Regex backreference without capture

QM has asked for the wisdom of the Perl Monks concerning the following question:

Is it possible to have a capture for the purposes of backreferences, but not returned as part of the match?

# Example
@x = 'bogus firstblahjokeblahthird bogus' =~ /(first)(?:blah).*?\2(thi
+rd)/; # doesn't work
# @x is empty
[download]

(In fact, I suspect the above doesn't work because, while there is eventually a 2nd capture group, it hasn't captured anything before the backreference. But that isn't really the point here.)

I don't really have a specific need, but it just seems interesting to say "this capture is only for backreference, and not for returning as a capture".

-QM
--
Quantum Mechanics: The dreams stuff is made of

Comment on Regex backreference without capture Download Code

Replies are listed 'Best First'.
Re: Regex backreference without capture by rsFalse (Hermit) on Mar 01, 2019 at 15:52 UTC
Hello, QM, Haven't heard about such capture groups for only backreferences. Interestingly, if any simple overcome exist? This overcome below looks ugly: `#!/usr/bin/perl -wl print map "[$_]", "bogus firstblahj?keblahthird bogus" =~ / (?\| (F) \| first ( (blah) .? \2 ) third (?{ $L = $1 }) (THEN) (F) # 1 2 2 1 \| (??{ defined $L ? "" : "(*F)" }) (first) (??{ print "mid-match:[$L]" ; quotemeta $L }) (third +) # 1 1 2 +2 ) /x;` [download] output: `mid-match:[blahj?keblah] [first][third]` [download] Idea here is to use conditional '(?\|...)' with 2 groups, save mid-match (if full match is successful), fail first branch anyway, then match same thing on alternative branch rewriting $1 and $2, and using previously defined mid-match to avoid captures.	[reply] [d/l] [select]
Re^2: Regex backreference without capture by QM (Parson) on Mar 05, 2019 at 10:31 UTC
Thanks. But I think that's too complex for any normal usage. (I'd hate to maintain that, even if I was the only coder.) But seriously, thanks. Maybe this will spur more ideas? -QM -- Quantum Mechanics: The dreams stuff is made of	[reply]
Re: Regex backreference without capture (?&NAME) NamedCapture by beech (Parson) on Mar 05, 2019 at 03:36 UTC
Hi ? So you want `@x = qw( first third ) ;` This "works" `dd( 'bogus firstblahjokeblahthird bogus' =~ m/ (first) (?&patblah) .*? (?&patblah) (third) (?(DEFINE) (?<patblah>blah) ) /sx ); __END__ ("first", "third", undef)` [download] The undef comes from the named capture `(?<patblah>blah)` which takes up slot $3 in this pattern, which is why (?(DEFINE) is placed at the end	[reply] [d/l] [select]
Re^2: Regex backreference without capture (?&NAME) NamedCapture by QM (Parson) on Mar 05, 2019 at 10:33 UTC
Thanks. But does this do backreferences? I see it matches the same pattern, but that's not the same as a backreference. But thanks, that makes me think about the problem differently. Edited to add: ------------------ I'm considering using named capture groups, where there's a need to have a well-defined return value. Which is something PCRE has, and some people (!) can't live without, apparently. -QM -- Quantum Mechanics: The dreams stuff is made of	[reply]
Re^2: Regex backreference without capture (?&NAME) NamedCapture by rsFalse (Hermit) on Mar 05, 2019 at 12:34 UTC
E.g. the input is `"bogus firstblAHj?keblOWthird bogus"`, and pattern is `'bl..'`, then it matches both, but OP would like to match only the same :)	[reply] [d/l] [select]


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