foreach (i; taskPool.parallel(0..2_000_000)

[Steven Schveighoffer]

On 4/4/23 11:34 AM, Salih Dincer wrote:
> On Tuesday, 4 April 2023 at 14:20:20 UTC, Steven Schveighoffer wrote:
>> parallel is a shortcut to `TaskPool.parallel`, which is indeed a 
>> foreach-only construct, it does not return a range.
>>
>> I think what you want is `TaskPool.map`:
>>
>> ```d
>> // untested, just looking at the
>> taskPool.map!(/* your map function here */)
>>    (s.iota(len)).writeln;
>> ```
> 
> 
> I tried, thanks but it goes into infinite loop. For example, the first 
> 50 of the sequence should have been printed to the screen immediately 
> without waiting.
> 
> ```d
> long[50] arr;
> RowlandSequence_v2 range;
> 
> auto next(long a)
> {
>    range.popFront();
>    return arr[a] = range.front();
> }
> 
> void main()
> {
>      range = RowlandSequence_v2(7, 2);
>      taskPool.map!next(iota(50))/*
>      s.iota(50)
>       .map!next
>       .parallel//*/
>       .writeln;
> }
> ```

Keep in mind that `arr` and `range` are thread-local, and so will be 
different states for different tasks.

Though I don't really know what you are doing there.

-Steve