swift calling async function without a return value

Question

Is there anyway in swift's new structured concurrency model to do the following without a dummy bool return?

func do() async -> Bool {
  something()
  return true
}
async let foo = do()

//do other stuff
stuff()

//now I need to know that "do" has finished
await foo

I know that I can do the following but it will not run concurrently:

func do() async {
  something()
}
await do()
stuff()

//cannot run "stuff" and "do" concurrently

I feel like I am missing a basic idea here because the top block of code does what I need but feels like a hack due to the Bool return.

Rob Napier · Accepted Answer · 2022-01-03 19:15:42Z

10

What you're describing is a Task. For example:

Task { await `do`() }
stuff()

This will run do() concurrently with stuff(). If you need to keep track of when do() completes, you can await the task's value:

let task = Task { await `do`() }
stuff()
await task.value // Doesn't actually return anything, but will block

This kind of Task runs in the context of the current Actor, which is usually what you want. If you want something independent of the current Actor, you can use Task.detached() instead.

If you've previously used DispatchQueues, in many of the places you would have written queue.async { ... }, you can now write Task { ... }. The new system is much more powerful, but it maps fairly nicely to the old system if you want it to.

edited Jan 3, 2022 at 19:15

answered Jan 3, 2022 at 19:04

Rob Napier

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Rico Crescenzio · Accepted Answer · 2022-01-03 11:13:14Z

9

Swift implicitly returns Void for non-returning function, so I guess this would be fine

func do() async {
  something()
}
async let foo: Void = do() // just explicit Void so the compiler doesn't emit a warning telling you that may not be expected

//do other stuff
stuff()

//now I need to know that "do" has finished
await foo

answered Jan 3, 2022 at 11:13

Rico Crescenzio

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