class Range
  def each(&block)
    current = @first
    while current < @last
      yield current
      current = current.succ
    end
  end
 
  def step(step_size,&block)
    counter = 0
    each do |o|
      yield o if counter % step_size = 0
      counter += 1
    end
  end
end

(见the Rubinius source code)

对于Time对象,#succ会在一秒后返回时间.因此,即使你每周都要求它,但无论如何它必须在两次之间逐步完成.

编辑：解决方案

构建一系列Fixnum,因为它们具有优化的Range#step实现.
就像是：

date_counter = Time.mktime(2011,"+05:00")
@weeks = Array.new
 
(date_counter.to_i..Time.now.to_i).step(1.week).map do |time|
  Time.at(time)
end.each do |week|
  logger.debug "WEEK: " + week.inspect
  @weeks << week
end