> 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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