How to sort a range

[Ali Çehreli via Digitalmars-d-learn]

On 03/09/2016 06:50 PM, rcorre wrote:

 > sort calls to quicksort (for unstable, at least) which uses
 > swapAt. swapAt takes the range by value, so it just swaps the values in
 > its local copy.

Remembering that a range is not the collection, swapAt takes the range 
by value but it does not copy the elements. So, sort() does sort the 
original array:

import std.algorithm;

void main() {
     auto a = [ 9, -1, 2, 0 ];
     a.sort();
     assert(a == [ -1, 0, 2, 9 ]);
}

 > The original OnlyResult is untouched. I guess a static
 > array slice maintains a pointer to the underlying array (which is why
 > returning one from a function errors with 'escaping reference to local
 > variable').

Yes: A static array is just a collection of elements, which implicitly 
converts to a slice and a slice is nothing but a pair of "pointer to the 
first element" and "number of elements".

Ali