On 22/08/05, Luke Palmer <[EMAIL PROTECTED]> wrote:
> Output?
>
> sub foo (+$a, *%overflow) {
> say "%overflow{}";
> }
>
> foo(:a(1), :b(2)); # b 2
> foo(:a(1), :overflow{ b => 2 }); # b 2
I would have thought:
overflow b 2
i.e. %overflow<overflow> = (b => 2)
Because :overflow() is an unrecognised named argument, so it goes in the slurpy
hash. The fact that the hash has the same name as the argument is a
coincidence--does it make sense to explicitly name a slurpy when you can just
splat *{ b => 2 } in directly?
> foo(:a(1), :overflow{ b => 2 }, :c(3)); # ???
overflow b 2
c 3
Of course, that's just my way of thinking.
(Also, last I heard you /could/ have multiple slurpy hashes, but any after the
first would always be empty.)
Stuart
