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pat = re.compile "{g <- .g / &'and' }"   -- lua pattern "(.*)and"
= lpeg.pcode( pat )                             -- using debug version of lpeg

i noticed its pcode has a "behind 3" instruction to not consume the last 'and'

there is a lpeg.B function to do look-behind, but how to go back to it if B matched ?

Is there a lpeg.U(n) (for undo n characters) or something similar ?

As an example of its usefulness, say # is lpeg re for undo 1 character

REDO above re pattern, but without backtrack stack overflow problem:
NOTE: I want to capture ALL except LAST 'and'

pat = re.compile "{ (g <- 'and' / . [^a]* g)+ ### }"

Without UNDO, I have to do this (likely much slower):

pat = re.compile( "(g <- 'and' / . [^a]* g)+ -> drop3", {drop3 = function(s) return s:sub(1,-4) end} )