Scala: Constructor taking either Seq or varargs

Question

I am guessing that, for compatibility reasons, the type of vararg parameters Any* is Array[Any] - please correct this if I\'m wrong. However, this does not expl

Answer 1

I suppose that you would like to make the method calls prettier and so explicit calling with _* is not an option. In that case you may solve the problem with method overloading.

class Api(api_url: String, params: Seq[(String, String)]) {
  def this(api_url: String, param : (String, String), params: (String, String)*)
    = this(api_url, param +: params)
  def this(api_url: String)
    = this(api_url, Seq())
}

Answer 2

No: actually, Any* is actually almost identical to Seq[Any], not to Array[Any].

To disambiguate between the two, you can use the technique to add a dummy implicit parameter to make the signature different:

class Api(api_url: String, params: Seq[(String, String)]) {
  def this(api_url: String, params: (String, String)*)(implicit d: DummyImplicit) =
    this(api_url, params)
}

Answer 3

A method taking varargs is also always taking a sequence, so there is no need to define an auxiliary constructor or overloaded method.

Given

class Api(api_url: String, params: (String, String)*)

you can call it like this

new Api("url", ("a", "b"), ("c", "d"))

or

val seq = Seq(("a", "b"), ("c", "d"))
new Api("url", seq:_*)

Also, in your question, you are calling method seq on the params parameter. This probably does not do what you intended. seq is used to ensure that operations on the resulting collection are executed sequentially instead of in parallel. The method was introduced with the parallel collections in version 2.9.0 of Scala.

What you probably wanted to use was toSeq, which returns the collection it is used on converted to a Seq (or itself if it is already a Seq). But as varargs parameters are already typed as Seq, that is a no-op anyway.