Do sorted ranges have any special properties?

[Tomek Sowiński]

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