Given a value in a monad, e.g. IO v, how can I remove v from the Monad?
Surely this would be required to seamlessly integrate stateful and
functional code?

Dave the ingenue.


-----Original Message-----
From: Jerome Vouillon [mailto:Jerome.Vouillon@inria.fr]
Sent: Friday, August 25, 2000 4:42 PM
To: John Max Skaller; Francois.Pottier@inria.fr
Cc: caml-list@inria.fr
Subject: Re: Language Design


On Fri, Aug 25, 2000 at 06:16:06AM +1000, John Max Skaller wrote:
> Francois Pottier wrote:
>  
> > On Wed, Aug 23, 2000 at 03:55:36PM +1000, John Max Skaller wrote:
> > > What is _actually_ required is a seamless way to integrate
> > > stateful and function code:
> > 
> > Have you thought about employing some kind of monadic type system?
> 
> 	Yes, but I don't know enough to do it at the moment.
> [Also, it turns out monads are not general enough to write
> web services in, which puts me off a bit]

I think you should really consider using monads. Here is an example.

We define a value of type void to be either a continuation expecting a
string or a final function that do not expect anything.

    type void = Cont of (string -> void)
              | Term of (unit -> unit)

This is a procedure that does nothing.

    let unit : void = Term (fun () -> ())

And here is a possible implementation for read: the input string is
assigned to v and then there is nothing more to do.

    let read (v : string ref) : void = Cont (fun s -> v := s; unit)

The following combinator takes two procedures p and p' and make them
be evaluated in order.

    let rec seq (p : void) (p': void) : void =
      match p, p' with
        Cont c, _ ->
          Cont (fun s -> seq (c s) p')
      | Term t, Cont c ->
          Cont (fun s -> t (); c s)
      | Term t, Term t' ->
          Term (fun () -> t (); t' ())

Finally, we have an operator to assign a value v to a reference x.

    let set (x : 'a ref) (v : unit -> 'a) : void = Term (fun () -> x := v
())

Now we can for instance define a procedure that reads to strings and
returns their concatenation. This procedure does not need to know the
actual definition of type void.

    let read2 x =
      let a = ref "" in let b = ref "" in
      seq (read a) (seq (read b) (set x (fun () -> !a ^ !b)))

You can also use some symbols to make it more readable:

    let ($) = seq
    let (-<-) = set
    let read2 x =
      let a = ref "" in
      let b = ref "" in
      read a $
      read b $
      x -<- (fun () -> !a ^ !b)

We can try this procedure. First we define an evaluator. It takes the
input stream and a procedure call as inputs.

    let rec eval l p =
      match l, p with
        _,      Term t -> t ()
      | s :: r, Cont c -> eval r (c s)
      | _              -> ((* Stuck evaluation *))

Then we evaluate read2 when two strings "a" and "b" are given as
input:

  let x = ref "" in eval ["a"; "b"] (read2 x); !x

-- Jerome