• Subject: Re: Matching a fixed number of repetitions in LPeG
• From: Patrick Donnelly <batrick@...>
• Date: Tue, 19 Jun 2012 15:21:28 -0400

```On Tue, Jun 19, 2012 at 3:04 PM, Hinrik Örn Sigurðsson
<hinrik.sig@gmail.com> wrote:
> I need to match a fixed number of repetitions of a pattern, so I tried
> the multiply_pattern() function listed in
> http://lua-users.org/wiki/LpegRecipes. However, for me it never
> matches. I also tried the function listed in the referenced blog post,
> with the same result. Has LPeG changed significantly since this was
> written, or am I using it wrong? Here is the code I'm using:
>
>    function multiply_pattern(item, count)
>      return lpeg.Cmt(lpeg.P(true), function(s, i)
>        local set, offset = {}, i
>        for j = 1, count do
>          set[j], offset = lpeg.match(item * lpeg.Cp(), s, offset)
>          if not offset then
>            return false
>          end
>        end
>        return offset, set
>      end)
>    end
>
>    print(lpeg.match(lpeg.C(multiply_pattern('a', 2)), 'aa'))    --> nil

I think you're overthinking this (you shouldn't need match-time captures). Try:

function power (patt, n)
local p = lpeg.P(true)
for i = 1, n do
p = p * patt
end
return p
end

> print(lpeg.match(lpeg.C(power('a', 2)), 'aa'))
aa

As is usual for a power function, you can achieve logarithmic time if
that optimization is needed.

--
- Patrick Donnelly

```