Andrej Mitrovic Wrote:

> I'm currently using this:
> 
> import std.algorithm;
> import std.array;
> import std.random;
> import std.range;
> 
> void main()
> {
>     auto arr2 = array(map!( (int){ return uniform(0, 1024); })(iota(0, 1024)));
> }
> 
> Is there a simpler way to do get an array of random values?

Untested:

auto arr = new int[1024];
fill(arr, uniform(0, 1024));

On 8/2/11 3:40 AM, Jesse Phillips wrote:
> Andrej Mitrovic Wrote:
>> Is there a simpler way to do get an array of random values?
>
> Untested:
>
> auto arr = new int[1024];
> fill(arr, uniform(0, 1024));

This does a great job of creating an array containing the same random value 1024 times. ;)

David

On Tue, 02 Aug 2011 03:48:03 +0200, David Nadlinger wrote:

> On 8/2/11 3:40 AM, Jesse Phillips wrote:
>> Andrej Mitrovic Wrote:
>>> Is there a simpler way to do get an array of random values?
>>
>> Untested:
>>
>> auto arr = new int[1024];
>> fill(arr, uniform(0, 1024));
> 
> This does a great job of creating an array containing the same random value 1024 times. ;)
> 
> David

Fine, let me provide a tested one:

import std.algorithm;
import std.array;
import std.random;
import std.range;

void main()
{
    auto arr = new int[1024];
    fill(arr, randomCover(iota(0,1024), rndGen));
}


Knew I had done something with fill before.

On 8/2/11 6:42 AM, Jesse Phillips wrote:
>      auto arr = new int[1024];
>      fill(arr, randomCover(iota(0,1024), rndGen));

Note that this produces a random permutation of the numbers 0 to 1023, which may or may not be what you want.

David