Do sorted ranges have any special properties?

[Andrei Alexandrescu]

Tomek Sowiński wrote:
> Philippe Sigaud wrote:
> 
>>> Yah but the type is a range.
>>
>> But then, so are Map and Filter.
>> Personally, I do not consider them algorithms per se, they are too
>> low-level: they're building blocks and, as ranges, should go in std.range.
>> LevenshteinDistance or bringToFront, those are algorithms.
>>
>> My own experience of this is that most of the time, I import std.range
>> only to get map/filter/reduce. That is, without reaching for a real
>> algorithm. They are in std.algo for 'historical' reasons, because std.algo
>> precedes std.range.
> 
> OK, range then.
> 
> From a different angle, I was thinking that assumptions should give more; 
> look at this:
> 
> auto range = [2, 3, 4];
> auto sorted = assumeSorted(r);
> range[0] = 666; // BANG! assumeSorted replaced range with Range.init
> sorted[0] = 666; // BANG! assertion in Sorted's setter went off
> range = sorted.dropAssumption;
> sorted[0] = 666;  // BANG! now sorted's range reference is nulled.
> range[0] = 666;  // OK... phew!
> 
> Assumption producers like sort() would take the range by ref to comply with 
> this idiom. Of course if you save() the range before it's eaten by 
> assumeSorted, you can twiddle away through the saved range. Still, it's 
> better than nothing. Oh, and the unique-like behavior can be expanded onto 
> assumeUnique and, I dare say, assumeWhatever.
> 
> Like it?
>  
> 
> Tomek

Like it, just one issue - destructive copy is unlikely to solve any 
problems. The big problems arise with long-distance aliasing, much less 
from messing with the source of the assignment directly.

Andrei