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

[Max Moiseev]

Hi Tim,

Thanks for bringing this up.
Here are my thoughts on the change you’re proposing.

func sequence<T>(first: T, next: (T) -> T?) -> UnfoldFirstSequence<T>

To me the type of the function as it is tells a clear story of what’s going to happen: take the `first`, make it a head of the resulting sequence, and then try to produce the tail by a series of applications of `next`. The only thing that controls when the sequence generation terminates is the result of `next`.

If we change the type of `first` to an Optional<T>, it would make the termination condition non-trivial. After all, the only thing it would do is try to unwrap the `first`, before doing what it needs to, but we already have a `map` for that. One should be able to simply do the `first.map { sequence(first: $0, next: next) } ?? []` but that won’t work with the types very well, unfortunately.

As an alternative, `let first: Int? = ...; sequence(first: first, next: next).flatMap({$0})` (or even `.lazy.flatMap({$0})`) will do the right thing without making an API more complex.

I see the point of `sequence(first:next:)` to be precisely the "generate the non-empty sequence using a seed and a simple producer", for anything more than that, there is `sequence(state:next:)`.

What do you think?

Max

> On Aug 14, 2016, at 4:27 PM, Tim Vermeulen via swift-evolution <swift-evolution at swift.org> wrote:
> 
> sequence(first:next:) takes a non-optional first argument. Is there a reason for that? sequence(state:next:) allows empty sequences, and I don’t see why sequence(first:next:) shouldn’t. The fix would be to simply add the `?` in the function signature; no other changes are required to make it work.
> 
> I considered just filing a bug report, but since this is a change of the public API...
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