Re: [LPeg] How can I make a recursive pattern?

[Sean Conner <sean@...>]

It was thus said that the Great Soni L. once stated:
> 
> 
> On 27/07/16 12:34 PM, Patrick Donnelly wrote:
> >Soni,
> >
> >On Wed, Jul 27, 2016 at 10:27 AM, Soni L. <fakedme@gmail.com> wrote:
> >>Sean Conner's incomplete solution uses lpeg.Carg(1) which doesn't work
> >>recursively.

  At the time, I wasn't sure of the output format and I had assumed (badly,
I see) that you wanted something like:

	{
	  key = value
	}

where both key and value are specified in the text you are parsing.  Doing
that in LPeg isn't straightforward and thus, lpeg.Carg() was used to pass in
a resulting table.  Given the example, it's both easy and hard.  Easy,
because collecting results into an array is easy in LPeg:

	a = lpeg.P"a"
	r = lpeg.Ct(lpeg.C(a)^0)
	x = r:match "aaaaaaaa"

Even recursively:

	nested = lpeg.P {
	        "start",
	        start = lpeg.Ct((lpeg.V"a" + lpeg.V"list")^0),
	        a     = lpeg.C(lpeg.P"a"),
	        list  = lpeg.P"("
	              * lpeg.V"start"
	              * lpeg.P")",
	}

	r = lpeg.Ct(nested)
	x = r:match "aaa(aaaa)aaa(a)aaa"

  Hard, because it appears you want to start with index [0], which makes it
... an interesting problem (there doesn't appear to be an easy way to get to
the table that lpeg.Ct() creates, which is why this is an interesting
problem).  Perhaps some combination of lpeg.Cg() with lpeg.P(true) divided
by a function that returns a new table and lpeg.Cb() but it's something
you'll have to experiment with.

  Or you could relax the [0] index requirement ... 

  -spc (You're in luck ... work is slow today ... )