scala splat 可以用于任何不是可变参数的东西吗？

发布于 2024-10-31 17:16:07 字数 677 浏览 1 评论 0原文

给定例如：

scala> def pipes(strings:String*) = strings.toList.mkString("|")

我可以正常调用：

scala> pipes("foo", "bar")
res1: String = foo|bar

或使用 splat：

scala> val args = List("a","b","c")
scala> pipes(args:_*)
res2: String = a|b|c

但是我可以使用 splat 为除 varargs 参数之外的任何参数提供参数吗？例如，我想做类似的事情：

scala> def pipeItAfterIncrementing(i:Int, s:String) = (i + 1) + "|" + s
scala> val args:Tuple2[Int, String] = (1, "two")
scala> pipeItAfterIncrementing(args:_*)

这不起作用，但是有什么方法可以达到从单个对象提供多个参数的相同效果，无论它是元组还是其他东西？鉴于元组的长度和类型在编译时已知，是否有任何原因无法对元组实现？

原文

given e.g:

scala> def pipes(strings:String*) = strings.toList.mkString("|")

which I can call normally:

scala> pipes("foo", "bar")
res1: String = foo|bar

or with a splat:

scala> val args = List("a","b","c")
scala> pipes(args:_*)
res2: String = a|b|c

But can I use a splat to provide arguments for anything but a varargs parameter? e.g I would like to do something like:

scala> def pipeItAfterIncrementing(i:Int, s:String) = (i + 1) + "|" + s
scala> val args:Tuple2[Int, String] = (1, "two")
scala> pipeItAfterIncrementing(args:_*)

That doesn't work, but is there any way to achieve the same effect of providing multiple arguments from a single object, whether it be a tuple or something else? Is there any reason this couldn't be implemented for tuples, given that both their length and types are known at compile-time?

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撞了怀 2024-11-07 17:16:07

您可以使用 Function.tupled 来完成此操作：将采用 n 个参数的函数转换为采用元数 n 的单个元组参数的函数>。正如所预料的，Function.untupled 执行相反的工作。

特殊类型归属 : _* 仅适用于重复参数（又名可变参数）。

scala> def pipeItAfterIncrementing(i:Int, s:String) = (i + 1) + "|" + s
pipeItAfterIncrementing: (i: Int,s: String)java.lang.String

scala> def tupledPipeItAfterIncrementing = Function.tupled(pipeItAfterIncrementing _)
tupledPipeItAfterIncrementing: ((Int, String)) => java.lang.String

scala> val args:Tuple2[Int, String] = (1, "two")
args: (Int, String) = (1,two)

scala> tupledPipeItAfterIncrementing(args)
res0: java.lang.String = 2|two

You can use Function.tupled to do exactly this: turn a function that takes n arguments into a function that takes a single tuple argument of arity n. As can be expected, Function.untupled does the reverse job.

The special type ascription : _* is only applicable for repeated parameter (a.k.a. varargs).

scala> def pipeItAfterIncrementing(i:Int, s:String) = (i + 1) + "|" + s
pipeItAfterIncrementing: (i: Int,s: String)java.lang.String

scala> def tupledPipeItAfterIncrementing = Function.tupled(pipeItAfterIncrementing _)
tupledPipeItAfterIncrementing: ((Int, String)) => java.lang.String

scala> val args:Tuple2[Int, String] = (1, "two")
args: (Int, String) = (1,two)

scala> tupledPipeItAfterIncrementing(args)
res0: java.lang.String = 2|two

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娇纵 2024-11-07 17:16:07

嗯……

scala> def pipeItAfterIncrementing(i:Int, s:String) = (i + 1) + "|" + s
scala> val args:Tuple2[Int, String] = (1, "two")
scala> (pipeItAfterIncrementing _).tupled(args)

会给你想要的2|two。

Well kind of ...

scala> def pipeItAfterIncrementing(i:Int, s:String) = (i + 1) + "|" + s
scala> val args:Tuple2[Int, String] = (1, "two")
scala> (pipeItAfterIncrementing _).tupled(args)

will give you the desired 2|two.

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