Yes, parenthesized expressions are primary expressions, which in turn are all postfix expressions as specified by the first grammar rule for postfix expressions and also the note you mentioned. I do think that swiftc's behavior is the most intuitive. I just think the spec needs to be clarified in that regard, that's all.
Cheers, Ingo On Mon, Aug 1, 2016 at 9:24 PM, Stephen Schaub <ssch...@gmail.com> wrote: > Ingo, > > Thanks for pointing me to the specification. The example given in that > section is helpful. > > You make a good point about what constitutes "the outermost expression." > Parenthesized expressions, according to the spec, are primary expressions, > not postfix expressions, but the spec also states that "Syntactically, every > primary expression is also a postfix expression." So, either I'm not > understanding the spec correctly, or the compiler is handling this case > differently for some reason. > > Stephen > > > > > On Mon, Aug 1, 2016 at 3:05 PM, Ingo Maier <ing...@gmail.com> wrote: >> >> 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 >> > > > > > > -- > Stephen Schaub _______________________________________________ swift-users mailing list swift-users@swift.org https://lists.swift.org/mailman/listinfo/swift-users
