Programmatically setting Repeated Parameters in Scala

Question

I'm trying to call Futures.awaitAll with a variable number of well... Futures. awaitAll is defined as awaitAll(timeout : Long, fts : Future[Any]*). I have tried passing in a List and an Array but both won't work:

list = future1 :: future2 :: Nil

Futures.awaitAll(1000, list)

found : List[scala.actors.Future[Any]] required: scala.actors.Future[Any]

EDIT: What I now want to do is call Futures.awaitAll programmatically with a variable number of arguments (1 to n). So using Futures.awaitAll(1000, future1, future2) is not an option.

Chapter 8.8 of Programming in Scala didn't give me any hints how to solve this either, so help is welcome :)

This does get mentioned somewhere in PinS, but very briefly. — Daniel C. Sobral
– Daniel C. Sobral, Commented Aug 11, 2009 at 22:02

Zack Marrapese · Accepted Answer · 2009-08-11 15:07:51Z

10

Using the * means that it's a vararg...it can take as many Future[Any] parameters as you add, but not a list/array of them.

So it's looking for a parameter list such as:

Futures.awaitAll(1000, future1, future2)

instead of

Futures.awaitAll(1000, list)

Edit: If you must have the ability to pass in Futures.awaitAll(1000, list), then try casting it.

So try this:

Futures.awaitAll(1000, list: _*)

edited Aug 11, 2009 at 15:07

answered Aug 11, 2009 at 14:46

Zack Marrapese

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starblue · Accepted Answer · 2009-08-11 15:09:24Z

1

Futures.awaitAll(1000, futures: _*)

should work (can't test it now). See 4.6.2 in the Scala Language Specification.

answered Aug 11, 2009 at 15:09

starblue

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