# Is this array operation correct? **URL:** \ **Category:** ruby-talk\ **Created:** [14 August 2002 01:59 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186 "2002-08-14T01:59:50Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![Mauricio](https://avatars.discourse-cdn.com/v4/letter/m/90db22/32.png) [@Mauricio](https://rubytalk.org/u/Mauricio)\ **Post date:** [14 August 2002 01:59 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/1 "2002-08-14T01:59:50Z")
Hi, ``` I've tried this: ``` a = Array.new(3,[]) =\> [[], [], []] a[1].push(3) =\> [[3],[3],[3]] Is this correct? a[1].push(3) is the same as a[2].push(3)? If the constructor creates the same copy for all array elements, shouldn’t the array create copies when writing? ``` Thanks, Maurício ``` ---
**Author:** ![David\_Alan\_Black1](https://avatars.discourse-cdn.com/v4/letter/d/278dde/32.png) [@David\_Alan\_Black1](https://rubytalk.org/u/David_Alan_Black1)\ **Post date:** [14 August 2002 02:38 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/2 "2002-08-14T02:38:47Z")
Hello – > Hi, > > ``` > I've tried this: > > ``` > > a = Array.new(3,) =\> [, ,] > a[1].push(3) =\> [[3],[3],[3]] > > Is this correct? a[1].push(3) is the same as a[2].push(3)? If the > constructor creates the same copy for all array elements, shouldn’t the > array create copies when writing? I’m not sure I follow the question, but the general thing going on is: a[0] and a[1] and a[2] are references to the same object, so if you change the object through one of those references, the change will be visible in all of them. It’s as if you said: ar = a = [ar, ar, ar] ar.push(3) and then ‘a’ would look like: [[3],[3],[3]] David > **···** > > On Wed, 14 Aug 2002, Maurício wrote: > > – > David Alan Black > home: [dblack@candle.superlink.net](mailto:dblack@candle.superlink.net) > work: blackdav@shu.edu > Web: [http://pirate.shu.edu/~blackdav](http://pirate.shu.edu/~blackdav) ---
**Author:** ![Christian\_Boos](https://avatars.discourse-cdn.com/v4/letter/c/ce7236/32.png) [@Christian\_Boos](https://rubytalk.org/u/Christian_Boos)\ **Post date:** [14 August 2002 08:23 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/3 "2002-08-14T08:23:36Z")
> From: [dblack@candle.superlink.net](mailto:dblack@candle.superlink.net) [[mailto:dblack@candle.superlink.net](mailto:dblack@candle.superlink.net)] > Sent: Wednesday, August 14, 2002 4:39 AM > To: ruby-talk ML > Subject: Re: Is this array operation correct? > > Hello – > > > Hi, > > > > ``` > > I've tried this: > > > > ``` > > > > a = Array.new(3,) =\> [, ,] > > a[1].push(3) =\> [[3],[3],[3]] … > It’s as if you said: > > ar = > a = [ar, ar, ar] > ar.push(3) > > and then ‘a’ would look like: [[3],[3],[3]] > > > Then, do something like: Array.new(3).map { } Is there something shorter/more efficient? – Christian > **···** > > > -----Original Message----- > > On Wed, 14 Aug 2002, Maurício wrote: ---
**Author:** ![Mrchameleon](https://avatars.discourse-cdn.com/v4/letter/m/8e7dd6/32.png) [@Mrchameleon](https://rubytalk.org/u/Mrchameleon)\ **Post date:** [14 August 2002 08:31 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/4 "2002-08-14T08:31:24Z")
> > ``` > > I've tried this: > > > > ``` > > > > a = Array.new(3,) =\> [, ,] > > a[1].push(3) =\> [[3],[3],[3]] You may be wanting this … a = (0…3).collect { Array.new } a[1].push(3) =\> [,[3],] Yours Chris ---
**Author:** ![ts1](https://avatars.discourse-cdn.com/v4/letter/t/a8b319/32.png) [@ts1](https://rubytalk.org/u/ts1)\ **Post date:** [14 August 2002 08:32 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/5 "2002-08-14T08:32:58Z")
> Array.new(3).map { } > Is there something shorter/more efficient? With 1.7 you can give it a block pigeon% ruby -ve 'a = Array.new(3) { }; a[1].push 1; p a' ruby 1.7.2 (2002-08-06) [i686-linux] [, [1], ] pigeon% Guy Decoux ---
**Author:** ![George\_Ogata1](https://avatars.discourse-cdn.com/v4/letter/g/d9b06d/32.png) [@George\_Ogata1](https://rubytalk.org/u/George_Ogata1)\ **Post date:** [15 August 2002 02:14 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/6 "2002-08-15T02:14:07Z")
> Then, do something like: > > Array.new(3).map { } > > Is there something shorter/more efficient? If you don’t like using Array.new, you can #map to a Range: (1…3).map{} (Dunno about efficiency, here.) If you’re going to map to a fresh Array, though, you may as well use the in-place version: Array.new(3).map!{} ---
**Author:** ![Gavin\_Sinclair](https://avatars.discourse-cdn.com/v4/letter/g/db5fbb/32.png) [@Gavin\_Sinclair](https://rubytalk.org/u/Gavin_Sinclair)\ **Post date:** [15 August 2002 03:12 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/7 "2002-08-15T03:12:24Z")
> > Then, do something like: > > > > Array.new(3).map { } > > > > Is there something shorter/more efficient? > > If you don’t like using Array.new, you can #map to a Range: > > (1…3).map{} > > (Dunno about efficiency, here.) > > If you’re going to map to a fresh Array, though, you may as well use the > in-place version: > > Array.new(3).map!{} Speaking from inexperience, I doubt there’s any significant efficiency difference between the two solutions mentioned above. –Gavin ---
**Author:** ![Christian\_Boos](https://avatars.discourse-cdn.com/v4/letter/c/ce7236/32.png) [@Christian\_Boos](https://rubytalk.org/u/Christian_Boos)\ **Post date:** [15 August 2002 07:08 UTC](https://rubytalk.org/t/is-this-array-operation-correct/1186/8 "2002-08-15T07:08:11Z")
# Really basic benchmarking (for 1.6.5 ) … def time\_it GC.start now = Time.now yield Time.now - now end time\_it { Array.new(100000).map {} } 0.611 time\_it { Array.new(100000).map {} } 0.631 time\_it { (1..100000).map {} } 0.471 time\_it { (1..100000).map {} } 0.471 time\_it { Array.new(100000).map! {} } 0.351 time\_it { Array.new(100000).map! {} } 0.361 Using map! seems the fastest way, thanks for the tip! – Christian > **···** > > > -----Original Message----- > > From: George Ogata [[mailto:g\_ogata@optushome.com.au](mailto:g_ogata@optushome.com.au)] > > Sent: Thursday, August 15, 2002 4:14 AM > > To: ruby-talk ML > > Subject: RE: Is this array operation correct? > > > > > Then, do something like: > > > > > > Array.new(3).map { } > > > > > > Is there something shorter/more efficient? > > > > If you don’t like using Array.new, you can #map to a Range: > > > > (1..3).map{} > > > > (Dunno about efficiency, here.) > > > > If you’re going to map to a fresh Array, though, you may as well use the > > in-place version: > > > > Array.new(3).map!{}