> Le 7 juin 2017 à 13:28, Adrian Zubarev <[email protected]> a
> écrit :
>
>> Xiaodi, Adrian, you are actively pushing so that something that was allowed,
>> well compiled (no runtime issue), and covered actual uses cases, becomes
>> forbidden. Without any developer advantage that would somehow balance the
>> change.
>>
>> That's called a regression.
>
>
> func foo(_: (Int, Int)) {}
> func bar(_: Int, _: Int) {}
>
> type(of: foo) == type(of: bar) //=> true - It's a BUG!
Then please push for the bug to be fixed. You are much better than am I at
that. This does not mean breaking Swift 3 ergonomics.
The bug you mention involves type comparison, which is nowhere to be seen in
the Swift 3 ergonomics checklist below. I'm sure we can all be happy.
func sum1(_ lhs: Int, _ rhs: Int) -> Int { return lhs + rhs }
func sum2(lhs: Int, rhs: Int) -> Int { return lhs + rhs }
func sum3(tuple: (Int, Int)) -> Int { return tuple.0 + tuple.1 }
func sum4(tuple: (lhs: Int, rhs: Int)) -> Int { return tuple.lhs +
tuple.rhs }
// two arguments
func f1(_ closure: (Int, Int) -> Int) { closure(1, 2) }
f1 { lhs, rhs in lhs + rhs }
f1 { (lhs, rhs) in lhs + rhs }
f1 { tuple in tuple.0 + tuple.1 }
f1 { (tuple) in tuple.0 + tuple.1 }
f1(+)
f1(sum1)
f1(sum2)
f1(sum3)
f1(sum4)
// two arguments, with documentation names
func f2(_ closure: (_ a: Int, _ b: Int) -> Int) { closure(1, 2) }
f2 { lhs, rhs in lhs + rhs }
f2 { (lhs, rhs) in lhs + rhs }
f2 { tuple in tuple.0 + tuple.1 }
f2 { (tuple) in tuple.0 + tuple.1 }
f2(+)
f2(sum1)
f2(sum2)
f2(sum3)
f2(sum4)
// one tuple argument
func f3(_ closure: ((Int, Int)) -> Int) { closure((1, 2)) }
f3 { lhs, rhs in lhs + rhs }
f3 { (lhs, rhs) in lhs + rhs }
f3 { tuple in tuple.0 + tuple.1 }
f3 { (tuple) in tuple.0 + tuple.1 }
f3(+)
f3(sum1)
f3(sum2)
f3(sum3)
f3(sum4)
// one keyed tuple argument
func f4(_ closure: ((a: Int, b: Int)) -> Int) { closure((a: 1, b: 2)) }
f4 { lhs, rhs in lhs + rhs }
f4 { (lhs, rhs) in lhs + rhs }
f4 { tuple in tuple.a + tuple.b }
f4 { (tuple) in tuple.a + tuple.b }
f4(+)
f4(sum1)
f4(sum2)
f4(sum3)
f4(sum4)
Gwendal
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