Can't use nil for optional generic parameter

Using Swift
1

Hello, is this a missing implementation, isn't it? I feel that it should be possible to use optional generic argument, but it is not. Is there a workaround for this?

here is an example of source code which doesn't compile.

protocol Dependency {}

protocol Interface {
    associatedtype D: Dependency
    
    func use(_ input: D)
}

struct SomeDependency: Dependency {}

class SomeImplementation: Interface {
    typealias D = SomeDependency
    
    func use(_ input: D) {
        
    }
}

func handle<I: Interface>(_ interface: I?) where I.D == SomeDependency {
    let dependency = SomeDependency()
    interface?.use(dependency)
}

let input = SomeImplementation()
handle(input)
handle<SomeImplementation>(nil) // Cannot explicitly specialize a generic function
handle(nil) // Generic parameter 'I' could not be inferred
2

try:

handle(nil as SomeImplementation?)
2 Likes
3

You could probably do something like:

enum InterfaceHandler<I: Interface> {
  static func handle<I>(_ interface: I?) where I.D == SomeDependency { 
   // ...
  }
}

InterfaceHandler<SomeImplementation>.handle(nil)

I think this should work, but I haven't tried it

1 Like
4

This and handle(SomeImplementation?.none) are how to do it, with the code you have.

But usually when I see code like this, it's a sign that the optionality should be moved outward.

if let implementation = SomeImplementation() as Optional {
  handle(implementation)
}

func handle(_ interface: some Interface<SomeDependency>) {
  interface.use(.init())
}
2 Likes
5

Thanks for the help

1 Like