Question

以下编译的代码失败，但如果我删除方法specialized中的dot注释，则会通过。

abstract class Op[@specialized Left, @specialized Right] {
  @specialized
  type Result

  def r: Numeric[Result]
  def times(left: Left, right: Right): Result
}


object Op {

  implicit object IntDoubleOp extends Op[Int, Double] {
    type Result = Double
    val r = implicitly[Numeric[Double]]
    def times(left: Int, right: Double): Double = left * right
  }
}


object calc {

  def dot[@specialized Left, @specialized Right](xs: Array[Left], ys: Array[Right])
          (implicit op: Op[Left, Right]): op.Result = {
    var total = op.r.zero
    var index = xs.length
    while(index > 0) {
      index -= 1
      total = op.r.plus(total, op.times(xs(index), ys(index)))
    }
    total
  }
}

test.scala:31: error: type mismatch;
 found   : op.Result
 required: op.Result
    total
    ^
one error found

这是另一个没有运气的尝试：

//cat Ops.scala 
import scala.{ specialized => sp }

trait OpsResult {
  type N
}

trait SymOps extends OpsResult {
  @sp override type N
  def zero: N
  def plus(left: N, right: N): N
}

trait AsyOps[@sp L, @sp R] extends OpsResult {
  @sp override type N
  def times(left: L, right: R): N
}

trait MixOps[@sp L, @sp R] extends AsyOps[L, R] with SymOps

object MixOps {
  trait DoubleOps extends SymOps {
    override type N = Double
    def zero: Double = 0.0
    override def plus(left: Double, right: Double): Double = left + right
  }
  trait IntDoubleOps extends AsyOps[Int, Double] {
    override type N = Double
    override def times(left: Int, right: Double): Double = left * right
  }

  implicit object MixIntDouble extends IntDoubleOps with DoubleOps
}

object Test {
  def dot[@sp L, @sp R](xs: Array[L], ys: Array[R])
         (implicit op: MixOps[L, R]): op.N = {
    op.zero
  }
}

$ scalac Ops.scala 
Ops.scala:36: error: type mismatch;
 found   : op.N
 required: op.N
    op.zero
       ^
one error found

Answer 1

这是一个错误（也在2.11.x上重现）。我已经联系过LightBend的人，这对于专业化和路径依赖类型的代码生成来说肯定是个怪癖。我把它缩小到了一个很小的重现：

trait M[@specialized T] {
  type Res
  def res: Res
}

object Test {
  def m[@specialized T](op: M[T]): op.Res = op.res
}

op的符号将不同步，不会像您在代码中看到的那样进行编译。 @AdriaanMoors has opened a bug关于这个问题。

Adriaan还向我指出了Aleksandar Prokopec的this blog post，它指出了@specilization代码生成中的一些怪癖。

这是scala的专业错误吗？

1 个答案: