type num = int -> int
type contents = Num of num | Op of (num -> num) | Op2 of (num -> num -> num)
type tree = Leaf of contents | Branch of tree * tree | Error
let mid x = fun _ -> x (* K combinator *)
let map f xx = fun n -> f (xx n) (* function composition, that is the B combinator *)
let mapply ff xx = fun n -> (ff n) (xx n) (* S combinator *)
let map2 f xx yy = fun n -> f (xx n) (yy n)
let rec eval (t : tree) = match t with
| Leaf _ -> t
| Branch (Leaf (Op f), right) -> (match (eval right) with
| Leaf (Num n) -> Leaf (Num (f n))
| _ -> Error)
| Branch (Leaf (Op2 f), right) -> (match (eval right) with
| Leaf (Num n) -> Leaf (Op (f n))
| _ -> Error)
| Branch (left, right) -> eval (Branch (eval left, eval right))
| _ -> Error
(* To use infix operators in ordinary prefix position, use (+).
Multiplication has to be handled a bit specially, because of how OCaml parses
its comment indicators. To use it in prefix position, make sure there is
space between it and the parentheses, like this: ( * ).
*)
(* Encoding of \n. (+ 1 ( * (/ 6 n) 4)) *)
let t1 = Branch ((Branch ((Leaf (Op2 (map2 (+)))),
(Leaf (Num (mid 1))))),
(Branch ((Branch ((Leaf (Op2 (map2 ( * ))),
(Branch ((Branch ((Leaf (Op2 (map2 (/)))),
(Leaf (Num (mid 6))))),
(Leaf (Num (fun n -> n)))))))),
(Leaf (Num (mid 4))))));;
(* Now evaluate:
match eval t1 with Leaf (Num f) -> f 2;;
The answer should be 13.
*)