Why does the pipe operator work?

Question

If the pipe operator is created like this:

let (|>) f g = g f

And used like this:

let result = [2;4;6] |> List.map (fun x         


        

Answer 1

The brackets around |> mean it is an infix operator so your example could be written

let result = (|>) [2;4;6] (List.map (fun x -> x * x * x))

Since |> applies its first argument to the second, this is equivalent to

let result = (List.map (fun x -> x * x)) [2;4;6]