Got it, I think! I didn't understand that the chaining of intermediate properties worked that way.
Thanks very much for the explanation. Stephen On Mon, Aug 1, 2016 at 2:25 PM, Saagar Jha <saa...@saagarjha.com> wrote: > When you write `(s?.characters).count`, the parentheses are evaluated > first; `(s?.characters)` gives an `String.CharacterView?`. Accessing > the `String.CharacterView?`’s `count` property requires a > `?`: `(s?.characters)?.count`. `s?.characters.count`, on the other hand, is > applying chaining, which only gives an Optional at the end, intermediate > properties don’t require a `?` unless they’re Optional themselves. > > Saagar Jha > > > > On Aug 1, 2016, at 11:17, Stephen Schaub <ssch...@gmail.com> wrote: > > I understand that the String.characters property is not optional. However, > I am puzzled as to why > > s?.characters.count > > is legal, but > > (s?.characters).count > > is not. This seems counterintuitive. Can someone explain the logic or > rules being used here? > > Stephen > > > > > On Mon, Aug 1, 2016 at 2:09 PM, Saagar Jha <saa...@saagarjha.com> wrote: > >> >> Saagar Jha >> >> This isn’t quite how optional chaining in Swift works; see the Swift >> Programming Guide >> <https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/OptionalChaining.html>, >> specifically “Linking Multiple Levels of Chaining". Basically, >> `s?.characters.count` works because `s.characters` isn’t Optional. You only >> use ? on properties that are Optional. >> >> On Aug 1, 2016, at 10:26, Stephen Schaub via swift-users < >> swift-users@swift.org> wrote: >> >> With optional chaining, if I have a Swift variable >> >> var s: String? >> >> s might contain nil, or a String wrapped in an Optional. So, I tried this >> to get its length: >> >> let count = s?.characters?.count ?? 0 >> >> However, the compiler wants this: >> >> let count = s?.characters.count ?? 0 >> >> or this: >> >> let count = (s?.characters)?.count ?? 0 >> >> My understanding of optional chaining is that, once you start using '?.' >> in a dotted expression, the rest of the properties evaluate as optional and >> are typically accessed by '?.', not '.'. >> >> So, I dug a little further and tried this in the playground: >> >> var s: String? = "Foo" >> print(s?.characters) >> >> The result indicates that s?.characters is indeed an Optional instance, >> indicating that s?.characters.count should be illegal. >> >> Why is s?.characters.count a legal expression? >> >> >> -- >> Stephen Schaub >> _______________________________________________ >> swift-users mailing list >> swift-users@swift.org >> https://lists.swift.org/mailman/listinfo/swift-users >> >> >> > > > -- > Stephen Schaub > > > -- Stephen Schaub
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