为什么在Swift中包装可选值？ 是否可以在不声明类型的情况下将变量设置为可选变量？

Question

If I want to make a variable an optional, I can do:

var exampleString: String? = nil

or

var exampleString: String? = "This is a string"

Is it possible to declare an optional value without assigning a type like String, Int, Bool, etc?

Something like (that doesn't return an error):

var exampleVar? = nil

I have no use case for why I want to do this, but I'm trying to better understand optionals. I just read through the Optionals section in Swift docs. If this can be done, would there be any advantage to doing this?

Also, why would I not want to use implicitly unwrapped optionals everywhere? What would be the disadvantages of this? Apple's Swift docs mentions that you shouldn't do this if you intend to reassign a variable to nil. Why?

Thanks so much for your help guys!

Answer 1

Is it possible to declare an optional value without assigning a type like String, Int, Bool, etc?

No. The reason is that Optional (without its generic parameter) is not a complete type — think of it as a type constructor , ie a way to produce a full type if you give it the missing information.

The compiler needs to know the full type — Optional<Int> or Optional<String> and so on — to allocate the right amount of memory etc. An Optional<String> will take up more memory than an Optional<Bool> , even if both are nil .

The same is true for other generic types like Array . Array is the type constructor, Array<Int> the complete type.

Edit: rmaddy makes a good point: var exampleVar: Any? = nil var exampleVar: Any? = nil allows you to store any value in this optional. Note that we're still dealing with a full type here — Optional<Any> . It's just that all types are compatible with Any . In that sense, Optional<Any> is not much different from Optional<SomeProtocol> , which can store any value that conforms to the SomeProtocol protocol.