Scala: how to specify varargs as type?

Question

Instead of

def foo(configuration: (String, String)*)

I'd like to be able to write:

type Configuration =  (String, String)*
def foo(configuration: Configuration)

The main use case is to provide an easy method signature when overriding in subclasses

UPDATE: I can come close by

type Param = (String, String)
def foo(configuration: Param*)

But is there a way of doing it better?

Answer 1

I think you could use

type Configuration =  ((String, String)*)
def foo(configuration: Configuration)

But it makes the compiler crash(2.8.0.r21161-b20100314020123). It seems a bug of the scala compiler.

Answer 2

No, the * is only allowed on a ParamType, that is the type of a parameter to a anonymous function or a method.

4.6.2 Repeated Parameters Syntax: ParamType ::= Type ‘’ The last value parameter of a parameter section may be suffixed by “”, e.g. (..., x:T *). The type of such a repeated parameter inside the method is then the sequence type scala.Seq[T ]. Methods with repeated parameters T * take a variable number of arguments of type T.

The compiler bug @Eastsun's example is in the first line, not the second. This should not be allowed:

scala> type VarArgs =  (Any*)
defined type alias VarArgs

I've raised a bug.

This is similar restriction to By-Name Parameters. In this case, the compiler prevents the creation of the type alias:

scala> type LazyString = (=> String) <console>:1: error: no by-name parameter type allowed here
       type LazyString = (=> String)

Your final attempt is the standard way to express this.

Answer 3

You could define it as

type Param = (String, String)
type Configuration = Seq[Param]

def foo(configuration : Configuration)

the user has to construct a Seq instance

foo(List("1"->"2"))

which is not optimal.