[swift-users] withoutActuallyEscaping example question

[Ray Fix]

Wow.  Thanks for the explanation! I didn’t realize I was toggling between different overloads but that makes sense now.

Thanks again,
Ray

PS: It is sort of a strange example in the first place (being lazy and then immediately eager.)  I don’t think it is a use case that is compelling enough to consider overload resolution rules over.

> On Mar 26, 2017, at 1:21 AM, Slava Pestov <spestov at apple.com> wrote:
> 
> Here’s another fun one: the following type checks despite the literal closure; we’ve constrained the return type of filter() to [Int], and not LazyFilterCollection<Int>, so again the type checker picks the overload that takes the non-escaping closure:
> 
> func myFilter(array: [Int], predicate: (Int) -> Bool) -> [Int] {
>   return array.lazy.filter { predicate($0) }
> }
> 
> Slava
> 
>> On Mar 26, 2017, at 1:14 AM, Slava Pestov via swift-users <swift-users at swift.org <mailto:swift-users at swift.org>> wrote:
>> 
>> Hi Ray,
>> 
>> There are two overloads of filter() available on ‘array.lazy’; the version that takes an escaping closure and returns a LazyFilterCollection, and the version that takes a non-escaping closure and returns [Int].
>> 
>> In the first example, we pick the LazyFilterCollection-returning overload, because the literal closure { predicate($0) } can be coerced to both an escaping or a non-escaping closure type, and in the absence of additional constraints we go with the overload from a concrete type over an overload in a protocol extension. After the overload has been picked we validate the body of the closure, and notice that it is invalid because whole the closure is already known to be @escaping, it references the non- at escaping ‘predicate’.
>> 
>> In the second example, ‘predicate’ is known to be non- at escaping, which rules out the first overload completely, so we go with the second overload and perform a non-lazy filter.
>> 
>> I would argue this is somewhat confusing, but it might be difficult to change the overload resolution rules in a way where the first overload is always chosen.
>> 
>> Slava
>> 
>>> On Mar 26, 2017, at 12:13 AM, Ray Fix via swift-users <swift-users at swift.org <mailto:swift-users at swift.org>> wrote:
>>> 
>>> 
>>> Hi,
>>> 
>>> One of the motivating examples for withoutActuallyEscaping looks like the following.  This predictably  doesn’t work because "use of non-escaping parameter 'predicate' may allow it to escape"
>>> 
>>> func myFilter(array: [Int], predicate: (Int) -> Bool) -> [Int] {
>>>   return Array(array.lazy.filter { predicate($0) })
>>> }
>>> 
>>> The solution is to use withoutActuallyEscaping.  This works and produces the expected results.
>>> 
>>> func myFilter(array: [Int], predicate: (Int) -> Bool) -> [Int] {
>>>   return withoutActuallyEscaping(predicate) { predicate in
>>>     Array(array.lazy.filter({predicate($0)}))
>>>   }
>>> }
>>> 
>>> What I find puzzling is the below example compiles and runs correctly. It seems like it should be the same compiler error as in the first example.
>>> 
>>> func myFilter(array: [Int], predicate: (Int) -> Bool) -> [Int] {
>>>   return Array(array.lazy.filter(predicate))
>>> }
>>> 
>>> If you understand why this is so, it would be very helpful.
>>> 
>>> Thank you and best wishes,
>>> Ray
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>> 
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