No, I think it's doing the right thing. To say "$x equals 1 if it's defined,
otherwise a whatever object," I think you need to say:
$x = 1 // {*}.()
Which is a bit ugly, but then Whatever isn't really meant to be used that
way. Amusingly it does work:
./perl6 -e 'my $x = Mu // {*}.(); for 1 ... $x -> $i { say $i }'
prints the positive integers.
On Mon, Aug 9, 2010 at 5:32 PM, Carl Mäsak via RT <
perl6-bugs-follo...@perl.org> wrote:
> On Mon Aug 09 06:44:14 2010, pawel.pab...@implix.com wrote:
> > [15:38] <bbkr> rakudo: my $x = 1 // *; $x.WHAT.say
> > [15:38] <p6eval> rakudo c1e19a: OUTPUT«WhateverCode()»
> >
> >
> > "1" is expected because it's first defined value in assignment.
>
> Is there something that makes the C<//> operator be treated specially, so
> that using it along with
> C<*> doesn't create a thunk/WhateverCode object? I'm not sure there is, at
> least not in the spec.
> Maybe there should be, but I'm not sure of that either.
>
--
Aaron Sherman
Email or GTalk: a...@ajs.com
http://www.ajs.com/~ajs
