Does the yin yang continuations puzzle make sense in a typed language?

Question

This question is related to "How the yin-yang puzzle works?". The yin yang example of continuations in scheme looks like this, according to Wikipedia article:

(let* ((yin
     ((lambda (cc) (display #\@) cc) (call-with-current-continuation (lambda (c) c))))
   (yang
     ((lambda (cc) (display #\*) cc) (call-with-current-continuation (lambda (c) c)))))
(yin yang))

I am trying to write an equivalent piece of code in a (edit: statically) typed language, such as SML/NJ, but it is giving me typing errors. So either the puzzle does not type, or I am misunderstanding the scheme syntax. What would the above piece of code look like in SML or Ocaml (with callcc extension)?

By the way, what is the source of the puzzle? Where did it come from?

Edit: I think I know the answer. We need a recursive type t satisfying t = t -> s for some type s.

Edit of edit: No it is not, the answer is a recursive type t satisfying t = t -> t.

Answer 1

I think I am going to answer my own question. I will show two solutions, one in eff and another in Ocaml.

eff

We are going to work with eff (I am blowing my own horn here, see below for another way in OCaml with Oleg's delimcc extension.) The solution is explained in the paper Programming with algebric effects and continuations.

First we define shift and reset in eff:

type ('a, 'b) delimited =
effect
  operation shift : (('a -> 'b) -> 'b) -> 'a
end

let rec reset d = handler
  | d#shift f k -> with reset d handle (f k) ;;

Here is the yin yang puzzle transcribed into eff:

let y = new delimited in
  with reset y handle
    let yin = (fun k -> std#write "@" ; k) (y#shift (fun k -> k k)) in
    let yang = (fun k -> std#write "*" ; k) (y#shift (fun k -> k k)) in
      yin yang

But eff complains about it that it can't solve the type equation α = α → β. At present eff cannot handle arbitrary recursive types, so we are stuck. As a way of cheating, we can turn off type checking to see if at the very least the code does what it is supposed to:

$ eff --no-types -l yinyang.eff
@*@**@***@****@*****@******@*******@********@*********@*******...

Ok, it's doing the right thing, but the types are not powerful enough.

OCaml

For this example we need Oleg Kiselyov's delimcc library. The code is as follows:

open Delimcc ;;

let y = new_prompt () in
  push_prompt y (fun () ->
    let yin = (fun k -> print_string "@" ; k) (shift y (fun k -> k k)) in
    let yang = (fun k -> print_string "*" ; k) (shift y (fun k -> k k)) in
      yin yang)

Again, Ocaml won't compile because it hits a recursive type equation. But with the -rectypes option we can compile:

ocamlc -rectypes -o yinyang delimcc.cma yinyang.ml

It works as expected:

$ ./yinyang
@*@**@***@****@*****@******@*******@********@*********@...

OCaml computes that the type of yin and yang is ('a -> 'a) as 'a, which is its way of saying "a type α such that α = α → α". This is precisely the type characteristic of the untyped λ-calculus models. So there we have it, the yin yang puzzle essentially uses features of the untyped λ-calculus.