[swift-users] Extending a generic type where T == Void

[Dave Abrahams]

on Tue Aug 09 2016, Benedict Cohen <swift-users-AT-swift.org> wrote:

> I have an enum that represents the states of a task:
>
> enum TaskState<InType, OutType>{
> case preparing
> case ready(InType)
> case executing(InType, TaskIdentifier)
> case success(InType, OutType)
> case failure(InType, Error)
>
> mutating func transition(toReady: InType) {...}
>
> }
>
> //Usage
> var locationSearchTaskState: TaskState<String, [Location]> = .preparing
> //...
> locationSearchTaskState.transition(toReady: text)
>
> This works as desired for most cases. Not all tasks, however, have an input:
>
> var topLocationsTaskState<Void, [Location]> = .ready( () )
>
> In this case we have to pass the empty tuple as the value which is ugly. I'd like to add some sugar to make this nicer. Something along the lines of:
>
> extension TaskState where InType: Void {...}
>
> Of course this doesn't work because Void is not a protocol (nor can it
> conform to a protocol because it's just a tuple). So my question: How
> can I extend a generic type when the generic parameter is Void? Are
> there alternative ways to model this problem?

There's a horrible hack you can use:

protocol TaskStateProtocol {
  associatedtype InType
  mutating func transition(toReady: InType)
}
extension TaskState : TaskStateProtocol {}

extension TaskStateProtocol where InType == Void {
  // ... <==== put your stuff here
}


HTH,
-- 
-Dave