Re: [Caml-list] Purity in ocaml

[Goswin von Brederlow <goswin-v-b@web.de>]

On Tue, Jan 28, 2014 at 10:21:20AM +0100, Arnaud Spiwack wrote:
> > Do you mean something like this?
> >
> > let g = function 0 -> raise G | n -> n
> > let f = function 1 -> raise F | n -> n
> >
> > map f (map g [1;0]) ===> raise G
> > map (f %> g) [1;0]  ===> raise F
> >
> > It is true, the exception prevents any optimization that would reorder
> > two functions that both throw an excepition.
> 
> 
> Precisely
> 
> 
> > But many other
> > optimizations are still possible, e.g. common subexpression
> > elimination:
> >
> > let n =
> >   let x = f (g 1) in
> >   let y = g 1 in
> >   x + y
> >
> > ==>
> >
> > let n =
> >   let t = g 1 in
> >   let x = f t in
> >   let y = t in
> >   x + y
> >
> 
> To some degree:
> 
> let n =
>   let x = f (g 1) (h ()) in
>   let y = g 1 in
>   x+y
> 
> is a different program than
> 
> let n =
>   let t = g 1 in
>   let x = f t (h ()) in
>   let y = t in
>   x+y
> 
> As it reorders (g 1) and (h ()). I think the rule is something like: when
> the program is in monadic form, and you have a call to e, you can factor in
> every subsequent calls to e. Which I guess is good enough. But I would tend
> to believe deforestation would be tremendously more useful than common
> expression elimination.

Does it reorder? I would expect these two to be equivalent

    let x = f (g 1) (h ()) in

and

    let f2 = f (g 1) in
    let x = f (h ()) in

and in the second case the order is clearly g before h.

If the order is indeed to evaluate the last argument first then

    let t1 = h () in
    let t2 = g 1 in
    let x = f t2 t1 in
    let y = t2 in
    x + y

It's not hard to preserve the order whatever it may be.

MfG
	Goswin