Assignment 5
Types and OCAML
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0. Recall that the S combinator is given by \x y z. x z (y z).
Give two different typings for this function in OCAML.
To get you started, here's one typing for K:
# let k (y:'a) (n:'b) = y;;
val k : 'a -> 'b -> 'a =
# k 1 true;;
- : int = 1
1. Which of the following expressions is well-typed in OCAML?
For those that are, give the type of the expression as a whole.
For those that are not, why not?
let rec f x = f x;;
let rec f x = f f;;
let rec f x = f x in f f;;
let rec f x = f x in f ();;
let rec f () = f f;;
let rec f () = f ();;
let rec f () = f () in f f;;
let rec f () = f () in f ();;
2. Throughout this problem, assume that we have
let rec omega x = omega x;;
All of the following are well-typed.
Which ones terminate? What are the generalizations?
omega;;
omega ();;
fun () -> omega ();;
(fun () -> omega ()) ();;
if true then omega else omega;;
if false then omega else omega;;
if true then omega else omega ();;
if false then omega else omega ();;
if true then omega () else omega;;
if false then omega () else omega;;
if true then omega () else omega ();;
if false then omega () else omega ();;
let _ = omega in 2;;
let _ = omega () in 2;;
3. The following expression is an attempt to make explicit the
behavior of `if`-`then`-`else` explored in the previous question.
The idea is to define an `if`-`then`-`else` expression using
other expression types. So assume that "yes" is any OCAML expression,
and "no" is any other OCAML expression (of the same type as "yes"!),
and that "bool" is any boolean. Then we can try the following:
"if bool then yes else no" should be equivalent to
let b = bool in
let y = yes in
let n = no in
match b with true -> y | false -> n
This almost works. For instance,
if true then 1 else 2;;
evaluates to 1, and
let b = true in let y = 1 in let n = 2 in
match b with true -> y | false -> n;;
also evaluates to 1. Likewise,
if false then 1 else 2;;
and
let b = false in let y = 1 in let n = 2 in
match b with true -> y | false -> n;;
both evaluate to 2.
However,
let rec omega x = omega x in
if true then omega else omega ();;
terminates, but
let rec omega x = omega x in
let b = true in
let y = omega in
let n = omega () in
match b with true -> y | false -> n;;
does not terminate. Incidentally, `match bool with true -> yes |
false -> no;;` works as desired, but your assignment is to solve it
without using the magical evaluation order properties of either `if`
or of `match`. That is, you must keep the `let` statements, though
you're allowed to adjust what `b`, `y`, and `n` get assigned to.
[[Hint assignment 5 problem 3]]
4. Baby monads. Read the lecture notes for week 6, then write a
function `lift` that generalized the correspondence between + and
`add`: that is, `lift` takes any two-place operation on integers
and returns a version that takes arguments of type `int option`
instead, returning a result of `int option`. In other words,
`lift` will have type
(int -> int -> int) -> (int option) -> (int option) -> (int option)
so that `lift (+) (Some 3) (Some 4)` will evalute to `Some 7`.
Don't worry about why you need to put `+` inside of parentheses.
You should make use of `bind` in your definition of `lift`:
let bind (x: int option) (f: int -> (int option)) =
match x with None -> None | Some n -> f n;;