From: Sylvester Keil <sylvester.keil@...>
Date: 2011-11-29T18:48:41+09:00
Subject: Re: Is high-speed sorting impossible with Ruby?


On Nov 29, 2011, at 12:02 AM, Ryan Davis wrote:

> 
> On Nov 28, 2011, at 06:34 , Sylvester Keil wrote:
> 
>> 
>> On Nov 28, 2011, at 3:17 PM, Gaurav C. wrote:
>> 
>>> Douglas Seifert wrote in post #1034017:
>>>>> 
>>>>>> a = Array.new(1e6+1) {|n| ""}
>>>>> 
>>>> 
>>>> The above creates an Array of 1,000,001 slots.  The block syntax lets
>>>> you
>>>> provide a default object dynamically.  The first time you reference a
>>>> particular index in the array, the block is called to allocate the
>>>> default
>>>> object at that position.
>>>> 
>>> 
>>> Hi Doug,
>>> 
>>> Thanks for the explanation.
>>> Didn't understand the Array initialization part.
>>> What's the difference between Array.new(1e6+1) {|n| ""} and
>>> Array.new(1e6+1, "")?
>> 
>> The former always returns a new string instance (this is what you want); the former creates one string and always returns a reference.
> 
> I don't think this really explains it adequately.

You're absolutely right; what I was describing was more like default blocks on Hashes and not the behavior of Array.new.


> The former initializes the array by running the block on every index. The latter assigns the same object to every index.
> 
> From ri:
> 
>  Array.new(size=0, obj=nil)
>  Array.new(array)
>  Array.new(size) {|index| block }
> 
> ------------------------------------------------------------------------------
> 
> Returns a new array. In the first form, the new array is empty. In the second
> it is created with size copies of obj (that is, size
> references to the same obj). The third form creates a copy of the array
> passed as a parameter (the array is generated by calling to_ary  on the
> parameter). In the last form, an array of the given size is created. Each
> element in this array is calculated by passing the element's index to the
> given block and storing the return value.
> 
>