1

If I have a Sendable type why aren't its methods also @Sendable func by default?

struct Foo: Sendable {
    func bar() { }
}
let foo = Foo()
let fn: @Sendable () -> Void = foo.bar
// ⚠️ Converting non-sendable function value to '@Sendable () -> Void' may introduce data races

The type can automatically fulfil the requirements of a protocol and then it appears to be a @Sendable func

protocol Barable: Sendable {
    @Sendable func bar()
}

extension Foo: Barable { }

let fn: @Sendable () -> Void = (foo as Barable).bar

So is it a @Sendable func or is the protocol conformance a bug?

I realise you can create a closure inferring @Sendable func from context but this is not what I am asking about.

let fn: @Sendable () -> Void = { foo.bar() }
2 Likes
2

This pitch proposes that @Sendable func should be inferred on types that conform to Sendable.

But methods of a nominal type cannot capture anything but the object instance itself. Thus, the proposed simplifications are:

  1. the inference of @Sendable on all methods of a type that conforms to Sendable.
  2. the inference of @Sendable on all non-local functions.
  3. the prohibition of marking a method @Sendable if the object type does not conform to Sendable.