Thank you Matthew, I will try to digest and incorporate your explanation.
On Sun, Dec 10, 2017 at 10:52 PM, Matthew Johnson <matthew@anandabits.com >> > wrote:
Sent from my iPad
On Dec 10, 2017, at 3:41 PM, Jens Persson via swift-users < >>> swift-users@swift.org> wrote:
I'm trying to get my head around the current behavior, but its very hard
to understand and remember, and judging by the comments here and on my bug
report (SR-6564), so does even people in the core team. It would be nice if
someone could present a complete set of rules (all the ones I've seen are
far to simplified and does not cover all cases).
Here's another example to consider:
protocol P {
associatedtype T
func f() // *
}
extension P {
func f() { print("T is unknown") }
}
extension P where T == Int {
func f() { print("T is Int") }
}
struct X<T> : P {
FWIW, this does not compile. You need to provide a typealias for T
which you can’t do when the generic parameter is named T.
func f() { print("this one is actually best") }
}
extension X where T == Int {
func f() { print("this one is actually better than best.") }
}
struct Y<U> where U: P, U.T == Int {
typealias T = U.T
var a: U
func g() { a.f() }
}
let x = X<Int>()
x.f() // What will this print?
This prints “this one is actually better than best.” because the method
is invoked on a concrete type. Overload resolution is used to identify the
most specific implementation which in this case is the method in the
concrete extension on X.
let y = Y(a: X<Int>())
y.g() // What will this print?
This prints ”this one is actually best”. This is because the method is
called in a generic context and is a protocol requirement. This means it
is dispatched through the protocol witness table. The methods in the
extensions on P are default implementations which are disregarded because X
provides its own implementation. The overload in the extension on X is not
visible at all in a generic context because it does not participate in X’s
conformance to P.
If anyone knows for sure what this program will print (without having to
run it), please enlighten me!
I hope the above helps. If you have further questions please ask!
/Jens
On Sat, Dec 9, 2017 at 1:54 AM, Jordan Rose <jordan_rose@apple.com> >>> wrote:
Consider this example:
protocol P {
associatedtype T
func f() // *
}
extension P {
func f() { print("T is unknown") }
}
extension P where T == Int {
func f() { print("T is Int") }
}
struct X<T> : P {
func f() { print("this one is actually best") }
}
struct Y<U> where U: P, U.T == Int {
typealias T = U.T
var a: U
func g() { a.f() }
}
let x = X<Int>()
x.f() // "this one is actually best"
let y = Y(a: X<Int>())
y.g() // "this one is actually best"
I can't think of any other choice for 'a.f()' that would preserve this
behavior, which means we're doing the best thing: prefer dynamic dispatch
over static dispatch when operating on a generic value. (Or at least the
"least bad" thing.) And of course this reasoning has to be local; Y.g can't
look and say "oh, I know nothing else conforms to P, and X {does, doesn't}
have a custom implementation, so I should pick the constrained extension
instead".
The real answer might be "we should have had a different syntax for
default implementations vs. mixin operations", but that's a much bigger can
of worms.
Jordan
On Dec 8, 2017, at 13:07, Jens Persson via swift-users < >>>> swift-users@swift.org> wrote:
Thanks Slava and Greg,
(
I'm aware that it prints "T is Int" from both calls if I remove func
f() from P itself, that's why I wrote "... unless * is commented out."
in the comment of the last line
Note that the "U.T == Int"-part of
struct Y<U> where U: P, U.T == Int {
is key here. If it had been only
struct Y<U> where U: P {
then I hadn't been surprised that it printed "T is unknown".
)
Filed [SR-6564] Compiler seems to ignore constraint on generic type parameter · Issue #49114 · apple/swift · GitHub since I think it is just
strange that the compiler should not use its knowledge of U.T == Int when
choosing between the two f()-implementations.
I think I will be a little disappointed if the solution is to deem it
an ambiguity
: )
/Jens
On Fri, Dec 8, 2017 at 9:19 PM, Greg Parker <gparker@apple.com> wrote:
Evidence in favor of Slava's analysis: if you remove `func f()` from P
itself, leaving it in the extensions only, then you get "T is Int" from
both calls.
> On Dec 8, 2017, at 12:12 PM, Slava Pestov via swift-users < >>>>> swift-users@swift.org> wrote:
>
> Hi Jens,
>
> I think the problem is that overload ranking always prefers a
protocol requirement to a protocol extension member, because usually you
want the dynamic dispatch through the requirement instead of calling the
default implementation. But it appears that this heuristic does not take
into account the fact that the protocol extension member could be more
constrained than the requirement.
>
> Please file a bug, but it is unclear what the desired behavior
actually is here. Perhaps it should just diagnose an ambiguity.
>
> Slava
>
>> On Dec 8, 2017, at 6:25 AM, Jens Persson via swift-users < >>>>> swift-users@swift.org> wrote:
>>
>> Hi all!
>>
>> Can someone please explain the rationale behind the last line
printing
>> "T is unknown"
>> rather than (what I would expect):
>> "T is Int"
>> in the following program?
>>
>>
>> protocol P {
>> associatedtype T
>> func f() // *
>> }
>> extension P {
>> func f() { print("T is unknown") }
>> }
>> extension P where T == Int {
>> func f() { print("T is Int") }
>> }
>>
>> struct X<T> : P {}
>>
>> struct Y<U> where U: P, U.T == Int {
>> // NOTE: The compiler/type-checker knows that U.T == Int here so
...
>> typealias T = U.T
>> var a: U
>> func g() { a.f() } // ... how/why could this print anything but
"T is Int"?
>> }
>>
>> let x = X<Int>()
>> x.f() // Prints "T is Int", no matter if * is commented out or not.
>>
>> let y = Y(a: X<Int>())
>> y.g() // Prints "T is unknown" unless * is commented out. Why?
>>
>>
>> IMHO this looks like the compiler simply ignores that struct Y<U>
has the constraint U.T == Int.
>> How else to explain this behavior?
>> /Jens
>>
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>
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