It looks intentional.
Function conversions can be performed on argument and return types that are
related via ‘Subtype’ constraints in the constraint solver, however array
conversions are only possible via the ‘Conversion’ relation.
Changing this would require some additional work in SILGen to support such
function conversions.
Slava
> On Feb 22, 2017, at 3:06 PM, Joe Groff via swift-users
> <swift-users@swift.org> wrote:
>
>>
>> On Feb 22, 2017, at 8:50 AM, Michael Roitzsch via swift-users
>> <swift-users@swift.org> wrote:
>>
>> Hi all,
>>
>> I am fairly new to Swift, so this may very well be a simple misunderstanding
>> on my part.
>>
>> I was exploring the subtyping rules of Swift, especially regarding
>> covariance. I came across two examples where the outcome puzzles me and I
>> would appreciate if someone could explain this to me.
>>
>> First I tried the covariance for the standard container types. Here is an
>> excerpt from a REPL session:
>>
>> Michael@carpo:~ > swift
>> Welcome to Apple Swift version 3.0.2 (swiftlang-800.0.63 clang-800.0.42.1).
>> Type :help for assistance.
>> 1> open class Super {}; class Sub: Super {}
>> 2> print([Sub()] is [Sub])
>> true
>> 3> print([Sub()] is [Super])
>> true
>> 4> print(Array<Sub>(arrayLiteral: Sub()) is [Sub])
>> true
>> 5> print(Array<Sub>(arrayLiteral: Sub()) is [Super])
>> error: repl.swift:5:39: error: 'Super' is not a subtype of 'Sub'
>> print(Array<Sub>(arrayLiteral: Sub()) is [Super])
>> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~
>>
>> Why does it matter for subtyping against [Super] whether I express an array
>> as [Sub] or Array<Sub>(arrayLiteral: Sub())?
>>
>> Then I tried combining array covariance with function argument type
>> contravariance:
>>
>> Michael@carpo:~ > swift
>> Welcome to Apple Swift version 3.0.2 (swiftlang-800.0.63 clang-800.0.42.1).
>> Type :help for assistance.
>> 1> open class Super {}; class Sub: Super {}
>> 2> func f(_: [Super]) {}
>> 3> let test: ([Sub]) -> Void = f
>> error: repl.swift:3:29: error: cannot convert value of type '([Super]) ->
>> ()' to specified type '([Sub]) -> Void'
>> let test: ([Sub]) -> Void = f
>> ^
>>
>> Why is this assignment not possible? It works fine (as expected) when using
>> plain Super and Sub instead of arrays.
>
> Those are both bugs. There's no fundamental reason these should be errors.
>
> -Joe
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