOfType {
  implicit def unit: SizeOfType[Unit] = new SizeOfType[Unit] {
    type Size = Short
    def getSize(): Short = 0
  }

  def test(implicit ev: SizeOfType[Unit]): Short = ev.getSize()
}

[error] type mismatch;
[error]  found   : ev.Size
[error]  required: Short
[error]   def test(implicit ev: SizeOfType[Unit]): Short = ev.getSize()

Answer 1

当使用类型成员而不是类型参数进行参数化时，您必须在请求类型类实例时提供类型细化，否则您只是在向 Scala 请求

SizeOfType[Unit]

而不是预期的

SizeOfType[Unit] { type Size = Short }

类似的东西

implicit val unit: SizeOfType[Unit] { type Size = Short } =
  new SizeOfType[Unit] {
    type Size = Short
    def getSize(): Short = 0
  }

def test(implicit ev: SizeOfType[Unit] { type Size = Short }): Short = ev.getSize()
val x: Short = test // ok

通常你会创建依赖类型的方法，然后你可以避免在方法定义中进行类型细化(但类型类实例的定义仍然需要它)

implicit val unit: SizeOfType[Unit] { type Size = Short } = ???
def test(implicit ev: SizeOfType[Unit]): ev.Size = ev.getSize()
val x: Short = test // ok

我建议学习 underscoreio/shapeless-guide 的“第 4 章，使用类型和隐式” ，特别是提到的部分

...If we define the return type as Second[L], the Out type member will be erased from the return type and the type class will not work correctly.