If you are looking for the actual rules, they are defined in the language reference, in the section on optional chaining expressions [1]. It states that if you have an optional chaining expression of the form `expression?` then
"If the value of the optional-chaining expression is nil, all of the other operations in the postfix expression are ignored and the entire postfix expression evaluates to nil. If the value of the optional-chaining expression is not nil, the value of the optional-chaining expression is unwrapped and used to evaluate the rest of the postfix expression. [...]. If a postfix expression that contains an optional-chaining expression is nested inside other postfix expressions, only the outermost expression returns an optional type." However, it seems to me that this is not quite a correct description of what swiftc does. Note the last sentence in particular. According to the grammar, given a postfix expression <e> and two identifiers x and y, the following are both postfix expressions: <e>?.x.y (<e>?.x).y Without further specification, I would consider "the outermost expression" to be `(<e>?.x).y` for the last case, when swiftc clearly stops at parentheses. The spec should distinguish between parenthesized-expressions and other postfix expressions or am I missing that part? [1] https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Expressions.html#//apple_ref/swift/grammar/optional-chaining-expression On Mon, Aug 1, 2016 at 8:17 PM, Stephen Schaub via swift-users <swift-users@swift.org> 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, 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 > > _______________________________________________ > swift-users mailing list > swift-users@swift.org > https://lists.swift.org/mailman/listinfo/swift-users > _______________________________________________ swift-users mailing list swift-users@swift.org https://lists.swift.org/mailman/listinfo/swift-users
