Resending due to BT doing bad things to good nameservers.
Damian Conway <[EMAIL PROTECTED]> writes:
> > > $val = (foo())[0];
> > >
> > > List?
> >
> > Scalar, obviously.
>
> How do you figure that? (Not a criticism: I'd really like to understand your
> thought process here so I can assess the relative DWIMity of the two
> alternatives).
I figure I'm going mad actually. Misread the initial post.
I'm now thinking that (foo())[0] is distinct from (foo()).[0], or
foo[0] or all the other options.
--
Piers
"It is a truth universally acknowledged that a language in
possession of a rich syntax must be in need of a rewrite."
-- Jane Austen?
