On Tue 05 Sep, Nathan Wiger wrote:
"normal" "reversed"
-- ---
103301
99aa99
(( ))
+ +
{{[!_ _!]}}
{__A1( )A1__}
That is, when a bracket is encountered, the "reverse" of
On Tue 05 Sep, David Corbin wrote:
Nathan Wiger wrote:
But, how about a new ?m operator?
/(?m|[).*?(?M|])/;
Let's combine yor operator with my example from above where everything
inside the (?m) or the ?(M)
fits the syntax of a RE.
/(?m()|\[).*?(?M()|(\]))
On Wed 06 Sep, Mark-Jason Dominus wrote:
I've been thinking the same thing. It seems to me that the attempts to
shoehorn parsers into regex syntax have either been unsuccessful
(yielding an underpowered extension) or illegible or both.
SNOBOL:
parenstring = '(' *parenstring ')'
On Fri 08 Sep, Kevin Walker wrote:
(This thread has been inactive for a while. See
http://www.mail-archive.com/perl6-language-regex@perl.org/index.html#0
0015 for it's short history.)
Long ago Tom Christiansen wrote:
This is useful in that it would stop being number dependent.
For
This list has gone a little quiet...
Hugo wrote:
I like this too. I'd suggest /t should mean a) return a scalar of
the number of matches and b) don't set any special variables. Then
/t without /g would return 0 or 1, but be faster since no extra
information need be captured (except
This RFC had three concepts, I propose dropping the "Not a pattern" from here
as it is now in RFC 198 and the null element. The List expansion might
benefit from a slight enhancement.
Hugo:
(?@foo) and (?Q@foo) are both things I've wanted before now. I'm
not sure if this is the right syntax,
On Mon 11 Sep, Mark-Jason Dominus wrote:
(?@foo) is sort of equivalent to (??{join('|',@foo)}), ie it expands into
a list of alternatives. One could possible use just @foo, for this.
It just occurs to me that this is already possible. I've written a
module, 'atq', such that if you
(proto RFC possibly, and some generalised ramblings)
Given that expansion of regexes could include (+...) and (*...) I have been
thinking about providing a general purpose way of adding functionality.
I propose that the entire (+...) syntax is kept free from formal
specification for this and
On Wed 13 Sep, Bart Lateur wrote:
On Tue, 12 Sep 2000 19:01:35 -0400, Mark-Jason Dominus wrote:
I don't know what you mean, but you're mistaken, because it means to
interpolate @foo as in a double-quoted string.
Which is precisely the meaning he wants for it, with $" set to '|'.
I
TomCs perl storm has:
Figure out way to do
/$e1 $e2/
safely, where $e1 might have '(foo) \1' in it.
and $e2 might have '(bar) \1' in it. Those won't work.
If e1 and e2 are qr// type things the answer might be to localise
the backref numbers in each qr// expression.
If they
On Sun 24 Sep, Hugo wrote:
In [EMAIL PROTECTED], Richard Proctor
writes
:
:TomCs perl storm has:
:
: Figure out way to do
:
: /$e1 $e2/
:
: safely, where $e1 might have '(foo) \1' in it.
: and $e2 might have '(bar) \1' in it. Those won't work.
:
:If e1 and e2 are qr// type
HI Tom,
Welcome to England (I presume)
This seems very complicated. Did you look at the Ram:6 recipe on
expressing AND, OR, and NOT in a regex? For example, to do
/FOO/ /BAR/ you need not write /FOO.*BAR|BAR.*FOO/ -- and in
fact, should not, as it doesn't work properly on some pairs!
On Wed 27 Sep, Dave Storrs wrote:
On Wed, 27 Sep 2000, Richard Proctor wrote:
Both \1 and $1 refer to what is matched by the first set of parens in a
regex. AFAIK, the only difference between these two notation is that
\1 is used within the regex itself and $1 is used outside
On Fri, 29 Sep 2000 01:02:40 +0100, Hugo wrote:
It also isn't clear what parts of the expression are interpolated at
compile time; what should the following leave in %foo?
%foo = ();
$bar = "one";
"twothree" =~ / (?$bar=two) (?$foo{$bar}=three) /x;
It's not just that. You act
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