From: Martin DeMello <martindemello@...>
Date: 2003-07-09T15:29:09+09:00
Subject: Re: Possible use for a continuation? [Generating all factors of a given number]

Harry Ohlsen <harryo@qiqsolutions.com> wrote:
 
> I have a feeling that some kind of recursive method that returns a
> continuation might be the way to go, but I'm open to any suggestions,
> whatsoever.

Overkill - this should do it:

class MultiCounter
  def initialize(a, b = nil)
    @max = a
    @min = b || a.map {0}
    raise "max and min arrays should have the same length" unless @max.length ==
 @min.length
    @cur = @min
  end

  def next
    inc (@cur.length - 1)
  end

  def inc(i)
    return nil if i < 0
    @cur[i] += 1
    if @cur[i] > @max[i]
      @cur[i] = 0
      return inc(i - 1)
    end
    @cur
  end
end

a = MultiCounter.new([3, 2, 2, 1])
while (b = a.next)
  p b
end

martin