std.range.iota enhancement: supporting more types (AKA issue 10762)

[monarch_dodra]

On Tuesday, 24 December 2013 at 11:57:04 UTC, Jakob Ovrum wrote:
> On Tuesday, 24 December 2013 at 11:47:12 UTC, Francesco 
> Cattoglio wrote:
>> Correct, but there's no way to compute "back" with less than 
>> O(n) complexity, unless division by increment is available. 
>> (Actually, I think we can achieve O(log n) with multiplication 
>> alone, but I think it might be lots of work with very little 
>> benefits)
>> I should have specified better: we need to provide `back', and 
>> Andrei suggested that no primitive should be more than complex 
>> than O(1).
>
> Implement `back` is really trivial.
>
> Simplified example:
> ---
> auto iota(T)(T start, T end)
> {
>     static struct Result
>     {
>         T start, end;
>
>         bool empty() @property { return start == end; }
>         T front() @property { return start; }
>         void popFront() { ++start; }
>         T back() @property { return end; }
>         void popBack() { --end; }
>         // etc
>     }
>
>     return Result(start, --end);
> }
> ---

I think you are missing the point of what happens if the step is 
not 1 (or if the passed in type can have fractional input). EG:

iota(0, 105, 10);
or
iota(0, 10.5);

In this case, "back" should be 100, and not 95. To compute back, 
you need to be able to evaluate length, and to add length*inc to 
front.