Question

我们说我有一个名为Marker的空标记特征和一些由Marker绑定的类型参数的函数：

trait Marker

object Marker {
  def works[M <: Marker](m:M):M = m
  def doesntWork[M <: Marker](f:M => String):String = "doesn't matter"
}

第一个功能正如我所料。也就是说，如果传递的参数不是Marker，则代码不会编译：

scala> works("a string")
<console>:14: error: inferred type arguments [String] do not conform to method works's type parameter bounds [M <: com.joescii.Marker]
       works("a string")
       ^
<console>:14: error: type mismatch;
 found   : String("a string")
 required: M
       works("a string")
             ^

但是，我能够将参数传递给不符合Marker的第二个函数。具体来说，我可以传递类型String => String的函数，并且代码很乐意编译并运行：

scala> doesntWork( (str:String) => "a string" )
res1: String = doesn't matter

我希望对doesntWork的调用无法编译。任何人都可以向我解释为什么它编译以及如何更改函数签名以防止类型在这种情况下进行检查？

完全披露：上述人为举例是this outstanding issue for lift-ng的简化版。

Answer 1

代码由于逆转而编译。您可以通过明确地提供推断类型参数来看到：

doesntWork[Nothing]((str: String) => "a string")

这是一个普遍的问题。有各种各样的技术可以解决这个问题，但它们通常归结为将T限制为某种类型的实例。

Answer 2

M => String实际上是Function1[M, String]。如果你看一下定义：

 trait Function1[-T1, +R]

因此M变为逆变，这意味着对于M1 >: M2，Function1[M1, String] <: Function1[M2, String]，我们说M1 = Any然后Function1[Any, String] <: Function1[Marker, String]。

doesntWork - f的输入也是逆变的，这意味着您可以传递小于M => String的内容，正如我刚才所示，Any => String是小于Marker => String，所以它完全没问题。

您也可以传递String => String，因为您的[M <: Marker]最终导致编译器将M解释为Nothing，因此即使String => String变得大于M => String {1}}。

要解决您的问题，只需引入包装器，它将使您的类型不变：

scala> case class F[M](f: M => String)
defined class F

scala> def doesntWork[M <: Marker](f:F[M]):String = "doesn't matter"
doesntWork: [M <: Marker](f: F[M])String

scala> doesntWork(F((str: String) => "a string"))
<console>:18: error: inferred type arguments [String] do not conform to method doesntWork's type parameter bounds [M <: Marker]
              doesntWork(F((str: String) => "a string"))
              ^
<console>:18: error: type mismatch;
 found   : F[String]
 required: F[M]
              doesntWork(F((str: String) => "a string"))
                          ^
scala> doesntWork(F((str: Any) => "a string"))
<console>:18: error: inferred type arguments [Any] do not conform to method doesntWork's type parameter bounds [M <: Marker]
              doesntWork(F((str: Any) => "a string"))
              ^
<console>:18: error: type mismatch;
 found   : F[Any]
 required: F[M]
Note: Any >: M, but class F is invariant in type M.
You may wish to define M as -M instead. (SLS 4.5)
              doesntWork(F((str: Any) => "a string"))

scala> doesntWork(F((str: Marker) => "a string"))
res21: String = doesn't matter

scala> trait Marker2 extends Marker
defined trait Marker2

scala> doesntWork(F((str: Marker) => "a string"))
res22: String = doesn't matter

scala> doesntWork(F((str: Marker2) => "a string"))
res23: String = doesn't matter

推荐这种隐式转换通常很糟糕，但这里似乎很好（如果你不会过度使用F）：

scala> implicit def wrap[M](f: M => String) = F(f)
warning: there was one feature warning; re-run with -feature for details
wrap: [M](f: M => String)F[M]

scala> doesntWork((str: Marker) => "a string")
res27: String = doesn't matter

scala> doesntWork((str: String) => "a string")
<console>:21: error: inferred type arguments [String] do not conform to method doesntWork's type parameter bounds [M <: Marker]
              doesntWork((str: String) => "a string")
              ^
<console>:21: error: type mismatch;
 found   : F[String]
 required: F[M]
              doesntWork((str: String) => "a string")
                                       ^

当参数不符合类型约束时，为什么Scala函数会编译？

2 个答案: