From: Magicloud Magiclouds <magicloud.magiclouds@...>
Date: 2008-03-24T19:21:58+09:00
Subject: Re: Question about Array#*

In source, I got this: MEMCPY(RARRAY(ary2)->ptr+i, RARRAY(ary)->ptr, 
VALUE, RARRAY(ary)->len);
So, I just got a bunch of pointer to the same object....


Stefano Crocco wrote:
> On Monday 24 March 2008, Magicloud Magiclouds wrote:
>   
>> Hello,
>> 	In `ri 'Array#*'`, it says that the result is the concatenation of
>> copies of self if the argument is an int.
>> 	So, it is "COPY".
>> 	Then why something like '[[0]] * 10', returns an array, that each
>> element is a pointer to the first element?
>> 	Does this mean "copy of an Array" is just a duplicated pointer?
>>
>> Thanks.
>>     
>
> I think that here the word "copy" has the same meaning it has in clone and 
> dup, that is it means a shallow copy: the array itself is copied, but its 
> contents aren't. In other words, what you get using Array#* with a number is 
> an array of size n times the original one, whose contents are the contents of 
> the original array repeated n times.
>
> I hope this helps
>
> Stefano
>
>
>
>