blocks|key|1330738|text|Scala+2.13引入了基于文字的单例类型，因此实际上您可以这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1330739|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)|code-block|syntax|javascript|1330740|而且它会汇编和工作。|1330741|当然，如果您愿意的话，您可以返回一个列表，而不是一个元组。|1330742|它也适用于类型的scala+(甚至在2.13之前)。|1330743|使用常规的亮点Scala，在2.13之前，您仍然可以这样做，但它要笨重得多。有必要使用一个额外的技巧，包括使用名为Witness的类型，但幸运的是，它是由不成形的|1330744|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)|1330745|当然，有必要添加无形状的依赖项：|1330746|libraryDependencies+%2B=+"com.chuusai"+%25%25+"shapeless"+%25+"2.3.3"|1330747|您可以使用的另一种方法就是为您的结果使用某种特殊的容器。我推荐元组：(String,+Option[String])。如果返回“常规”结果，则返回(String,+None)，如果返回10，则可以返回(String,+Some[String])。|CODE|1330748|entityMap|0|LINK|mutability|MUTABLE|url|https://scastie.scala-lang.org/4s68LKHdTcGRe8pxTtJjvQ^0|0|A|0|0|4|5|0|0|0|5|8|I|4|0|5|7|E|4|0|0|0|0|Y|O|21|E|2S|M|0^^$0|@$1|2|3|4|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1D|8|@]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|M|3|N|5|6|7|1H|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1N|8|@$9|1O|A|1P|B|C]|$9|1Q|A|1R|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1S|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|1T|8|@]|D|@]|E|$]]|$1|W|3|X|5|H|7|1U|8|@]|D|@]|E|$I|J]]|$1|Y|3|Z|5|6|7|1V|8|@$9|1W|A|1X|B|10]|$9|1Y|A|1Z|B|10]|$9|20|A|21|B|10]]|D|@]|E|$]]|$1|11|3|-4|5|6|7|22|8|@]|D|@]|E|$]]]|12|$13|$5|14|15|16|E|$17|18]]]]

Scala 2.13 introduced literal-based singleton types, so actually you can do a crazy thing like this:

<pre><code>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)
</code></pre>

and it will <a href="https://scastie.scala-lang.org/4s68LKHdTcGRe8pxTtJjvQ" rel="noreferrer">compile and work</a>.

Of course, you can return a list instead of a tuple for the 10 case if you wish.

It should also work for typelevel scala (even before 2.13).

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:

<pre><code>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)
</code></pre>

And of course it is necessary to add shapeless as dependency:

<pre><code>libraryDependencies += "com.chuusai" %% "shapeless" % "2.3.3"
</code></pre>

<hr>

Another approach you could use is just the use of some kind of special container for your result. I would recommend tuple: <code>(String, Option[String])</code>. In case you're returning "regular" result, you would return <code>(String, None)</code> and in case of 10 you can return <code>(String, Some[String])</code>.

blocks|key|2769959|text|由于结果空间被清晰地分割成两面(不相交)，请考虑Either如下|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2769960|def+foo(i:+Int):+Either[(String,+String),+String]+=
++if+(i+==+10)+Left("one",+"two")+else+Right("one")

val+Left(Tuple2(x,+y))+=+foo(10)
val+Right(a)+=+foo(2)|code-block|syntax|javascript|2769961|这是受@Krzysztof关于“特殊容器”的编辑的启发。|2769962|entityMap^0|O|6|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

Since the result space is cleanly split into two sides (disjointed) consider <code>Either</code> like so

<pre><code>def foo(i: Int): Either[(String, String), String] =
 if (i == 10) Left("one", "two") else Right("one")

val Left(Tuple2(x, y)) = foo(10)
val Right(a) = foo(2)
</code></pre>

which is inspired by @Krzysztof's edit regarding "special container".

I have a method 

<pre><code>def foo(num: Int): String
</code></pre>

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:

<pre><code>def foo(num: Int): List[String]
</code></pre>

Where each time I call <code>foo</code> 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 <code>foo(10)</code>, I get 2 strings, and for the rest - only one). 

Is there a more idiomatic scala/functional for this?

Creating a method which returns one or two parameters generically

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有个方法def foo(num: Int): String我在代码中的某些地方调用了一些东西，一切都很好。最近，我遇到了一种情况，我需要调用相同的方法，但是使用一些参数int值，我需要返回2 Strings，而不仅仅是一个。我目前的实施方式是：def foo(num: Int): List[String]每次调用fo...

问创建一个一般返回一个或两个参数的方法
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