On Fri, May 13, 2005 at 01:07:20PM -0700, Larry Wall wrote:
> On Fri, May 13, 2005 at 11:54:47AM -0500, Patrick R. Michaud wrote:
> : $r1 = rx / abc :: def | ghi :: jkl | mn :: op /;
> : $r2 = rx / abc ::: def | ghi ::: jkl | mn ::: op /;
> : $r3 = rx / [ abc :: def | ghi :: jkl | mn ::
On Sat, May 14, 2005 at 04:26:44AM +, Luke Palmer wrote:
> On 5/14/05, Larry Wall <[EMAIL PROTECTED]> wrote:
> > I want ::: to break out of *that* dynamic scope (or the equivalent
> > "matchrighthere" scope), but not ::.
>
> I'm not sure that's such a good idea. When you say:
>
> rule fo
On 5/14/05, Larry Wall <[EMAIL PROTECTED]> wrote:
> On Sat, May 14, 2005 at 01:15:36AM +, Luke Palmer wrote:
> : I think the misunderstanding is rather simple. You keep talking like
> : you prepend a .*? to the rule we're matching. I think that's wrong
> : (and this is where I'm making a desi
On Sat, May 14, 2005 at 01:15:36AM +, Luke Palmer wrote:
: I think the misunderstanding is rather simple. You keep talking like
: you prepend a .*? to the rule we're matching. I think that's wrong
: (and this is where I'm making a design call, so we can dispute on this
: once we're clear that
On 5/13/05, Patrick R. Michaud <[EMAIL PROTECTED]> wrote:
> First, I'm quite certain that $r2 and $r3 are different. For
> illustration, let's use a variation like:
>
> $q2 = rx / \w [ abc ::: def | ghi ::: jkl | mn ::: op ] /;
> $q3 = rx / \w [ [ abc :: def | ghi :: jkl | mn :: op ] ]/;
Larry wrote:
I'm still not sure I believe in booleans to that extent. I suppose
we could go as far as to make it :p(0 but true). Actually, it's more
like "undef but true", if you want to be able to distinguish
sub foo (+$p = 0) { # no :p at all
say "true" if $p; # :p with
On Fri, May 13, 2005 at 11:54:47AM -0500, Patrick R. Michaud wrote:
: $r1 = rx / abc :: def | ghi :: jkl | mn :: op /;
: $r2 = rx / abc ::: def | ghi ::: jkl | mn ::: op /;
: $r3 = rx / [ abc :: def | ghi :: jkl | mn :: op ] /;
I would prefer that $r1 work like $r3, not like $r2, for t
On Fri, May 13, 2005 at 11:43:42AM +0300, Markus Laire wrote:
: Perhaps spec should be changed so that :p means :p(bool::true) or :p(?1)
: and not :p(1)
I'm still not sure I believe in booleans to that extent. I suppose
we could go as far as to make it :p(0 but true). Actually, it's more
like "
On Fri, May 13, 2005 at 03:36:50PM +, Luke Palmer wrote:
> I'm basically saying that you should treat your:
> $str ~~ /abc :: def | ghi :: jkl | mn :: op/;
> As:
> $rule = rx/abc :: def | ghi :: jkl | mn :: op/;
> $str ~~ /^ .*? <$rule>/;
> Which means that you fail the rule, your .
On 5/13/05, Patrick R. Michaud <[EMAIL PROTECTED]> wrote:
> To use the phrase from later in your message, there's still
> the "implicit .*? followed by the rule call." Since the rule
> itself hasn't failed (only the group failed), we're still free to
> try to match the pattern at later positions.
On 5/12/05, Patrick R. Michaud <[EMAIL PROTECTED]> wrote:
> I have a couple of questions regarding C< :: > in perl 6 rules.
> First, a question of verification -- in
>
> $rule = rx :w / plane :: (\d+) | train :: (\w+) | auto :: (\S+) / ;
>
> "travel by plane jet train tgv today" ~~ $rule
On Fri, 2005-05-13 at 00:26, Patrick R. Michaud wrote:
> On Thu, May 12, 2005 at 08:56:39PM -0700, Larry Wall wrote:
> > On Thu, May 12, 2005 at 09:33:37AM -0500, Patrick R. Michaud wrote:
> > : Also, A05 proposes incorrect alternatives to the above
> > :
> > : /[:w[]foo bar]/
> > I would ju
Markus Laire skribis 2005-05-13 11:43 (+0300):
> Perhaps spec should be changed so that :p means :p(bool::true) or :p(?1)
> and not :p(1)
Agreed
Juerd
--
http://convolution.nl/maak_juerd_blij.html
http://convolution.nl/make_juerd_happy.html
http://convolution.nl/gajigu_juerd_n.html
TSa (Thomas Sandlaß) kirjoitti:
Larry Wall wrote:
Speaking of which, it seems to me that :p and :c should allow an
argument that says where to start relative to the current position.
