Why can't a vararg also be a call by name param ?

To get a good error message:

scala> def f[T](a: (=> T)*) : List[T] = Nil
<console>:1: error: no by-name parameter type allowed here
       def f[T](a: (=> T)*) : List[T] = Nil

instead of

scala> def f[T](a: => T*) : List[T] = Nil 
<console>:1: error: ')' expected but identifier found.
       def f[T](a: => T*) : List[T] = Nil


2009/11/24 Maxime Lévesque <maxime.levesque@gmail.com>

Scalac does not like this mix :

  def funk[T](a: =>T*)

Are there any good reason why ?

Cheers

Max

--
Viktor Klang
| "A complex system that works is invariably
| found to have evolved from a simple system
| that worked." - John Gall

Blog: klangism.blogspot.com
Twttr: twitter.com/viktorklang
Code: github.com/viktorklang

On Tue, Nov 24, 2009 at 09:54:04AM -0500, Maxime Lévesque wrote:
> Scalac does not like this mix :
>
> def funk[T](a: =>T*)
>
> Are there any good reason why ?

https://lampsvn.epfl.ch/trac/scala/ticket/237
"Parameters should be allowed as by-name and repeated"

(Opened 2 years ago, so don't hold your breath)

2009/11/24 Viktor Klang :
> scala> def f[T](a: (=> T)*) : List[T] = Nil
> :1: error: no by-name parameter type allowed here
>        def f[T](a: (=> T)*) : List[T] = Nil
>
> instead of
>
> scala> def f[T](a: => T*) : List[T] = Nil
> :1: error: ')' expected but identifier found.
>        def f[T](a: => T*) : List[T] = Nil

I think the second case could be interesting as well... It would have
to be of type Function0[Seq[T]]. The question would be what the exact
semantics of such a call from the view of the call-site would be.