type lambda_term = Var of identifier | Abstract of identifier * lambda_term | App of lambda_term * lambda_term
+<a id="occurs_free"></a>
+
16. Write a function `occurs_free` that has the following type:
occurs_free : identifier -> lambda_term -> bool
OCAML ANSWERS:
- type ('a) sysf_bool = 'a -> 'a -> 'a
- let sysf_true : ('a) sysf_bool = fun y n -> y
- let sysf_false : ('a) sysf_bool = fun y n -> n
+ type 'a sysf_bool = 'a -> 'a -> 'a
+ let sysf_true : 'a sysf_bool = fun y n -> y
+ let sysf_false : 'a sysf_bool = fun y n -> n
- type ('a) sysf_nat = ('a -> 'a) -> 'a -> 'a
- let sysf_zero : ('a) sysf_nat = fun s z -> z
+ type 'a sysf_nat = ('a -> 'a) -> 'a -> 'a
+ let sysf_zero : 'a sysf_nat = fun s z -> z
- let sysf_iszero (n : 'a sysf_nat) : 'b sysf_bool = n (fun _ -> sysf_false) sysf_true
+ let sysf_iszero (n : ('a sysf_bool) sysf_nat) : 'a sysf_bool = n (fun _ -> sysf_false) sysf_true
(* Annoyingly, though, if you just say sysf_iszero sysf_zero, you'll get an answer that isn't fully polymorphic.
This is explained in the comments linked below. The way to get a polymorphic result is to say instead
`fun next -> sysf_iszero sysf_zero next`. *)
let sysf_succ (n : 'a sysf_nat) : 'a sysf_nat = fun s z -> s (n s z)
- (* Again, to use this you'll want to call `fun next -> sysf_succ sysf_zero next` *)
-
- let sysf_pred (n : 'a sysf_nat) : 'a sysf_nat = (* NOT DONE *)
-
-<!--
- (* Using a System F-style encoding of pairs, rather than native OCaml pairs ... *)
- let pair a b = fun f -> f a b in
- let snd x y next = y next in
- let shift p next = p (fun x y -> pair (sysf_succ x) x) next in (* eta-expanded as in previous definitions *)
- fun next -> n shift (pair sysf_zero sysf_zero) snd next
-
- let pair = \a b. \f. f a b in
- let snd = \x y. y in
- let shift = \p. p (\x y. pair (succ x) x) in
- let pred = \n. n shift (pair 0 err) snd in
--->
-
+ (* Again, to get a polymorphic result you'll want to call `fun next -> sysf_succ sysf_zero next` *)
+
+ And here is how to get `sysf_pred`, using a System-F-style encoding of pairs. (For brevity, I'll leave off the `sysf_` prefixes.)
+
+ type 'a natpair = ('a nat -> 'a nat -> 'a nat) -> 'a nat
+ let natpair (x : 'a nat) (y : 'a nat) : 'a natpair = fun f -> f x y
+ let fst x y = x
+ let snd x y = y
+ let shift (p : 'a natpair) : 'a natpair = natpair (succ (p fst)) (p fst)
+ let pred (n : ('a natpair) nat) : 'a nat = n shift (natpair zero zero) snd
+
+ (* As before, to get polymorphic results you need to eta-expand your applications. Witness:
+ # let one = succ zero;;
+ val one : '_a nat = <fun>
+ # let one next = succ zero next;;
+ val one : ('a -> 'a) -> 'a -> 'a = <fun>
+ # let two = succ one;;
+ val two : '_a nat = <fun>
+ # let two next = succ one next;;
+ val two : ('a -> 'a) -> 'a -> 'a = <fun>
+ # pred two;;
+ - : '_a nat = <fun>
+ # fun next -> pred two next;;
+ - : ('a -> 'a) -> 'a -> 'a = <fun>
+ *)
Consider the following list type, specified using OCaml or Haskell datatypes:
match b with true -> y () | false -> n ()
that would arguably still be relying on the special evaluation order properties of OCaml's native `match`. You'd be assuming that `n ()` wouldn't be evaluated in the computation that ends up selecting the other branch. Your assumption would be correct, but to avoid making that assumption, you should instead first select the `y` or `n` result, _and then afterwards_ force the result. That's what we do in the above answer.
+
+ Question: don't the rhs of all the match clauses in `match b with true -> y | false -> n` have to have the same type? How can they, when one of them is `blackhole` and the other is `blackhole ()`? The answer has two parts. First is that an expression is allowed to have different types when it occurs several times on the same line: consider `let id x = x in (id 5, id true)`, which evaluates just fine. The second is that `blackhole` will get the type `'a -> 'b`, that is, it can have any functional type at all. So the principle types of `y` and `n` end up being `y : unit -> 'a -> 'b` and `n : unit -> 'c`. (The consequent of `n` isn't constrained to use the same type variable as the consequent of `y`.) OCaml can legitimately infer these to be the same type by unifying the types `'c` and `'a -> 'b`; that is, it instantiates `'c` to the functional type had by `y ()`.