Why is amap implemented as a member function of TaskPool?

[Atila Neves via Digitalmars-d-learn]

The point is I _want_ a delegate.

Atila

On Thursday, 18 September 2014 at 20:51:30 UTC, Jared wrote:
> On Thursday, 18 September 2014 at 19:49:00 UTC, Atila Neves 
> wrote:
>> Or what I really want to ask: why can't I call amap from 
>> std.parallelism with a lambda? I assume it's because it's a 
>> member function but I'm not 100% sure.
>>
>> Atila
>
> You have to tell DMD that the lambda is not in fact a delegate.
>
> import std.stdio;
> import std.range;
> import std.parallelism;
>
> void main()
> {
>     auto w = iota(0,1_000_000);
>     int[] foo;
>
>     // Not OK, dmd can't infer lambda isn't a delegate
>     // foo = taskPool().amap!(a => a + 1)(w);
>
>     // OK:
>     foo = taskPool().amap!`a+1`(w); // string lambdas, yeah!
>     foo = taskPool().amap!(function int(int a) => a + 1)(w);
>     static int func(int a) { return a + 1; }
>     foo = taskPool().amap!func(w);
> }