Pattern matching: Why no \1 in [ ]?

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

I thought, naively, that the easiest way of matching a string such as 'XOX' or 'TNT' but not 'XXX' or 'TTT' would be:

/(.)[^\1]\1/

...but this doesn't work:

$_ = 'BBB';
print /(.)[^\1]\1/ ? 'match' : 'no match';
[download]

This prints match. Indeed, it seems to match if I replace the middle 'B' with any single character (including '1' and '\').

Even more strangely, to me, if I 'un-negate' the character class:

/(.)[\1]\1/

...I don't seem to match anything (BTW, I get no complaints with strictures and warnings).

Context: I'm trying to code a simple substitution cipher solver (eg where 'ABCABC' will match 'murmur' and 'tsetse' but not 'booboo'). I'm aware that there are other ways of doing it, notably merlyn's post here. However, he uses (unless I'm mistaken) negative loook-behind magic, and I haven't got that far in Mastering Regular Expressions yet :-). I'm not looking for a solution, I'm just curious as to what /[^\1]/ and /[\1]/ actually match? I've scoured perlre and done an index search for the (admittedly large :-) part of Mastering Regular Expressions that I still haven't read, but found nothing that appears relevant to my question (that's not to say that there is nothing relevant, just that I might not have understood its relevance...).

TIA for your enlightenment.

dave

Comment on Pattern matching: Why no \1 in [ ]? Select or Download Code

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Re: Pattern matching: Why no \1 in [ ]? by japhy (Canon) on Aug 07, 2003 at 20:25 UTC
As has been stated, a character class is determined at a regex's compile-time, because it's turned into a big bitwise array, more or less. Having a backref in there would make it dynamic, and dynamic char classes don't exist (yet?). If `$1` only has one character in it, then sure, you can do `/(.)(?!\1)(?s:.)\1/` as was suggested. But if it's got more than one character, I think the "easiest" way is with the dynamic-regex assertion: `/(.)(??{ "[^\Q$1\E]" })\1/`. _____________________________________________________ Jeff`[japhy]`Pinyan: Perl, regex, and perl hacker, who'd like a job (NYC-area) `s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;`	[reply] [d/l] [select]
Re: Pattern matching: Why no \1 in [ ]? by MarkM (Curate) on Aug 07, 2003 at 19:46 UTC
\1 provides the ability to substitute the literal text string of a previously matched parameter into the regular expression. It doesn't give you the ability to recompile the regular expression. Character class (`[...]`) are a compile time directive. \1 is a run time directive. `[\1]` is interpretted as 'the character with value 1'.	[reply] [d/l] [select]
Re: Pattern matching: Why no \1 in [ ]? by artist (Parson) on Aug 07, 2003 at 19:49 UTC
Look at Backreferences Also for your prolem can be solved easily without character class. `$_ = 'abcab'; if(/^(.)(?!\1)(.)(?:(?!\1\|\2).)\1\2$/){ print "$&\n"; }` [download] artist	[reply] [d/l]
Re: Pattern matching: Why no \1 in [ ]? by japhy (Canon) on Aug 07, 2003 at 20:31 UTC
In fact, merlyn's code uses the approach Juerd showed you. It's just a bunch of negative look-aheads for the backreferences you've already matched. For instance, text matching "ABCADB" would be the regex `/^(.)(?!\1)(.)(?!\1\|\2)(.)\1(?!\1\|\2\|\3)(.)\2$/s`. _____________________________________________________ Jeff`[japhy]`Pinyan: Perl, regex, and perl hacker, who'd like a job (NYC-area) `s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;`	[reply] [d/l]
Re: Pattern matching: Why no \1 in [ ]? by Juerd (Abbot) on Aug 07, 2003 at 20:09 UTC
`[^\1]` `(?!\1)(?s:.)` [download] Juerd # { site => 'juerd.nl', plp_site => 'plp.juerd.nl', do_not_use => 'spamtrap' }	[reply] [d/l] [select]
Re: Pattern matching: Why no \1 in [ ]? by errr (Initiate) on Aug 08, 2003 at 06:55 UTC
To see what the regexp is doing, you can use the re pragma. use re 'debug' to enable debugging: perl -Mre=debug -le 'print /(.)[^\1]\1/ ? "match" : "no match"' Freeing REx: `","' Compiling REx `(.)[^\1]\1' size 19 Got 156 bytes for offset annotations. first at 3 1: OPEN1(3) 3: REG_ANY(4) 4: CLOSE1(6) 6: ANYOF[\0\2-\377{unicode_all}](17) 17: REF1(19) 19: END(0) [download] from this you can see that the \1 is being interpretted as a unicode 1 to check you can: `perl -le '$_ = "B\1B"; print /(.)[^\1]\1/ ? "match" : "no match"'` [download] and you'll see that that is a "no match"	[reply] [d/l] [select]


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