In other words, :p means :p(0) and :c means :c(0). I could also see
uses for :p(-1) and :p(+1).
Isn't that slightl
Larry Wall wrote:
Speaking of which, it seems to me that :p and :c should allow an
argument that says where to start relative to the current position.
In other words, :p means :p(0) and :c means :c(0). I could also see
uses for :p(-1) and :p(+1).
Isn't that slightly inconsistent with :p meaning :p
On Thu, May 12, 2005 at 08:56:39PM -0700, Larry Wall wrote:
> On Thu, May 12, 2005 at 09:33:37AM -0500, Patrick R. Michaud wrote:
> : Also, A05 proposes incorrect alternatives to the above
> :
> : /[:w[]foo bar]/# null pattern illegal, use
> : /[:w()foo bar]/# null capture illega
On Thu, May 12, 2005 at 09:33:37AM -0500, Patrick R. Michaud wrote:
: Also, A05 proposes incorrect alternatives to the above
:
: /[:w[]foo bar]/# null pattern illegal, use
: /[:w()foo bar]/# null capture illegal, and probably undesirable
: /[:w\bfoo bar]/# not exactly the
On Thu, May 12, 2005 at 05:15:55PM -0400, Aaron Sherman wrote:
> On Thu, 2005-05-12 at 15:41, Patrick R. Michaud wrote:
> > False. In the first case the group is the whole rule. In the second
> > case the group would not include the (implied) '.*?' at the start of
> > the rule.
>
> This was a ve
On Thu, 2005-05-12 at 15:41, Patrick R. Michaud wrote:
> $rule = rx :w / plane ::: (\d+) | train ::: (\w+) | auto ::: (\S+) / ;
> $rule = rx :w /[ plane :: (\d+) | train :: (\w+) | auto :: (\S+) ]/ ;
>
> On Thu, May 12, 2005 at 02:29:24PM -0400, Aaron Sherman wrote:
> > On Thu, 2005-05-12
$rule = rx :w / plane ::: (\d+) | train ::: (\w+) | auto ::: (\S+) / ;
$rule = rx :w /[ plane :: (\d+) | train :: (\w+) | auto :: (\S+) ]/ ;
On Thu, May 12, 2005 at 02:29:24PM -0400, Aaron Sherman wrote:
> On Thu, 2005-05-12 at 13:44, Patrick R. Michaud wrote:
> > On Thu, May 12, 2005 at 1
On Thu, 2005-05-12 at 13:44, Patrick R. Michaud wrote:
> On Thu, May 12, 2005 at 12:53:46PM -0400, Aaron Sherman wrote:
> > > In other words, it acts as though one had written
> > >
> > > $rule = rx :w / plane ::: (\d+) | train ::: (\w+) | auto ::: (\S+) / ;
> > >
> > > and not
> > >
>
> "PRM" == Patrick R Michaud <[EMAIL PROTECTED]> writes:
PRM> On Thu, May 12, 2005 at 12:33:59PM -0500, Jonathan Scott Duff wrote:
>>
>> > > /[:w\bfoo bar]/# not exactly the same as above
>> >
>> > No, I think that's exactly the same.
>>
>> What does \b mean again? I
On Thu, May 12, 2005 at 12:48:16PM -0500, Patrick R. Michaud wrote:
> On Thu, May 12, 2005 at 12:33:59PM -0500, Jonathan Scott Duff wrote:
> >
> > > > /[:w\bfoo bar]/# not exactly the same as above
> > >
> > > No, I think that's exactly the same.
> >
> > What does \b mean again? I assum
On Thu, May 12, 2005 at 12:33:59PM -0500, Jonathan Scott Duff wrote:
>
> > > /[:w\bfoo bar]/# not exactly the same as above
> >
> > No, I think that's exactly the same.
>
> What does \b mean again? I assume it's no longer backspace?
For as long as I can remember \b has meant "word boun
On Thu, May 12, 2005 at 12:53:46PM -0400, Aaron Sherman wrote:
> My take, based on S05:
>
> > In other words, it acts as though one had written
> >
> > $rule = rx :w / plane ::: (\d+) | train ::: (\w+) | auto ::: (\S+) / ;
> >
> > and not
> >
> > $rule = rx :w /[ plane :: (\d+) | tr
On Thu, May 12, 2005 at 12:53:46PM -0400, Aaron Sherman wrote:
> On Thu, 2005-05-12 at 10:33, Patrick R. Michaud wrote:
> > Next on my list, S05 says "It is illegal to use :: outside of
> > an alternation", but A05 has
> >
> > /[:w::foo bar]/
>
> I can't even figure out what that means. :w t
My take, based on S05:
On Thu, 2005-05-12 at 10:33, Patrick R. Michaud wrote:
> I have a couple of questions regarding C< :: > in perl 6 rules.
> First, a question of verification -- in
>
> $rule = rx :w / plane :: (\d+) | train :: (\w+) | auto :: (\S+) / ;
>
> "travel by plane jet train
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