f# - 如何忽略异步 block 中异步函数的返回值？

Question

以下函数中的m1和m2存在编译错误。

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 在最后一行被调​​用。如何忽略它的返回值？以下是唯一的方法吗(编译器会优化它的执行吗？)？

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

Answer 1

您可以使用 Async.Ignore为此:

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

来自文档:

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