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Home/ Questions/Q 6553111
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T12:32:31+00:00 2026-05-25T12:32:31+00:00

So I just wanted to ask why this works : let internal X th

  • 0

So I just wanted to ask why this works :

let internal X th =
    foo()
    th()
    bar()

let Start() = 
    X <| fun () -> ( foo(); bar(); etc... )

And this doesn’t work :

let internal XD A =
    let X th =
        foo()
        th()
        bar()
    (X <| fun () -> A)

let Start() = 
    XD ( foo(); bar(); etc... )

it’s looking like the same for me but first variant works as wrapper and I completely can’t understand how second variant works.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T12:32:32+00:00Added an answer on May 25, 2026 at 12:32 pm

    The below is the correct code for 2nd version for what you want to achieve (without lambda using lazy values).

    let internal XD (A:Lazy<unit>) =
        let X th =
            foo()
            th()
            bar()
        X <| (fun () -> A.Force())
    
    let Start() = 
        XD ( lazy(foo(); bar();) )
    
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