[英]Why does Option[Try[_]] not conform to F[_]?
So having something like this: 所以有这样的事情:
@ trait IntWrapper[F[_]] { def apply(i: Int): F[Int] }
defined trait IntWrapper
@ class OptWrapper extends IntWrapper[Option] { def apply(i: Int) = Option(i) }
defined class OptWrapper
I now want to do something like this: 我现在想做这样的事情:
@ class TryOptWrapper extends IntWrapper[Try[Option]] { def apply(i: Int) = Try(Option(i)) }
cmd19.sc:1: scala.util.Try[Option] takes no type parameters, expected: one
class TryOptWrapper extends IntWrapper[Try[Option]] { def apply(i: Int) = Try(Option(i)) }
^
Compilation Failed
(Same thing if I declare the trait extension as class TryOptWrapper extends IntWrapper[Try[Option[_]]]
) (如果我在
class TryOptWrapper extends IntWrapper[Try[Option[_]]]
声明了特征扩展,则与此相同)
Now, perhaps the most interestingly, this works: 现在,也许最有趣的是,这可行:
@ type Topt[T] = Try[Option[T]]
@ class ToptWrapper extends IntWrapper[Topt] { def apply(i: Int) = Try(Option(i)) }
defined class ToptWrapper
Now, is it possible to do the same thing – ie implement a trait with a type parameter being a nested parametrized type – without having to explicitly declare the type alias? 现在,是否可以做同样的事情-即使用类型参数为嵌套参数化类型的类型参数实现特征-而无需显式声明类型别名? It definitely feels like I'm missing some syntax here.
确实感觉就像我在这里缺少一些语法。
Try
expects type parameter of kind *
and Option
has kind * => *
so you can't write Try[Option]
, only Try[Option[Int]]
, Try[Option[String]]
, Try[Option[_]]
... Try
期望类型*
类型参数,而Option
具有类型* => *
因此您不能编写Try[Option]
,只能编写Try[Option[Int]]
, Try[Option[String]]
, Try[Option[_]]
...
Try type lambda 尝试输入lambda
class TryOptWrapper extends IntWrapper[({ type λ[A] = Try[Option[A]] })#λ] { def apply(i: Int) = Try(Option(i)) }
Or with kind-projector 或带投影仪
class TryOptWrapper extends IntWrapper[Lambda[A => Try[Option[A]]]] { def apply(i: Int) = Try(Option(i)) }
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