How to Async.AwaitTask on plain Task (not Task<T>)?

You can use ContinueWith:

let awaitTask (t: Task) = t.ContinueWith (fun t -> ()) |> Async.AwaitTask

Or AwaitIAsyncResult with infinite timeout:

let awaitTask (t: Task) = t |> Async.AwaitIAsyncResult |> Async.Ignore

Update:

The FSharp.Core library for F# 4.0 now includes an Async.AwaitTask overload that accepts a plain Task. If you're using F# 4.0 then you should use this core function instead of the code below.

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Original answer:

If your task could throw an exception then you probably also want to check for this. e.g.

let awaitTask (task : Task) =
    async {
        do! task |> Async.AwaitIAsyncResult |> Async.Ignore
        if task.IsFaulted then raise task.Exception
        return ()
    }

Update:

The FSharp.Core library for F# 4.0 now includes an Async.AwaitTask overload that accepts a plain Task. If you're using F# 4.0 then you should use this core function instead of the code below.

Original answer:

I really liked Ashley's suggestion using function composition. Additionally, you can extend the Async module like this:

module Async =
    let AwaitTaskVoid : (Task -> Async<unit>) =
        Async.AwaitIAsyncResult >> Async.Ignore

Then it appears in Intellisense along with Async.AwaitTask. It can be used like this:

do! Task.Delay delay |> Async.AwaitTaskVoid

Any suggestions for a better name?