Partial application

We could define an addition function as follows:

let add x y = x + y

Here's a rather similar function:

let addx x = fun y -> x + y

Function addx takes an integer x as input and returns a function of type int -> int that will add x to whatever is passed to it.

The type of addx is int -> int -> int. The type of add is also int -> int -> int. So from the perspective of their types, they are the same function. But the form of addx suggests something interesting: we can apply it to just a single argument.

# let add5 = addx 5;;
add5 : int -> int = <fun>

# add5 2;;
- : int = 7

It turns out the same can be done with add:

# let add5 = add 5;;
add5 : int -> int = <fun>

# add5 2;;
- : int = 7

What you just did is called partial application: we partially applied the function add to one argument, even though you would normally think of it as a multi-argument function. This works because the following three functions are syntactically different but semantically equivalent. That is, they are different ways of expressing the same computation:

let add x y = x+y
let add x = fun y -> x+y
let add = fun x -> (fun y -> x+y)

So add is really a function that takes an argument x and returns a function (fun y -> x+y). Which leads us to a deep truth...

Function associativity

Are you ready for the truth? Here goes...

Every OCaml function takes exactly one argument.

Why? Consider add: although we can write it as let add x y = x + y, we know that's semantically equivalent to let add = fun x -> (fun y -> x+y). And in general,

let f x1 x2 ... xn = e

is semantically equivalent to

let f =
  fun x1 ->
    (fun x2 ->
       (...
          (fun xn -> e)...))

So even though you think of f as a function that takes n arguments, in reality it is a function that takes 1 argument and returns a function.

The type of such a function

t1 -> t2 -> t3 -> t4

really means the same as

t1 -> (t2 -> (t3 -> t4))

That is, function types are right associative: there are implicit parentheses around function types, from right to left. The intuition here is that a function takes a single argument and returns a new function that expects the remaining arguments.

Function application, on the other hand, is left associative: there are implicit parenthesis around function applications, from left to right. So

e1 e2 e3 e4

really means the same as

((e1 e2) e3) e4

The intuition here is that the left-most expression grabs the next expression to its right as its single argument.