在Scala中输入Nothing的类型

Question

当我尝试编译小例子时:

trait Foo[A,B] {
  type F[_,_]
  def foo(): F[A,B]
}

class Bar[A,B] extends Foo[A,B] {
  type F[D,E] = Bar[D,E]
  def foo() = this
}

object Helper {
  def callFoo[A,B,FF <: Foo[A,B]]( f: FF ): FF#F[A,B] =
    f.foo()
}

object Run extends App {
  val x = new Bar[Int,Double]
  val y = Helper.callFoo(x)
  println( y.getClass )
}

我收到错误:

[error] src/Issue.scala:20: inferred type arguments
[Nothing,Nothing,issue.Bar[Int,Double]] do not conform to method callFoo's type
parameter bounds [A,B,FF <: issue.Foo[A,B]]
[error]       val y = Helper.callFoo(x)

显然,类型推断机制无法推断Bar [A,B]中的A和B. 但是,如果我手动传递所有类型,它会起作用:

val y = Helper.callFoo[Int,Double,Bar[Int,Double]](x)

我有办法避免明确传递类型吗？

Answer 1

您必须将签名更改为callFoo:

def callFoo[A, B, FF[A, B] <: Foo[A, B]](f: FF[A, B]): FF[A, B]#F[A, B] =

您必须告诉编译器FF实际上是参数化类型.