This was a deliberate change between Swift 3 beta 1 and beta 2 after a friend
of mine pointed the following inconsistency out to me:
struct Foo {
func bar() {}
}
let foo: Foo? = Foo()
foo?.bar() // Does not create a warning
true ? foo?.bar() : foo?.bar() // expression of type '()?' is unused
After some offline discussion at WWDC with the Swift team we decided to move to
a consistent model where ()?, ()??, … is always discardable since we didn't
want to take the convenience of foo?.bar() away (something that regularly
occurs with weak variables, e.g. captures in closures).
So much for the history of this feature.
– Alex
> On 30 Jan 2017, at 22:58, Daniel Duan via swift-evolution
> <[email protected]> wrote:
>
> Hi all,
>
> Right now, expressions that evaluates to Optional<()>,
> Optional<Optional<()>>… gets special treatment when it’s unused. For example:
>
> func f(s: String) {}
> let s: String = “”
> s.map(f) // no warning here, even tho the resulting type is `Optional<()>`
> and unused.
>
> func g() throws {}
> try? g() // no warnings here neither.
>
> This is convenient, but encourages composing map/filter/reduce, etc with
> side-effect-ful functions, which we have found a few cases of in our
> production code recently. Granted, these cases could’ve been caught with more
> careful code reviews. But we wouldn’t have missed them if this “feature”
> didn’t exist.
>
> I think we should remove the special treatment so that code in the example
> above would generate a warning about `()?` being unused. Users can silence it
> manually by assigning the result to `_`.
>
> OTOH, this would undermine the convenience of `try?` when the throwing
> function don’t return anything.
>
> What do y’all think?
>
> Daniel Duan
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> [email protected]
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