[swift-evolution] allowing to specify optionality with type inference

[Hugues Bernet-Rollande]

Oh my,

  

I totally miss this one.

The generic one is pretty cool too.

  

You think the `?` shortcut could still make sense?

  

let y: ? = 123  

  

Thanks!

  

  

Hugues BERNET-ROLLANDE

  

> On Mar 24 2016, at 11:41 am, Andrew Bennett <cacoyi at gmail.com> wrote:  

>

> Hi Huges,

>

>  

>

> You can achieve a very similar result like this:

>

>  

>

> let x = 123

>

> let y: Optional = x

>

>  

>

> The generic arguments can be inferred in more complicated cases too, for
example:

>

>  

>

> let z: Set = [1, 2, 3]

>

>  

>

> Which creates a Set<Int> rather than an Array<Int>.  

>

>  

>

> I hope this helps!

>

> Andrew Bennett  

>

>  

>

>  

>

> On Thu, Mar 24, 2016 at 9:22 PM, Hugues Bernet-Rollande via swift-evolution
<[swift-evolution at swift.org](mailto:swift-evolution at swift.org)> wrote:  

>

>> In a strongly typed language, such as Swift is, type inference is greatly
appreciated as it declutter the instance type definition.

>>

>>  

>>

>> For example, in the following statement, the compiler easily deduct the
type of `aString` by the return type of `String.init()`

>>

>>  

>>

>> ```

>>

>> let aString = String()

>>

>> ```

>>

>>  

>>

>> Optional are generic enum which can contain a value of a generic type or
nil (<https://github.com/apple/swift/blob/master/stdlib/public/core/Optional.s
wift>)

>>

>>  

>>

>> Sometime you may want specify that the instance is of type optional in
which even if you are a the same time assigning a value to this instance.

>>

>> In which case you loose the type inference mechanism and you have to define
both (the optionality as well as the type).

>>

>> For example, you may want specify an optional String with a default value.

>>

>>  

>>

>> ```

>>

>> var a:String? = String()

>>

>> // or

>>

>> var b = String() as String?

>>

>> // or

>>

>> var c:Optional&lt;String&gt; = String()  

>>

>> ```

>>

>>  

>>

>> But the compiler can already infer the variable type from this assignment,
it just miss the "optionality" of the variable.

>>

>> It would be nice to be able to express this.

>>

>>  

>>

>> Then for, I propose the following syntax evolution:

>>

>>  

>>

>> ```

>>

>> var a:? = String()

>>

>> // and/or (not recommended because more prone to typo and unclear...)

>>

>> var a = String() as?

>>

>> ```

>>

>>  

>>

>> This would allow for more synthetic optional declaration.

>>

>>  

>>

>> I've draft a proposal on my `swift-evolution` fork:

>>

>> <https://github.com/huguesbr/swift-evolution/blob/optionality-type-with-
type-inference/proposals/0057-optionality-type-with-type-inference.md>

>>

>>  

>>

>> Let me know what you think.

>>

>>  

>>

>>  

>>

>> Hugues BERNET-ROLLANDE

>>

>>  

>>

>> \--

>>

>> [hugues at xdev.fr](mailto:hugues at xdev.fr)

>>

>> <http://www.xdev.fr>

>>

>> <http://www.linkedin.com/in/huguesbr>

>>

>>  
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