OCaml evaluation of expression, order?

Question

add is a function that is built in to OCaml.

Exp:

# add 5 1;;
- : int = 6

My question: what is the order of evaluation for add 3 (add 5 3)?

# add 3 (add 5 3);;
- : int = 11

Is (add 5 3) evaluated first to 8, then add 3 (8) is evaluated to 11? Or is add 3 evaluated first into a function, and then (add 5 3) is used as a parameter for the function (add 3)?

Thanks.

Answer 1

For what it's worth, there's no function named add that's built in to OCaml. (I.e., there's no such symbol in the Pervasives module.)

$ ocaml
        OCaml version 4.06.0

# add;;
Error: Unbound value add

The order of evaluation of a function and its arguments in OCaml is unspecified. So there is no guaranteed order.

This is documented in Section 7.7.1 of the OCaml manual.

If you want an evaluation to happen in a certain order, you can use let to evaluate each subexpression:

# let add a b = a + b;;
val add : int -> int -> int = <fun>
# let temp = add 5 3 in
  add 3 temp;;
- : int = 11
# let tempf = add 3 in
  tempf (add 5 3);;
- : int = 11

Answer 2

Just to complete Jeffrey's answer, here is a version of add which lets you know when it is executed:

# let add a b =
    Printf.printf "Adding %d and %d...\n" a b;
    a + b;;

val add : int -> int -> int = <fun>

Let's see it in action.

# add (add 1 2) (add 5 8);;

Adding 5 and 8...
Adding 1 and 2...
Adding 3 and 13...

- : int = 16

So here, the second add is evaluated first, but you can't really count on it since it is not specified.

Here is the same example using a local let to force the order:

# let x = add 1 2 in
  add x (add 5 8);;

Adding 1 and 2...
Adding 5 and 8...
Adding 3 and 13...

- : int = 16

And now using partial application.

# let add_3 = add 3 in
  add_3 (add 5 8);;

Adding 5 and 8...
Adding 3 and 13...

- : int = 16

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Of course, with (pure) functional programming, order of execution does not matter since everything is immutable. But OCaml is't purely functional, so these little tricks are good to know. :)