Comment by emih

You can still do this though:

  let result = (barbalyze(c, d, $) . foobinade(a, b, $)) input

Or if you prefer left-to-right:

  let result = input
    |> foobinade(a, b, $)
    |> barbalyze(c, d, $)

Maybe what isn't clear is that this hole operator would bind to the innermost function call, not the whole statement.