Question

Is piping parameter into line is working only for functions that accept one parameter?
If we look at the example at Chris Smiths’ page,

// Using the Pipe-Forward operator (|>)
let photosInMB_pipeforward =
    @"C:\Users\chrsmith\Pictures\"
    |> filesUnderFolder
    |> Seq.map fileInfo
    |> Seq.map fileSize
    |> Seq.fold (+) 0L 
    |> bytesToMB

where his filesUnderFolder function was expecting only rootFolder parameter,
what if the function was expecting two parameters, i.e.
let filesUnderFolder size rootFolder

Then this does not work:

// Using the Pipe-Forward operator (|>)
let size= 4
let photosInMB_pipeforward =
    @"C:\Users\chrsmith\Pictures\"
    |> filesUnderFolder size
    |> Seq.map fileInfo
    |> Seq.map fileSize
    |> Seq.fold (+) 0L 
    |> bytesToMB

Since I can define
let inline (>>) f g x y = g(f x y)
I think I should be able to use pipeline operator with functions having multiple input parameters, right? What am I missing?

Answer 1

When mixing pipeline operators and curried arguments be aware of the order you pass arguments with.

let size = 4
let photosInMB_pipeforward =
    size, @"C:\Users\chrsmith\Pictures\"
    ||> filesUnderFolder
    |> Seq.map fileInfo
    |> Seq.map fileSize
    |> Seq.fold (+) 0L 
    |> bytesToMB

Think about it as if the compiler is putting parentheses around the function and its parameters like this.

@"C:\Users\chrsmith\Pictures\" |> filesUnderFolder size
becomes
@"C:\Users\chrsmith\Pictures\" |> (filesUnderFolder size)
or
(filesUnderFolder size) @"C:\Users\chrsmith\Pictures\"

Out of order example

let print2 x y = printfn "%A - %A" x y;;

(1, 2) ||> print2;;
1 - 2

1 |> print2 2;;
2 - 1

With three arguments

let print3 x y z = printfn "%A - %A - %A" x y z;;

(1, 2, 3) |||> print3;;
1 - 2 - 3

(2, 3) ||> print3 1;;
1 - 2 - 3

3 |> print3 1 2;;
1 - 2 - 3

Definitions

let inline (|>) x f = f x

let inline (||>) (x1,x2) f = f x1 x2

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