Scala中方法的默认泛型类型 [英] Default generic type on method in scala

问题描述

我正在尝试找出一种方法来摆脱重载方法.目前，我正在使用这种方法向用户提供不错的API:

I'm trying to figure out a way to get away from overloaded methods. Currently I'm using this approach to provide nice APIs to the user:

  def get(id: String): Option[JsonDocument]

  def get(id: String, timeout: Duration): Option[JsonDocument]

  def get[D <: Document[_]](id: String, target: Class[D]): Option[D]

  def get[D <: Document[_]](id: String, target: Class[D], timeout: Duration): Option[D]

现在，由于scala具有默认参数，因此我想将其浓缩为一种方法.但是因为D是通用的，所以如果没有提供，我需要使用默认值，而不是"Nothing"，而是"JsonDocument".

Now since scala has default parameters I want to condense this into one method. But because D is generic, I need the default if not provided not to be "Nothing", but rather "JsonDocument".

我目前的做法是:

  def get[D <: Document[_]](id: String, target: Class[D] = classOf[JsonDocument], timeout: Duration = null): Option[D]

事实证明，编译器对此感到非常满意，但是IDE存在问题.如果未显式提供目标(例如target = JsonDocument或任何其他目标)，则它认为返回类型为Option [Nothing]，因此会使用户感到困惑.

And as it turns out the compiler is super happy with it, but the IDE has problems. if the target is not explicitly provided (like with target = JsonDocument or any other) it thinks the return type is Option[Nothing] and therefore its confusing to the user.

所以我的问题是:对于这些类型，如果用户未提供要覆盖的D类型的JsonDocument，是否可以提供默认"类型的JsonDocument?

So my question is: with these types, is it possible to provide a "default" type of JsonDocument for type D if it is not provided by the user to be overridden?

推荐答案

在

There is an solution to the default generic parameters problem given at http://www.cakesolutions.net/teamblogs/default-type-parameters-with-implicits-in-scala. Applying to your case, you'd get something like (untested):

trait DefaultsTo[Type, Default]

object DefaultsTo {
  implicit def defaultDefaultsTo[T]: DefaultsTo[T, T] = null
  implicit def fallback[T, D]: DefaultsTo[T, D] = null  
}

// use target.runtimeClass in the implementation
def get[D <: Document[_]](id: String, timeout: Duration)(implicit target: scala.reflect.ClassTag[D], default: DefaultsTo[D, JsonDocument]): Option[D]

要使用: JsonDocument 的 get(id，timeout)或 get [OtherDocument](id，timeout)的 get(id，timeout).当然，IDE(IntelliJ?)是否能够正确推断类型是另一个问题！

To use: get(id, timeout) for JsonDocument or get[OtherDocument](id, timeout). Of course, whether the IDE (IntelliJ?) is going to infer the type correctly is a different question!

这篇关于Scala中方法的默认泛型类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！