`((Int, Int)) -> Void` will be same type as `(Int, Int) -> Void`
2017-06-07 18:09 GMT+08:00 Adrian Zubarev <adrian.zuba...@devandartist.com>:
> Keep in mind there is also SE–0111 cometary which promises sugar for
> parameter labels for closures:
>
> // **
> let foo(tuple:): ((Int, Int)) -> Void
>
> // Sugar for **
> let foo: (tuple: (Int, Int)) -> Void
>
> What will happen if you’d always flatten here?
>
>
>
> --
> Adrian Zubarev
> Sent with Airmail
>
> Am 7. Juni 2017 um 12:03:08, Adrian Zubarev (adrian.zubarev@devandartist.
> com) schrieb:
>
> Well please no:
>
> let fn2: ((Int, Int)) -> Void = { lhs, rhs in }
>
> Instead use destructuring sugar pitched by Chris Lattner on the other
> thread:
>
> let fn2: ((Int, Int)) -> Void = { ((lhs, rhs)) in }
>
> That’s a correct error:
>
> let fn3: (Int, Int) -> Void = { _ in }
>
> This should be allowed, because we might want to work with the whole tuple
> and not a desctructured elements only:
>
> let fn4: ((Int, Int)) -> Void = { tuple in }
>
>
>
> --
> Adrian Zubarev
> Sent with Airmail
>
>
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