Re: two things about defs

Alexander,

Well, I always thought that method is a function that is associated to
particular object. In this case, it's not associated to any object since
I'm writing definition in toplevel (or maybe it's associated to some
artificial one).

mhm, that's right. the interpreter puts this stuff in an object. here's more info http://article.gmane.org/gmane.comp.lang.scala/3242/match=interpreter

It seems I still don't understand the difference between methods and
functions, could you tell me?

Sure! here's a post that has lots of examples
http://article.gmane.org/gmane.comp.lang.scala/3101/match=closure

cheers,
B

> BTW I think you could have omitted the & in definition of xxx, given
> that you gave a function type in the annotation.
>
> hope this helps,
> Burak

-- WBR, Alexander Sergeev.

> On 1/27/07, Alexander Sergeev <as.beta@xxxxxxxxx> wrote:
>>
>> I've posted one of these features into buglist, another one is probably
>> waiting it's turn.
>>
>> First one:
>>
>> scala> def sum(a: Int)(b: Int): Int = a+b sum:
>> (scala.Int)(scala.Int)scala.Int
>>
>> scala> val sum2 = (x: int) => (y: int) => x + y sum2: ( scala.Int) =>
>> (scala.Int) => scala.Int = <function>
>>
>> Could interpreter print out only one version of type output? Personally
>> I think the best one is: scala.Int => scala.Int => scala.Int =
>> <function>
>>
>> Second one:
>>
>> scala> def yyy = &(List(1,2) foldLeft) yyy: => (scala.Nothing) =>
>> ((scala.Nothing , scala.Int) => scala.Nothing) => scala.Nothing
>>
>> scala> def xxx[a] : (a => ((a, int) => a) => a) =  &(List(1,2)
>> foldLeft) xxx: [a](a) => ((a, scala.Int) => a) => a
>>
>> Is it possible for type inference mechanism to abstract over
>> scala.Nothing when deciding type for yyy? It would be great to have xxx
>> automatically. Though, this problem may be undecidable.
>>
>> -- WBR, Alexander Sergeev.
>>
>>
>
>

-- 
Burak Emir
Research Assistant / PhD Candidate
Programming Methods Group
EPFL, 1015 Lausanne, Switzerland
http://lamp.epfl.ch/~emir