Hi Hooman,
Since the protocol P is not class-bounded, the requirement can be witnessed by
a protocol extension method which re-assigns ‘self’:
protocol Initable {
init()
}
extension P where Self : Initable {
mutating func f(_ x: Int) -> Int {
self = Self()
return x
}
}
class C : P, Initable {
required init() {}
}
Now imagine you could do this,
let x: P & AnyObject
x.f(12)
This would be invalid because ‘x’ is a let binding but the requirement ‘f’ is
witnessed by the protocol extension method, which performs a mutating access of
‘self’.
Slava
> On Dec 21, 2017, at 6:01 PM, Hooman Mehr via swift-evolution
> <swift-evolution@swift.org> wrote:
>
> The title is confusing, let me clarify by example:
>
> We have this protocol with a mutating method:
>
> protocol P { mutating func f(_ x: Int) -> Int }
>
> And a conforming class (which has to conform with a non-mutating method):
>
> class C: P { func f(_ x: Int) -> Int { return x } }
>
> An instance of this class can be used with a let constant:
>
> let c = C()
> c.f(1) // OK
>
> If we make it an existential object conforming to P, the immutability of the
> method will be erased:
>
> let c: AnyObject & P = C()
> c.f(1) // Cannot use mutating member on immutable value: 'c' is a 'let'
> constant
>
> A generic context has the same issue:
>
> func f<T: AnyObject & P>(_ arg: T)-> Int { return arg.f(1) } // Cannot use
> mutating member on immutable value: ‘arg' is a 'let' constant
>
> My question:
>
> Is it too much work to preserve method non-mutability in in these cases?
>
> The workaround I am using is this:
>
> protocol Q: class, P { func f(_ x: Int) -> Int } // 'Refine' it to be
> non-mutating.
> extension C: Q {}
>
> // Now these work:
> let c: Q = C()
> c.f(1) // OK
> func f<T: Q>(_ arg: T)-> Int { return arg.f(1) } // OK
>
> This workaround creates a lot of duplication and is hard to maintain. It is
> not something that I do often, but when I do, it is pretty annoying.
>
> Supplemental questions:
>
> Have you guys ever ran into this?
> Is there already a bug tracking this?
>
>
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