Re: how to translate lua pattern "(.*)and(.*)" with lpeg re ?
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- Subject: Re: how to translate lua pattern "(.*)and(.*)" with lpeg re ?
- From: Sean Conner <sean@...>
- Date: 2018年1月22日 12:33:15 -0500
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