创建一个通常返回一个或两个参数的方法 [英] Creating a method which returns one or two parameters generically
问题描述
我有一种方法
def foo(num: Int): String
我在代码中的某些位置进行了调用，一切都很好.
最近，我遇到一种情况，我需要调用相同方法，但是由于具有一些参数int值，我需要获取返回的 2个字符串，而不仅仅是一个.我目前的实现方式是:
Where I call some in some places in my code, and everything was good.
Lately, I encountered a situation where I need to call the same method, but with some parameter int value, I need to get in return 2 Strings, and not only one. My current way of implementing it is:
def foo(num: Int): List[String]
每次调用foo并获得 1 字符串时，我都将成为列表的头，并且每次调用并返回2个字符串时，我都将得到其中的元素[0，1](我知道当我调用foo(10)时，我得到2个字符串，其余的-只有一个).
Where each time I call foo and get 1 String, I will get the head of the list, and each time I call for and it returns 2 strings, I will get the elements in [0, 1] (I know that when I call foo(10), I get 2 strings, and for the rest - only one).
是否有更多惯用的scala/功能?
Is there a more idiomatic scala/functional for this?
推荐答案
Scala 2.13 引入了基于文字的单例类型，因此实际上您可以做这样的疯狂事情:
Scala 2.13 introduced literal-based singleton types, so actually you can do a crazy thing like this:
def foo(num: 10): (String, String) = ("Hello", "World")
def foo(num: Int): String = s"Hello $num"
val (left, right) = foo(10)
val single = foo(2)
，它将编译并运行.
当然，您可以根据需要返回列表，而不是 10 的元组.
Of course, you can return a list instead of a tuple for the 10 case if you wish.
它也应该适用于 typelevel scala (甚至在 2.13 之前).
It should also work for typelevel scala (even before 2.13).
在 2.13 之前使用常规的 Lightbend Scala ，您仍然可以做到这一点，但这要笨拙得多.有必要使用一个额外的技巧，涉及使用称为 Witness 的类型，但幸运的是，它是由 shapeless
With regular Lightbend Scala before 2.13 you could still do that, but it was a lot clunkier. It was necessary to use an additional trick involving using type called Witness, but fortunately, it is provided by shapeless:
import shapeless.Witness
import shapeless.syntax.singleton._
def foo(num: Witness.`10`.T): (String, String) = ("Hello", "World")
def foo(num: Int): String = s"Hello $num"
val (left, right) = foo(10)
val single = foo(2)
当然，有必要添加 shapeless 作为依赖项:
And of course it is necessary to add shapeless as dependency:
libraryDependencies += "com.chuusai" %% "shapeless" % "2.3.3"
您可以使用的另一种方法是使用某种特殊的容器来获得结果.我会推荐元组:(String, Option[String]).如果返回的是常规"结果，则返回(String, None)；如果返回的是 10 ，则可以返回(String, Some[String]).
Another approach you could use is just the use of some kind of special container for your result. I would recommend tuple: (String, Option[String]). In case you're returning "regular" result, you would return (String, None) and in case of 10 you can return (String, Some[String]).
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