scala – 使用重复参数的成本
发布时间：2020-12-16 09:56:15 所属栏目：安全 来源：网络整理
导读：我考虑重构一些方法签名,这些签名当前采用List类型的参数或具体类的集合–List [Foo] – 来改为使用重复的参数：Foo *. Update : Following reasoning is flawed,move along… This would allow me to use the same method name and overload it based on th
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我考虑重构一些方法签名,这些签名当前采用List类型的参数或具体类的集合–List [Foo] – 来改为使用重复的参数：Foo *.
在我的情况下,重构的方法适用于scala.Seq [Foo],它来自重复的参数.我必须更改所有调用并将序列参数类型注释添加到所有集合参数：baz.doStuffWith(foos：_ *). 鉴于从集合参数切换到重复参数在语义上是等效的,这个更改是否会产生一些我应该注意的性能影响？ scala 2.7._和2.8的答案是否相同？ 解决方法
当
Scala调用
Scala varargs方法时,该方法将接收一个扩展Seq的对象.当使用：_ *进行调用时,对象将按原样传递,而不进行复制.以下是此示例：
scala> object T {
| class X(val self: List[Int]) extends SeqProxy[Int] {
| private val serial = X.newSerial
| override def toString = serial.toString+":"+super.toString
| }
| object X {
| def apply(l: List[Int]) = new X(l)
| private var serial = 0
| def newSerial = {
| serial += 1
| serial
| }
| }
| }
defined module T
scala> new T.X(List(1,2,3))
res0: T.X = 1:List(1,3)
scala> new T.X(List(1,3))
res1: T.X = 2:List(1,3)
scala> def f(xs: Int*) = xs.toString
f: (Int*)String
scala> f(res0: _*)
res3: String = 1:List(1,3)
scala> f(res1: _*)
res4: String = 2:List(1,3)
scala> def f(xs: Int*): Seq[Int] = xs
f: (Int*)Seq[Int]
scala> def f(xs: Int*) = xs match {
| case ys: List[_] => println("List")
| case _ => println("Something else")
| }
f: (Int*)Unit
scala> f(List(1,3): _*)
List
scala> f(res0: _*)
Something else
scala> import scala.collection.mutable.ArrayBuffer
import scala.collection.mutable.ArrayBuffer
scala> def f(xs: Int*) = xs match {
| case ys: List[_] => println("List")
| case zs: ArrayBuffer[_] => zs.asInstanceOf[ArrayBuffer[Int]] += 4; println("Array Buffer")
| case _ => println("Something else")
| }
f: (Int*)Unit
scala> val ab = new ArrayBuffer[Int]()
ab: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer()
scala> ab + 1
res11: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1)
scala> ab + 2
res12: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1,2)
scala> ab + 3
res13: scala.collection.mutable.Buffer[Int] = ArrayBuffer(1,3)
scala> f(ab: _*)
Array Buffer
scala> ab
res15: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1,3,4)
注意 > Array作为WrappedArray传递.但是,不会复制所涉及的元素,并且对WrappedArray的更改将反映在数组中. （编辑：李大同） 【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容! |