[swift-evolution] [swift-evolution-announce] [Review] SE-0184: Unsafe[Mutable][Raw][Buffer]Pointer: add missing methods, adjust existing labels for clarity, and remove deallocation size

[Dave Abrahams]

> On Sep 29, 2017, at 3:48 PM, Taylor Swift <kelvin13ma at gmail.com> wrote:
> 
> 
> 
> On Fri, Sep 29, 2017 at 4:13 PM, Andrew Trick <atrick at apple.com <mailto:atrick at apple.com>> wrote:
> 
> 
>> On Sep 29, 2017, at 1:23 PM, Taylor Swift <kelvin13ma at gmail.com <mailto:kelvin13ma at gmail.com>> wrote:
>> 
>> Instead of
>> 
>>   buf.intialize(at: i, from: source)
>> 
>> We want to force a more obvious idiom:
>> 
>>   buf[i..<n].intialize(from: source)
>> 
>> 
>> The problem with subscript notation is we currently get the n argument from the source argument. So what would really have to be written is 
>> 
>> buf[i ..< i + source.count].initialize(from: source) 
>> 
>> which is a lot more ugly and redundant. One option could be to decouple the count parameter from the length of the source buffer, but that opens up the whole can of worms in which length do we use? What happens if n - i is less than or longer than source.count? If we enforce the precondition that source.count == n - i, then this syntax seems horribly redundant. 
> 
> Sorry, a better analogy would have been:
> 
>  buf[i...].intialize(from: source)
> 
> Whether you specify the slice’s end point depends on whether you want to completely initialize that slice or whether you’re just filling up as much of the buffer as you can. It also depends on whether `source` is also a buffer (of known size) or some arbitrary Sequence.
> 
> Otherwise, point  taken.
> 
> -Andy
> 
> After thinking about this more, one-sided ranges might provide just the expressivity we need. What if:
> 
> buf[offset...].initialize(from: source) // initializes source.count elements from source starting from offset
> 
> buf[offset ..< endIndex].initialize(from: source) // initializes up to source.count elements from source starting from offset
> 
> 
> The one sided one does not give a full initialization guarantee. The two sided one guarantees the entire segment is initialized.

In every other context, x[i...] is equivalent to x[i..<x.endIndex]

I don't think breaking that precedent is a good idea.

> For move operations, the one sided one will fully deinitialize the source buffer while the two sided one will only deinitialize endIndex - offset elements. 

—
-Dave





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