Re: how to translate lua pattern "(.*)and(.*)" with lpeg re ?

[Sean Conner <sean@...>]

It was thus said that the Great albertmcchan once stated:
> during testing, lpeg backtracking can only handle max string length of 199:
> is this a lpeg bug ? only 199 chars limit ?

  Nope.  It's documented.  From the LPeg documentation:

	lpeg.setmaxstack (max)

		Sets a limit for the size of the backtrack stack used by
		LPeg to track calls and choices. (The default limit is 400.)
		Most well-written patterns need little backtrack levels and
		therefore you seldom need to change this limit; before
		changing it you should try to rewrite your pattern to avoid
		the need for extra space. Nevertheless, a few useful
		patterns may overflow. Also, with recursive grammars,
		subjects with deep recursion may also need larger limits.

And this is exactly what you are hitting:

	lua: a.lua:18: backtrack stack overflow (current limit is 400)

You need to rethink your solution.  Nick Gamon's email [1] has a solution
that can work, albeit with a slight tweak:

c = re.compile [[
   parse     <-  {noDelim} lastDelim   -- look for all up to the last
                                       -- delimiter followed by the last part
   delim     <- 'and'                  -- our delimiter
   noDelim   <- (!lastDelim .)*        -- zero or more characters
                                       -- without the last delimiter
   lastDelim <- delim {(!delim .)*} !. -- the delimiter without any more
                                       -- delimiters and then end of subject
]]

print(c:match( str:rep(50) ))

  -spc

[1]	http://lua-users.org/lists/lua-l/2018-01/msg00294.html