Re: [swift-users] Optional chaining and String properties

[Rimantas Liubertas via swift-users]

>  
> 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?

See print(s?.characters.count) — you get the optional despite count not being 
defined as optional
  
Also, try this:

struct Foo {  
    let bar: Int
}

var foo: Foo? = Foo(bar: 42)  
print(foo?.bar)

Then try this:

struct Foo {
    let bar: Int
    let baz: Int?
}

var foo: Foo? = Foo(bar: 42, baz: 69)
print(foo?.bar)

print(foo?.baz?)



This may give you some ideas.

Best regards,
Rimantas


_______________________________________________
swift-users mailing list
swift-users@swift.org
https://lists.swift.org/mailman/listinfo/swift-users