Re: lpeg.U ?
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- Subject: Re: lpeg.U ?
- From: Albert Chan <albertmcchan@...>
- Date: 2018年1月26日 08:56:20 -0500
> On Jan 25, 2018, at 11:01 PM, Sean Conner <sean@conman.org> wrote:
>
> It was thus said that the Great albertmcchan once stated:
>>
>>> On Jan 25, 2018, at 8:03 PM, Sean Conner <sean@conman.org> wrote:
>>>
>>> If you are looking for a final "and" (which ends the input), then this
>>> works:
>>>
>>> last_and = P"and" * P(-1)
>>> char = R("0円96円","b255円")^1
>>> + -last_and * P"a"
>>> pat = C((char)^0) * last_and
>>>
>>> print(pat:match(string.rep("this and that land",400) .. "and"))
>>
>> your example also shows usefulness of undo lpeg.U (if exist):
>>
>> pat = C( P(1)^3 * U(3) * #P'and' )
>
> Even if lpeg.U() existed, I don't think this would do what you expect
Sorry for the mis-understanding.
I was referring to YOUR example, a final "and" (which ends the input).
The natural way of matching is to move to position -3, then check for 'and'
If I use lua string library, i would do this:
function string_match_and_at_endofstring(s)
if string.sub(s, -3) ~= 'and' then return nil end
return string.sub(s, 1, -4)
end
Above code, literally translate to lpeg (if #s is available)
function lpeg_match_and_at_endofstring(s)
local pat = C(P(#s - 3) * #P'and')
return lpeg.match(pat, s)
end
Without #s and if U exist then pat = C(P(1)^3 * U(3) * #P'and')