jaytea

Just checking back in to say hey guys :) The last few months have been a blur: planned a wedding, got married, had a honeymoon - whew! It's great to see there's still some activity on the regex front. I'm hoping to rekindle the old flame and start participating in challenges, updating my blog, and finally completing the quiz I've had in the works for a long ass time

What an awesome display of resourcefulness :P

Grimy is brilliant and I wouldn't put it past him to have discovered it independently.. but what a cool coincidence. And you discovered it in response to the very same challenge (find a unique character) if I'm not mistaken?

I just got your e-mail and was about to post that here

Hauke :D

debtsm.art/rgx-english.html this is hilarious

Oh, of course. Didn't even think that it might be considering it a variable name in the middle of a pattern like that

Looks like Perl is doing something with "$x". Maybe some sort of optimization. Very puzzling. "x^" is fine though, so the expression can be touted for both Perl/PCRE now with that :)

I mean (?=(?'a'[\s\S]*))(?'b'X(?=\k'a'\z)|(?<=(?=(?&b)).))

er

Wait, the opposite, Perl doesn't need those: (?=(?'a'[\s\S]*))(?'b'X(?=\k'a'\z)|(?<=(?=$x|(?&b)).)) works fine haha

Found it :P I think "$x" isn't good enough of a non-empty contradiction for Perl

(?1) isn't matching empty there, at least

The result implies "(?&b)" matches an empty string, so (?<=(?=$x|)[\s\S]) matches at the first opportunity, just after "b"

the result contains the value of \1 and \2 (a and b)

Also, I'm getting a very puzzling result from this perl bot on IRC (running v 5.24)

Yay

But that appears to have changed in 5.20

Grimy, has that changed? Just quickly trying to find info about that, and I see on regular-expressions.info/recursecapture.html it says "When Perl 5.10 [through to 5.18]'s regex engine enters recursion, all capturing groups appear as they have not participated in the match yet."

God I hate HTML

drregex.com/2019/02/variable-length-lookbehinds-actually.html I banged out a quick post about it :D Gotta head home right now, but I'll have to play with that damn explanation tomorrow because the tabs/lines are all screwed

Yeah.. so just re-write the whole expression so it works in tandem with the whole "go back one character at a time" framework

I feel like I should leave it as (?=[\s\S]*) for demo purposes to avoid further complicating things

Did all this testing maybe 8 years ago

Does ^\C*$ hold true for arbitrary strings in any environment? Because I could've sworn I've seen it fail.. but I might be confusing that with it throwing runtime errors when encountering stray surrogates

For the first occurrence? I seem to recall it failing to match under certain conditions

But yes, agreed, I'll go back to that mess

I should've said, in case the arbitrary subexpression relies on //s being one way or the other :P

just to simplify ".*" and "." at the end. Really, I shouldn't use that at all, in case the arbitrary subexpression 'X' relies on //s being disabled

Do your worst! :p

I moved 'X' into the parent capture group in case it was zero-width, such as ^

(?s:(?=(?<a>.*))(?<b>X(?=\k<a>\z)|(?<=(?=$x|(?&b)).))) you guys fancy trying to break that, claimed to be a plug-n-play equivalent of (?<=X) (aside from the fact that it creates two capture groups)

Any personal webpage or anything you want linked?

@Grimy, how would one introduce you? :p

Yeah

By all means post about it on SO :P Let's get the word out!

Ah but "$x" fails regardless of the set status of (?m)

Haha. Mime it

how the devil are you

hey David :P

Golf addict!

no matter how you cut it, it's going to look a little "wtf" at first glance :p

And $x should be a contradiction in all contexts, if I'm not mistaken. in place of [^\s\S]

Good idea!

@Grimy if you're not planning to publish this formally anywhere, would you mind terribly if I wrote it up and put it on drregex.com? With profuse crediting to yourself, of course. The regex world needs to see this!

Fun new toy haha

Longest word in a string fully functional: \b(\w++)(?=([\s\S]*))(?!.*\b\w++(?=([\s\S]*))((?<=(?=(?=\1\2\z)(?:(?=.*\b(\5?+\‌​w)\w*+\3\z)\w)++\b|(?4)|[^\s\S]).)))

(oh, that should be [\s\S]* at the start there*)