How to ignore the return value of an Async function in an Async block?

Question

The m1 and m2 in the following functions have compiling errors.

let m p = async { return p * 2 }
let m1 () = async { do! m 2 } // ERR: was expected 'int' but here has type 'unit'
let m2 () = async { do! m 2 |> ignore } // ERR: expecting 'Async<int>->Async<'a>' but given 'Async<int>->unit'

m is called at the last line. How to ignore its return value? Is the following the only way (will executing of it be optimized by the compiler?)?

let m1 () = 
    async { 
      let! x = m 2 
      () 
    }

Are you sure you want to do this? It smells C#

Szer
– Szer

2018-06-08 17:20:02 +00:00
Commented Jun 8, 2018 at 17:20 — Szer
– Szer, Commented Jun 8, 2018 at 17:20

Chad Gilbert · Accepted Answer · 2018-06-08 14:11:22Z

6

You can use Async.Ignore for this:

let m1 () = async { do! m 2 |> Async.Ignore }

From the documentation:

Async.Ignore Creates an asynchronous computation that runs the given computation and ignores its result.

answered Jun 8, 2018 at 14:11

Chad Gilbert

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2 Comments

Szer Over a year ago

btw, it is literally the same as let m1 () = m 2 |> Async.Ignore

ca9163d9 Over a year ago

The signature of m1 (the return type) are different?

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