From: Benoit Daloze <eregontp@...>
Date: 2010-08-02T16:02:07+09:00
Subject: Re: Lazy array

On 2 August 2010 05:22, jeremy Ruten <jeremy.ruten@gmail.com> wrote:
> Ruby 1.9 has the Enumerator class, which can act like a lazy list I
> think. A good example is given in the documentation:
>
>  fib = Enumerator.new { |y|
>    a = b = 1
>    loop {
>      y << a
>      a, b = b, a + b
>    }
>  }
>
>  p fib.take(10) #=> [1, 1, 2, 3, 5, 8, 13, 21, 34, 55]
>
> Also, this is a nice trick:
>
>  Infinity = 1.0/0
>
>  range = 5..Infinity
>  p range.take(10) #=> [5, 6, 7, 8, 9, 10, 11, 12, 13, 14]
>
> This one only works for consecutive values though.
>
> jeremy
>
> On Sun, Aug 1, 2010 at 8:56 PM, skim <skim.la@gmail.com> wrote:
>> Ruby Arrays dynamically expand as needed.  You can apply blocks to them to
>> return things like even numbers.
>>
>> array = []
>> array.size # => 0
>> array[0] # => nil
>> array[9999] # => nil
>> array << 1
>> array.size # => 1
>> array << 2 << 3 << 4
>> array.size # => 4
>>
>> array = (0..9).to_a
>> array.select do |e|
>>  e % 2 == 0
>> end
>>
>> # => [0,2,4,6,8]
>>
>> Does this help?
>>
>> On Sun, Aug 1, 2010 at 17:45, Andrew Wagner <wagner.andrew@gmail.com> wrote:
>>
>>> How can I get a lazy array in Ruby? E.g., in Haskell, I can talk about
>>> [1..], which is an infinite list, lazily generated as needed. I can also do
>>> things like "iterate (+2) 0", which applies whatever function I give it to
>>> generate a lazy list. In this case, it would give me all even numbers.
>>> Anyway, I'm sure I can do such things in Ruby, but can't seem to work out
>>> how.
>>>
>>
>

So, with Enumerator it should be:

even = Enumerator.new { |y|
  i = 0
  loop {
    y << i
    i += 2
  }
}

p even.take(5) # => [0, 2, 4, 6, 8]

But that is not really common practice in Ruby (Enumerator for so
simple things).

You would probably need something like (0..max).step(2) { |even_n| p even_n }

B.D.