将 F# 管道运算符( <|、>>、<<)转换为 OCaml [英] Converting F# pipeline operators ( <|, >>, << ) to OCaml

问题描述

我正在将一些 F# 代码转换为 OCaml，并且我看到了这个管道运算符 <| 的很多用途，例如:

I'm converting some F# code to OCaml and I see a lot of uses of this pipeline operator <|, for example:

let printPeg expr =
    printfn "%s" <| pegToString expr

<| 运算符显然定义为:

# let ( <| ) a b = a b ;;
val ( <| ) : ('a -> 'b) -> 'a -> 'b = <fun>

我想知道为什么他们费心在 F# 中定义和使用这个运算符，是不是为了避免像这样输入括号?:

I'm wondering why they bother to define and use this operator in F#, is it just so they can avoid putting in parens like this?:

let printPeg expr =
    Printf.printf "%s" ( pegToString expr )

据我所知，这就是将上面的 F# 代码转换为 OCaml，对吗?

As far as I can tell, that would be the conversion of the F# code above to OCaml, correct?

另外，我将如何在 Ocaml 中实现 F# 的 << 和 >> 运算符?

Also, how would I implement F#'s << and >> operators in Ocaml?

( |> 运算符似乎只是: let ( |> ) a b = b a ;; )

( the |> operator seems to simply be: let ( |> ) a b = b a ;; )

推荐答案

直接来自 F# 源代码:

let inline (|>) x f = f x
let inline (||>) (x1,x2) f = f x1 x2
let inline (|||>) (x1,x2,x3) f = f x1 x2 x3
let inline (<|) f x = f x
let inline (<||) f (x1,x2) = f x1 x2
let inline (<|||) f (x1,x2,x3) = f x1 x2 x3
let inline (>>) f g x = g(f x)
let inline (<<) f g x = f(g x)

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