Two chunks but No allocation

[Salih Dincer]

On Wednesday, 27 March 2024 at 20:50:05 UTC, rkompass wrote:
> This works:

I decided to give the full code. Maybe then it will be better 
understood what I mean. I actually pointed out the indirect 
solution above but it's a bit ugly and I'm sure there must be a 
better way?

```d
import std.datetime.stopwatch;
import std.stdio, std.algorithm,
        std.range;

auto cube(T)(T n)
   => n * n * n;

auto S1(T, R)(R a, T n)
   => n.cube - n; //

auto s1(T)(T first, size_t skip = 3)
   => recurrence!S1(0, first).drop(skip);

enum ITERATIONS = 14_000;
enum BENCHMARKS = 100;

void main()
{
     real result;
     long total_time = 0;

     for(int i = 0; i < BENCHMARKS; i++)
     {
       auto sw = StopWatch(AutoStart.no);
       sw.start();

       result = s1(3.0).take(ITERATIONS)
                       .chunks(2)
                       .map!(r => 4/r.front)
                       .array // <- can't it be done without this?
                       .chunks(2)
                       .map!"a[0] - a[1]"
                       .fold!"a + b"(3.0);
       sw.stop();
       total_time += sw.peek.total!"nsecs";
     }
     writefln("%f (nsecs)", total_time / BENCHMARKS / 1e9);
     writefln("%0.16f (result)", result);```
} /*
0.000456 (nsecs)
3.1415926535890670 (result)
//*/
```

SDB at 79