Jonathan Scott Duff wrote:
> How can you be sure that is
> implemented as a character class instead of being some other arbitrary
> rule? An answer is that perl should know how these things are
> implemented and if you try arithmetic on something that's not a
> character class, it should carp a
On Wed, 2002-09-04 at 09:55, Markus Laire wrote:
> On 4 Sep 2002 at 0:22, Aaron Sherman wrote:
>
> > On Wed, 2002-09-04 at 00:01, Sean O'Rourke wrote:
> >
> > > None, I think. Of course, if we ignore internals, there's no
> > > difference bewteen that and "rx / | 1 | 7/".
> >
> > Then, why is
On 4 Sep 2002 at 0:22, Aaron Sherman wrote:
> On Wed, 2002-09-04 at 00:01, Sean O'Rourke wrote:
>
> > None, I think. Of course, if we ignore internals, there's no
> > difference bewteen that and "rx / | 1 | 7/".
>
> Then, why is there a C<+>? Why not make it C<|>?
>
> $foo = rx/ <||[cde
On Tue, 2002-09-03 at 23:57, Luke Palmer wrote:
> On Tue, 3 Sep 2002, Brent Dax wrote:
> >
> > How can you be sure that is implemented as a character
> > class, as opposed to (say) an alternation?
>
> What's the difference? :)
>
> Neglecting internals, semantically what I the difference?
>
O
On Wed, 2002-09-04 at 00:01, Sean O'Rourke wrote:
> On Tue, 3 Sep 2002, Luke Palmer wrote:
>
> > On Tue, 3 Sep 2002, Brent Dax wrote:
> >
> > > Damian Conway:
> > > # $roundor7 = rx /<+[17]>/
> > > #
> > > # That is: the union of the two character classes.
> > >
> > > How can you be sur
On Tue, Sep 03, 2002 at 09:57:31PM -0600, Luke Palmer wrote:
> On Tue, 3 Sep 2002, Brent Dax wrote:
>
> > Damian Conway:
> > # Neither. You need:
> > #
> > # $roundor7 = rx /<+[17]>/
> > #
> > # That is: the union of the two character classes.
> >
> > How can you be sure that is impl
On Tue, 3 Sep 2002, Luke Palmer wrote:
> On Tue, 3 Sep 2002, Brent Dax wrote:
>
> > Damian Conway:
> > # Neither. You need:
> > #
> > # $roundor7 = rx /<+[17]>/
> > #
> > # That is: the union of the two character classes.
> >
> > How can you be sure that is implemented as a character
>
On Tue, 3 Sep 2002, Brent Dax wrote:
> Damian Conway:
> # Neither. You need:
> #
> # $roundor7 = rx /<+[17]>/
> #
> # That is: the union of the two character classes.
>
> How can you be sure that is implemented as a character
> class, as opposed to (say) an alternation?
What's the d
Damian Conway:
# Neither. You need:
#
# $roundor7 = rx /<+[17]>/
#
# That is: the union of the two character classes.
How can you be sure that is implemented as a character
class, as opposed to (say) an alternation?
--Brent Dax <[EMAIL PROTECTED]>
@roles=map {"Parrot $_"} qw(embeddin
At 9:24 PM -0400 8/31/02, Ken Fox wrote:
>Damian Conway wrote:
>>No. It will be equivalent to:
>>
>> <[\x0a\x0d...]>
>
>I don't think \n can be a character class because it
>is a two character sequence on some systems. Apoc 5
>said \n will be the same everywhere, so won't it be
>something li
Damian Conway wrote:
> No. It will be equivalent to:
>
> <[\x0a\x0d...]>
I don't think \n can be a character class because it
is a two character sequence on some systems. Apoc 5
said \n will be the same everywhere, so won't it be
something like
rule \n { \x0d \x0a | \x0d | \x0a }
Hmm.
On Sat, 2002-08-31 at 07:07, Damian Conway wrote:
> Aaron Sherman wrote:
>
> > Is C<\n> going to be a rule (e.g. C<< >>)
>
> There might be an named rule like that. But C<\n> will certainly
> still be available.
>
> > or is it implicitly translated to:
> >
> > <[\x0a\x0d...]>+
>
> No. It
> > $roundor7 = rx /<+[17]>/
> > That is: the union of the two character classes.
>
> Thank you; that wasn't in A5, E5 or S5. Will there be <-> as
> well?
>From A5:
The outer <...> also naturally serves as a container
for any extra syntax we decide to come up with for
charac
[EMAIL PROTECTED] (Damian Conway) writes:
> Neither. You need:
> $roundor7 = rx /<+[17]>/
> That is: the union of the two character classes.
Thank you; that wasn't in A5, E5 or S5. Will there be <-> as
well?
--
I wish my keyboard had a SMITE key
-- J-P Stacey
Aaron Sherman wrote:
> Is C<\n> going to be a rule (e.g. C<< >>)
There might be an named rule like that. But C<\n> will certainly
still be available.
> or is it implicitly translated to:
>
> <[\x0a\x0d...]>+
No. It will be equivalent to:
<[\x0a\x0d...]>
(no repetition)
> Al
Is C<\n> going to be a rule (e.g. C<< >>) or is it implicitly
translated to:
<[\x0a\x0d...]>+
If it's the latter, then what does this do?
\n?
Do I get
[<[\x0a\x0d...]>+]?
Or do I get
<[\x0a\x0d...]>+?
If the former (which I assume is the case), how do I get
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