On Thu, Jun 13, 2002 at 03:48:25PM -0700, Larry Wall wrote:
But the most straightforward way to match longest is probably to use
:any to get a superposition of matches, and then pull out the longest
match.
So, does :any return a list of the substrings that matched or a list
of match
I'm still unclear as to how you implement lex-like longest token rule with
P6 regexes. If the | operator grabs the first one it matches, how do I
match bacamus out of this?:
bacamus =~ / b.*a | b.*s /
Luke
I'm still unclear as to how you implement lex-like longest token rule with
P6 regexes. If the | operator grabs the first one it matches, how do I
match bacamus out of this?:
bacamus =~ / b.*a | b.*s /
Borrow this trick from Parse::RecDescent:
rule max (*@candidates) {{
Damian Conway:
# I'm still unclear as to how you implement lex-like longest
# token rule
# with P6 regexes. If the | operator grabs the first one it matches,
# how do I match bacamus out of this?:
#
# bacamus =~ / b.*a | b.*s /
#
# Borrow this trick from Parse::RecDescent:
#
#
I figured that (I actually did it, in a less-pretty form, in my early
Perl days when I wrote a syntax highlighter for my website). So there's
no elegant way the new regexes support it? That's a shame.
But I see now how state objects are a very cool idea.
Oh, and I'd just thought I'd let
On Thu, 13 Jun 2002, David Whipp wrote:
: Second, we should eliminate as much of the syntactic noise as possible:
:
: max b.*a b.*s
:
: would be nice -- with parenthesis, or the like, needed only when things
: become ambiguous. I think, though am not sure, that having whitespace act as
: an
