В F#, что делает>> средний оператор?

Я заметил в некотором коде в этом образце, который содержал>> оператор:

let printTree =
  tree >> Seq.iter (Seq.fold (+) "" >> printfn "%s")

Что делает>> оператор, mean/do?


Спасибо очень, теперь это намного более ясно. Вот мой пример, который я генерировал для приобретения навыка его:

open System
open System.IO

let read_lines path = File.ReadAllLines(path) |> Array.to_list

let trim line = (string line).Trim()
let to_upper line = (string line).ToUpper()

let new_list = [ for line in read_lines "myText.txt" -> line |> (trim >> to_upper) ]

printf "%A" new_list

задан jkdev 6 August 2019 в 11:19

4 ответа

It's the function composition operator.

More info on Chris Smith's blogpost.

Introducing the Function Composition operator (>>):

let inline (>>) f g x = g(f x)

Which reads as: given two functions, f and g, and a value, x, compute the result of f of x and pass that result to g. The interesting thing here is that you can curry the (>>) function and only pass in parameters f and g, the result is a function which takes a single parameter and produces the result g ( f ( x ) ).

Here's a quick example of composing a function out of smaller ones:

let negate x = x * -1 
let square x = x * x 
let print  x = printfn "The number is: %d" x
let square_negate_then_print = square >> negate >> print 
asserdo square_negate_then_print 2

When executed prints ‘-4’.

ответ дан 27 November 2019 в 17:44

According to F# Symbol and Operator Reference it is Forward Function Composition operator.

ответ дан 27 November 2019 в 17:44

That is function composition, used for partial application

ответ дан 27 November 2019 в 17:44

The >> operator composes two functions, so x |> (g >> f) = x |> g |> f = f (g x). There's also another operator << which composes in the other direction, so that (f << g) x = f (g x), which may be more natural in some cases.

ответ дан 27 November 2019 в 17:44
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