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Home/ Questions/Q 667267
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T23:54:34+00:00 2026-05-13T23:54:34+00:00

I consider refactoring few method signatures that currently take parameter of type List or

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I consider refactoring few method signatures that currently take parameter of type List or Set of concrete classes —List[Foo]— to use repeated parameters instead: Foo*.

Update: Following reasoning is flawed, move along…
This would allow me to use the same method name and overload it based on the parameter type. This was not possible using List or Set, because List[Foo] and List[Bar] have same type after erasure: List[Object].

In my case the refactored methods work fine with scala.Seq[Foo] that results from the repeated parameter. I would have to change all the invocations and add a sequence argument type annotation to all collection parameters: baz.doStuffWith(foos:_*).

Given that switching from collection parameter to repeated parameter is semantically equivalent, does this change have some performance impact that I should be aware of?

Is the answer same for scala 2.7._ and 2.8?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T23:54:34+00:00Added an answer on May 13, 2026 at 11:54 pm

    When Scala is calling a Scala varargs method, the method will receive an object that extends Seq. When the call is made with : _*, the object will be passed as is*, without copying. Here are examples of this:

    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, 2, 3)
    
    scala> new T.X(List(1,2,3))
    res1: T.X = 2:List(1, 2, 3)
    
    scala> def f(xs: Int*) = xs.toString
    f: (Int*)String
    
    scala> f(res0: _*)
    res3: String = 1:List(1, 2, 3)
    
    scala> f(res1: _*)
    res4: String = 2:List(1, 2, 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(1,2,3): _*)
    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, 2, 3)
    
    scala> f(ab: _*)
    Array Buffer
    
    scala> ab
    res15: scala.collection.mutable.ArrayBuffer[Int] = ArrayBuffer(1, 2, 3, 4)
    

    Note

    • An Array is passed as a WrappedArray. There’s no copying of elements involved, however, and changes to the WrappedArray will be reflected in the Array.
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