[swift-evolution] Passing an optional first argument to sequence(first:next:)

[Xiaodi Wu]

On Tue, Aug 16, 2016 at 7:34 PM, Braeden Profile via swift-evolution <
swift-evolution at swift.org> wrote:

> Okay, I’m actually confused about the current state of things.
>
> Earlier, this was introduced:
>
> // This function is silly,
> // but useful functions may take a similar form
> func foo(x: Int?) -> Int? {
>   guard let x = x else { return 0 }
>   return x > 12 ? nil : x + 1
> }
>
> let a: Int? = nil
> for i in sequence(first: a, next: { foo($0) })
> {
>   // This is a pretty useless thing to do,
>   // but there are useful things that can be done
>   // without checking whether `i == nil`
>   print(i)
> }
>
> …and it returns 14 lines of output.  But this doesn’t make sense.  I
> expected the anonymous closure for next in sequence(first: Int?, next:
> (Int?) -> Int??) to raise (foo($0)) to (Optional(foo($0))) to avoid having
> the type signature end up being (Int?) -> Int?.  In that case, the result
> of (foo($0)) would always be encapsulated in Optional.Some(), so at the
> 15th run, it would return (Optional.Some(nil)).  Yet it stops when foo
> returns nil.  Why is this?
>

Yeah, I think that's a bug :) Which means it should be fixed whether we
change the function signature or not.


> And if you replace "next: { foo($0) }" with "next: foo", then it compiles
> with the same result.  Replacing it with “next: { Optional(foo($0)) }”
> gives the result I originally expected.
>
> This actually would have made more sense to me if the signature was “func
> sequence<T>(first: T?, next: (T) -> T?) -> UnfoldFirstSequence<T>”, and
> “let a = nil" would have caused an empty sequence.  I understand that this
> is a situation that would change unexpectedly if this stdlib change
> occurred, but I sure think the changed sequence(first:next:) function makes
> for a more understandable result.
>
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