Optional chaining with Swift strings

问题

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

My understanding of optional chaining is that, once you start using ?. in a dotted expression, the rest of the properties are made 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)
// Output: Optional(Swift.String.CharacterView(_core: Swift._StringCore(_baseAddress: 0x00000001145e893f, _countAndFlags: 3, _owner: nil)))

The result indicates that s?.characters is indeed an Optional instance, indicating that s?.characters.count should be illegal.

Can someone help me understand this state of affairs?

回答1:

When you say:

My understanding of optional chaining is that, once you start using ?. in a dotted expression, the rest of the properties are made optional and are typically accessed by ?., not ..

I would say that you are almost there.

It’s not that all the properties are made optional, it’s that the original call is optional, so it looks like the other properties are optional.

characters is not an optional property, and neither is count, but the value that you are calling it on is optional. If there is a value, then the characters and count properties will return a value; otherwise, nil is returned. It is because of this that the result of s?.characters.count returns an Int?.

If either of the properties were optional, then you would need to add ? to it, but, in your case, they aren’t. So you don’t.

---

Edited following comment

From the comment:

I still find it strange that both s?.characters.count and (s?.characters)?.count compile, but (s?.characters).count doesn't. Why is there a difference between the first and the last expression?

I’ll try and answer it here, where there is more room than in the comment field:

s?.characters.count

If s is nil, the whole expression returns nil, otherwise an Int. So the return type is Int?.

(s?.characters).count // Won’t compile

Breaking this down: if s is nil, then (s?.characters) is nil, so we can’t call count on it.

In order to call the count property on (s?.characters), the expression needs to be optionally unwrapped, i.e. written as:

(s?.characters)?.count

---

Edited to add further

The best I can get to explaining this is with this bit of playground code:

let s: String? = "hello"

s?.characters.count
(s?.characters)?.count
(s)?.characters.count
((s)?.characters)?.count

// s?.characters.count
func method1(s: String?) -> Int? {
    guard let s = s else { return nil }

    return s.characters.count
}

// (s?.characters).count
func method2(s: String?) -> Int? {
    guard let c = s?.characters else { return nil }

    return c.count
}

method1(s)
method2(s)

回答2:

On the Swift-users mailing list, Ingo Maier was kind enough to point me to the section on optional chaining expressions in the Swift language spec, which states:

If a postfix expression that contains an optional-chaining expression is nested inside other postfix expressions, only the outermost expression returns an optional type.

It continues with the example:

var c: SomeClass?
var result: Bool? = c?.property.performAction()

This explains why the compiler wants s?.characters.count in my example above, and I consider that it answers the original question. However, as @Martin R observed in a comment, there is still a mystery as to why these two expressions are treated differently by the compiler:

s?.characters.count
(s?.characters).count

If I am reading the spec properly, the subexpression

(s?.characters) 

is "nested inside" the overall postfix expression

(s?.characters).count

and thus should be treated the same as the non-parenthesized version. But that's a separate issue.

Thanks to all for the contributions!

来源：https://stackoverflow.com/questions/38699734/optional-chaining-with-swift-strings