rary/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
> wrote:
> > I understand that the String.characters property is not o
ally,
>> `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,
“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
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:
