Question

假设我们有一个提供高阶函数applyTest的库。

是否可以将其与异步功能asyncFunction一起使用，同时保留异步代码的优点？

是否可以将库设计为更好地支持异步应用程序而无需专门提供异步版本？

let applyTest f =
    f 2 > 0

let syncFunction x =
    x - 1

let asyncFunction x =
    x - 2 |> async.Return

async {
    let a = applyTest syncFunction
    let b = applyTest (asyncFunction >> Async.RunSynchronously)
    printfn "a = %b, b = %b" a b
}
|> Async.RunSynchronously

Answer 1

如果您不想丢失强类型检查或像示例中那样同步运行异步计算，则需要提供单独的异步版本。这两件事都应尽可能避免。

如果要避免重复实际测试部分（f 2 > 0），可以将其拆分为一个函数，该函数将参数2传递给该函数，并检查该值是否更大大于零：

// LIBRARY CODE

let checkValue x = x > 0

// This function is generic so it can return a value or an async value
// (int -> 'a) -> 'a
let runTestFunction f = f 2

// (int -> int) -> bool
let applyTest f = f |> runTestFunction |> checkValue

// (int -> Async<int>) -> Async<bool>
let applyTestAsync f = async {
    let! value = runTestFunction f // use let! to await the value
    return checkValue value }

// USAGE

let syncFunction x = x - 1
let asyncFunction x = x - 2 |> async.Return

async {
    let a = applyTest syncFunction
    let! b = applyTestAsync asyncFunction // use let! to await the test result
    printfn "a = %b, b = %b" a b
}

另一个选择是使用重载方法。这是建立在上面定义的功能之上的：

type Test =
    static member Apply f = applyTest f
    static member Apply f = applyTestAsync f

// USAGE

async {
    let a = Test.Apply syncFunction
    let! b = Test.Apply asyncFunction // We still need to consume this differently with a let!
    printfn "a = %b, b = %b" a b
}

异步不可知的高阶函数

1 个答案: