> On 7 Jul 2016, at 02:06, Dmitri Gribenko <[email protected]> wrote:
> 
> On Wed, Jul 6, 2016 at 4:21 AM, Karl <[email protected]> wrote:
>> I had a PR open for this which added a Collection specialisation, but you
>> said this could be handled in a more general way which allows for more
>> optimised mutations. I’m curious how this would work; how does
>> `RangeReplaceableCollection.replaceSubrange<S:Sequence>(range:, with: S)`
>> call a more optimised implementation if S also conforms to Collection, if
>> not by adding a specialisation?
> 
> The RRC can call into S.someCustomizationPoint(), which will
> initialize a region of memory in the most efficient way possible,
> since it has complete knowledge about the memory layout of S.
> 
> Dmitri
> 
> -- 
> main(i,j){for(i=2;;i++){for(j=2;j<i;j++){if(!(i%j)){j=0;break;}}if
> (j){printf("%d\n",i);}}} /*Dmitri Gribenko <[email protected]>*/

Sorry, it’s been a little while since I looked at it 
(https://github.com/apple/swift/pull/3067 
<https://github.com/apple/swift/pull/3067>). Actually, the issue is with the 
append() function specifically - it only has a Sequence specialisation. That’s 
the point; we were losing type information and not calling the more optimised 
replaceSubrange — which is exactly the specialisation point you are talking 
about, Dmitry (I think everything funnels down in to replaceSubrange).

I must have misunderstood what you were saying at the time. I’ll have to test, 
but I think it’s still an issue. I thought there was some more general work on 
RRC planned, so when I saw the bullet I thought maybe that was it.

@Dave:

I was trying to initialise a String.CharacterView - depending on whether I 
initialised it with another CharacterView, or created it empty and appended the 
other CharacterView, I would get huge performance differences, as the other 
CharacterView would get split up and a new Character created and appended for 
every character in the string (which had like 10,000 characters). It’s a subtle 
bug and a bit nasty when it hits you.

Karl


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