Concatenate 2 ranges

[RazvanN via Digitalmars-d-learn]

On Friday, 11 November 2016 at 13:33:20 UTC, Vladimir Panteleev 
wrote:
> On Friday, 11 November 2016 at 13:30:17 UTC, RazvanN wrote:
>> I know that I can use the .array property, but I think that 
>> this iterates through all of my elements. Using 
>> assumeSorted(chain(r1, r2).array) will return a SortedRange, 
>> but I am not sure what the complexity for this operation is.
>
> .array allocates, so it's going to be O(n), but the allocation 
> will probably be more expensive.
>
>> Is there a way to concatenate 2 ranges (SortedRange in my 
>> case) in O(1) time?
>
> assumeSorted(chain(a, b)) ?
>
> This works for me:
>
> auto r = assumeSorted(chain([1, 2, 3].sort(), [1, 2, 
> 3].sort()));

It does work, the problem is that [1, 2, 3].sort() is of type 
SortedRange(int[], "a < b") while r is of type 
SortedRange(Result, "a < b"). This is a problem if you want to 
return r in a function which has return type SortedRange(int[], 
"a < b").