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 roundascii is implemented as a character
class, as opposed to (say) an alternation?
What's the difference? :)
Neglecting internals, semantically what Iis the difference?
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 /roundascii | 1 | 7/.
Then, why is there a C+? Why not make it C|?
$foo = rx/ a|b|[cde]|f
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 /roundascii | 1 | 7/.
Then, why is there a
Jonathan Scott Duff wrote:
How can you be sure that roundascii 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
Damian Conway:
# Neither. You need:
#
# $roundor7 = rx /roundascii+[17]/
#
# That is: the union of the two character classes.
How can you be sure that roundascii is implemented as a character
class, as opposed to (say) an alternation?
--Brent Dax [EMAIL PROTECTED]
@roles=map {Parrot
On Tue, 3 Sep 2002, Brent Dax wrote:
Damian Conway:
# Neither. You need:
#
# $roundor7 = rx /roundascii+[17]/
#
# That is: the union of the two character classes.
How can you be sure that roundascii is implemented as a character
class, as opposed to (say) an alternation?
On Tue, 3 Sep 2002, Luke Palmer wrote:
On Tue, 3 Sep 2002, Brent Dax wrote:
Damian Conway:
# Neither. You need:
#
# $roundor7 = rx /roundascii+[17]/
#
# That is: the union of the two character classes.
How can you be sure that roundascii is implemented as a character
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 /roundascii+[17]/
#
# That is: the union of the two character classes.
How can you be sure that roundascii is
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 /roundascii+[17]/
#
# That is: the union of the two character classes.
How can you be sure that
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 like
rule
Aaron Sherman wrote:
Is C\n going to be a rule (e.g. C eol )
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)
Along those
[EMAIL PROTECTED] (Damian Conway) writes:
Neither. You need:
$roundor7 = rx /roundascii+[17]/
That is: the union of the two character classes.
Thank you; that wasn't in A5, E5 or S5. Will there be foo-bar as
well?
--
I wish my keyboard had a SMITE key
-- J-P Stacey
$roundor7 = rx /roundascii+[17]/
That is: the union of the two character classes.
Thank you; that wasn't in A5, E5 or S5. Will there be foo-bar as
well?
From A5:
The outer ... also naturally serves as a container
for any extra syntax we decide to come up with for
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 eol )
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
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. Now
Is C\n going to be a rule (e.g. C eol ) 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 the
16 matches
Mail list logo
