Scala 模式匹配推断“Any"而不是存在类型，打破了类型安全? [英] Scala pattern match infers `Any` instead of an existential type, breaks type safety?

问题描述

我遇到了一个令人费解的案例类类型推断问题.这是一个最小的例子:

I ran into a puzzling type inference problem with case classes. Here's a minimal example:

trait T[X]
case class Thing[A, B, X](a: A, f: A => B) extends T[X]

def hmm[X](t: T[X]) = t match {
  case Thing(a, f) => f("this really shouldn't typecheck")
}

Scala 决定 a: Any 和 f: Any =>任何，但这是不合适的；他们真的应该有类型 a: SomeTypeA 和 f: SomeTypeA =>SomeTypeB，其中 SomeTypeA 和 SomeTypeB 是未知类型.

Scala decides that a: Any and f: Any => Any, but that's inappropriate; they really ought to have types a: SomeTypeA and f: SomeTypeA => SomeTypeB, where SomeTypeA and SomeTypeB are unknown types.

另一种说法是，我认为假设的 Thing.unapply 方法应该看起来像

Another way of saying this is that I think the hypothetical Thing.unapply method should look something like

def unapply[X](t: T[X]): Option[(A, A => B)] forSome { type A; type B } = {
  t match {
    case thing: Thing[_, _, X] => Some((thing.a, thing.f))
  }
}

这个版本在f("this真的不应该类型检查")处正确地给出了一个类型错误.

This version correctly gives a type error at f("this really shouldn't typecheck").

这看起来像是编译器中的一个错误，还是我遗漏了什么?

Does this seem like a bug in the compiler, or am I missing something?

这是在 Scala 2.10.3 上.

This is on Scala 2.10.3.

推荐答案

Mark Harrah 在 Freenode 的 #scala 频道中指出了这一点:是的，这是一个错误.

Mark Harrah pointed this out in the #scala channel on Freenode: yes, this is a bug.

https://issues.scala-lang.org/browse/SI-6680

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