in reply to Regex fun

I would assume that quantifier cannot be a '\x' variable, but don't really know.

I think it's important to note that \1 is not a variable * (which is why you can't use it outside of a regex); the variable that contains the contents of the first capture group is $1, but that 's empty doesn't take on its new value *** until the capture has completed **.

I think that the reason that ($rx){\1} isn't allowed is that the regex engine wants to compile the regex before running it. Since the contents of \1, hence the number of times that $rx is supposed to be captured, aren't known until run-time, this interferes with the compilation. For example, /\+32767.{32767}/ is rejected at compile time, but a '+32767' =~ /\+([0-9]*).{\1}/ construct would circumvent this restriction. (“Why, then,” you ask, “is something like /(.)\1/, which suffers from the same compilation problem, OK?” I dunno. :-) )

* Not a Perl variable, anyway. See Re^3: Regex fun, and probably Re^2: Regex fun as well.
** Except that (?{ print $1 }) works correctly, which is somewhat miraculous to me and very very helpful for debugging regexes.
UPDATE: *** Still false (see Re^6: Regex fun for where realisation finally dawns). It takes on its new value as soon as the capture completes (which explains the miracle referenced above); it's just that the interpolation in the text of the regex has already happened, so that the quantifier doesn't ‘see’ the new value.