What is the Scala syntax for calling a function with variadic arguments but with named arguments?

Question

Say I have a function

def f(a:Int = 0, b:String = "", c:Float=0.0, foos: Foo*) { ... }

Notice the use of default arguments for some parameters. Typically, to make use of default values, you invoke a function with named parameters like so:

val foo = Foo("Foo!")
f(foos = foo)

This syntax works because I'm only invoking the method with one foo. However, if I have two or more

val foo1 = Foo("Foo!")
val foo2 = Foo("Bar!")
f(foos = ...)

it's not so obvious what should be fed in here. Seq(foo1,foo2) and Seq(foo1,foo2):_* do not type check.

What's more, how can I call it with no foos?

// All out of foos!
f(foos = ...)

For this case, calling the function with empty parentheses (f()) does not work.

Thanks!

Answer 1

For the default parameters, see my comment to your question. For how to call the variadic part with a named argument, see below (scala 2.9.2):

scala> case class Foo(a:Int)
defined class Foo

scala> def f(a:Int, foos: Foo*) = foos.length
f: (a: Int, foos: Foo*)Int

scala> f(a=2, foos=Foo(2))
res0: Int = 1

// I'm not sure if this is cheating...
// am I just naming the first of the two variadic arguments?
scala> f(a=2, foos=Foo(2),Foo(3))
res1: Int = 2

//anyway you can do ....
scala> f(a=2, foos=Seq(Foo(1), Foo(2)):_*)
res2: Int = 2

// with no foos...
scala> f(a=2, foos=Nil:_*)
res3: Int = 0