On Tue, Oct 30, 2001 at 12:27:32PM +1100, Damian Conway wrote:
>
> > > ($obj1, $obj2)->foo(@args);
> >
> > Is that merely sugar for:
> >
> > # errr, $_.foo(@args) ?
> > $_->foo(@args) foreach($obj1, $obj2);
>
> No. What you showed would be achieved with either a hyperoperation:
>
> ($obj1, $obj2)^.foo(@args);
[ ... ]
> method foo ( $me, $again : $big, $boy ) {...}
>
> should be able to be called via either of:
>
> foo $obj1, $obj2 : @args;
> or:
> ($obj1, $obj2).foo(@args);
This stuff brings to mind all sorts of questions:
* If foo() is declared as above, what values do $me and $again have when
the method is called as $obj.foo(@args);? Is $again undef?
* What happens when foo is declared as "method foo ($a: *@_) { ... }"
and called as "foo $a, $b: @args;"? Is $b ignored?
* Would "foo $a, $b ^: @args" work as "($a,$b)^.foo(@args)"?
* Would (($o1,$o2),($o3,$o4))^.foo(@args); be the same as
($o1,$o2).foo(@args); ($o3,$o4).foo(@args); ?
* "method foo ($a, $b: *@_) { ... }; foo $o1, $o2: @args;" seems like a
binding that doesn't mention :=. i.e., ($a,$b) := ($o1,$o2); Will
there be other "hidden" uses of the := operator?
-Scott
--
Jonathan Scott Duff
[EMAIL PROTECTED]
