[swift-users] Optional chaining and String properties

[Stephen Schaub]

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 <saagar at 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 <sschaub at 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 <saagar at 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 at 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 at swift.org
>> https://lists.swift.org/mailman/listinfo/swift-users
>>
>>
>>
>
>
> --
> Stephen Schaub
>
>
>


-- 
Stephen Schaub
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