role of the |> operator vs partial application

Question

I have the following code:

type transaction = Withdraw of int | Deposit of int | Checkbalance

(* Bank account generator. *)
let make_account(opening_balance: int) =
    let balance = ref opening_balance in
    fun (t: transaction) ->
      match t with
        | Withdraw(m) ->  if (!balance > m)
                          then
                            ((balance := !balance - m);
                            (Printf.printf "Balance is %i" !balance))
                          else
                            print_string "Insufficient funds."
        | Deposit(m) ->
           ((balance := !balance + m);
            (Printf.printf "Balance is %i\n" !balance)) 
        | Checkbalance -> (Printf.printf "Balance is %i\n" !balance)
;;

whenever I try to run the following command: make_account(100) Deposit(50) ;; I get the following error: This function has type int -> transaction -> unit. It is applied to too many arguments; maybe you forgot a `;'.

However the following command works just fine Deposit(50) |> make_account(100) ;;

But why aren't these two line of codes equivalent? shouldn't (make_account 100) be replaced by (fun (t: transaction) -> ...) ? Because then I don't understand why my first attempt did not work.

Thank you!

Bergi Bergi · Accepted Answer · 2020-03-25T18:50:51

Notice that Ocaml doesn't use parenthesis for function calls. In the statement

make_account (100) Deposit (50);;

you have two unnecessary grouping parenthesis, it is the same as

make_account  100  Deposit  50 ;;

where make_account is applied with three arguments. What you meant to write was

make_account 100 (Deposit 50);;

which is equivalent to the parenthesis-less

Deposit 50 |> make_account 100;;

role of the |> operator vs partial application

1 Answers