Larry Wall wrote:
>
> : In summary: assuming Perl 6 allows user-defined while-ish structures, how
> : would it be done?
>
> I think the secret is to allow easy attachment of regex rules to sub
> and parameter declarations. There's little point in re-inventing
> regex syntax using declarations. The whole point of making Perl 6
> parse itself with regexes is to make this sort of stuff easy.
>
> Larry
Ah. I'd been wondering how Damian's
sub if (bool $condition, &block);
was going to be extended to handle if-else, if-elsif, if-elsif-else, if-elsif-elsif,
if-elsif-elsif-else, etc.
I *think* what you're saying is that user defined while-ish and if-ish constructs
should really be created by altering the Perl 6 parser at compile time. Of course,
that's not really news: you already said that in Apocalypse N (for some value of N
(I'm to lazy to determine which, since it doesn't matter (and besides, laziness is one
of my cardinal virtues :-) (parse *that* if you can -- don't smileys just play havoc
with parenthesis-counting?)))).
I was provoked to ask my question about C<while> by Damian's message which said, in
part:
> So C<if>'s signature is:
>
> sub if (bool $condition, &block);
Now, if I'm understanding correctly, *that* turns out not to be the case. Of course,
Damian's point was to talk about semicolons after blocks, not C<if> structures in all
their glorious complexity. Nonetheless, the signature of C<if> in Perl 6 will have to
be more complicated than that, or it wouldn't be able to handle zero-or-more elsif's
and zero-or-one else's.
Again quoting Damian:
> Likewise I could write my own C<perhaps> subroutine:
>
> sub perhaps (bool $condition, num $probability, &block) {
This, on the other hand, would be true. The simplicity of this hypothetical control
structure makes it possible to use the built-in syntactic features of Perl 6
(especially the rules concerning blocks and how they interact with commas and
semicolons), so it wouldn't be necessary to extend the grammar to implement C<perhaps>.
Am I getting this straight?
=thom
