Because generic uses `Car` instead of `Position` when running the code, so
there is no casting as `car as Position` as in your original code. The
`T:Position`
part restricts that the type conforms `Position`, but it won't use `Position`,
it uses the type.
Zhaoxin
On Sat, May 27, 2017 at 4:58 PM, Седых Александр via swift-users <
swift-users@swift.org> wrote:
> Thanks. Why generic function don not require memory allocate for inout
> variable, even if it is protocol type?
>
>
> Пятница, 26 мая 2017, 19:35 +03:00 от Guillaume Lessard <
> gless...@tffenterprises.com>:
>
> In your example, the compiler needs a parameter of type Position. Car is a
> type of Position, but they are not interchangeable. See below:
>
> > On May 26, 2017, at 00:33, Седых Александр via swift-users <
> swift-users@swift.org> wrote:
> >
> > protocol Position {
> > var x: Double { getset }
> > }
> >
> > struct Car: Position {
> > var x: Double
> > }
> >
> > func move(item: inout Position) {
> > item.x += 1
> > }
> >
> > var car = Car(x: 50)
>
> var pos: Position = car
>
> move(item: &pos) // this works.
> assert(pos.x == 51) // works
>
> The move function as you wrote it requires the memory representation of a
> Position variable, which Car does not have; when you assign it to a
> Position variable, the Car struct gets accessed through an indirection
> layer. (There was a WWDC talk about this last year or the year before.)
>
> You may want a generic function instead:
>
> func move<P: Position>(item: inout P) {
> item.x += 1
> }
>
> move(item: &car) // this works, since it’s now calling the generic
> function.
> assert(car.x == 51) // works
>
> Cheers,
> Guillaume Lessard
>
>
>
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