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Home/ Questions/Q 6359723
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T23:33:24+00:00 2026-05-24T23:33:24+00:00

Let’s say I have this code let identifier = spaces_surrounded (many1Satisfy isLetter) I was

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Let’s say I have this code

let identifier = spaces_surrounded (many1Satisfy isLetter)

I was wondering if it there was any native F# function that allowed me to refactor it to

let identifier = spaces_surrounded $ many1Satisfy isLetter

that is, something such as

let ($) f1 f2 = f1 (f2)

(that is if I am not mistaken, my Haskell skills are not too sharp..).

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  1. Editorial Team
    Editorial Team
    2026-05-24T23:33:25+00:00Added an answer on May 24, 2026 at 11:33 pm

    The standard F# idiom for this is the forward pipe operator |> were you would rewrite

    let identifier = spaces_surrounded (many1Satisfy isLetter)
    

    as

    let identifier = many1Satisfy isLetter |> spaces_surrounded 
    

    you can also use the backward pipe operator <| if you want to maintain the original order, but this tends to be a little less common

    let identifier = spaces_surrounded <| many1Satisfy isLetter
    
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