From: MenTaLguY <mental@...>
Date: 2005-12-12T21:56:54+09:00
Subject: Re: increasing counter whithin loop?

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On Mon, 2005-12-12 at 21:07 +0900, Patrick Gundlach wrote:
> Your two solutions are very nice, thank you. I think I'll go for the
> each_with_next.

Cool.  Note that (per the recent thread with Matz) I was wrong about
needing to program the iterator defensively for redo/next.  Those
confine their effects to the block, and you'd handle retry/break with
'ensure' like you would other exits (e.g. exceptions), but normally you
don't need to care.

Anyway, this is sufficient for each_with_next:

 module Enumerable
   def each_with_next(last=3Dnil)
     prev =3D nil
     first =3D true
     each do |elt|
       if first
         first =3D false
         next
       else
         yield prev, elt
       end
       prev =3D elt
     end
     yield prev, last unless first
   end
 end

> There are other items in the schedule that I need to output. I collect
> all these parts in an array:
>=20
> @schedule << ListOfMember.new.parse(the source)
> ....
>=20
> and output all like
>=20
> @schedule.each { |elt|  elt.render }
>=20
> but in case of two 'ListOfMembers' following each other, they should be
> output next to each other, so I wrap them in a 'Group of two ListOfMember=
s'

Hmm.  What happens if you have three in a row?

Seems to me you could do something like this:

 collapsed =3D []
 @schedule.each do |elt|
   prev =3D collapsed.last
   if LOM =3D=3D=3D prev && LOM =3D=3D=3D elt
     grp =3D GOLM.new
     grp.add( prev )
     grp.add( elt )
     collapsed[-1] =3D grp
   elsif GOLM =3D=3D=3D prev && LOM =3D=3D=3D elt
     prev.add( elt )
   else
     collapsed << elt
   end
 end
 collapsed.each { |elt| elt.render }

Lookbehind + a little mutation.  No magic iterators required...

-mental

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