Re: Page 76 of OCaml by John Whitington ???

I also gave something like this a try, but I ended up with a couple of
error messages:

*let pack list = match list with*
*|[ ] -> [ ]*
*|head :: tail -> match tail with*
*|[ ] -> [ ]*
*headOfTail :: tailOfTail -> if head = headOfTail then..... ( more code )*

I thought it was clever, but either I just ended up with an empty list, a
one-element list, or the compiler just rejected it.  I'm starting to feel a
little dumb!  But I'll keep at it.



On Mon, Mar 28, 2016 at 7:12 AM, Kenneth Miller [email protected]
[ocaml_beginners] <[email protected]> wrote:

>
>
> Right - almost always, if you're using references in a strongly typed
> functional language, you're going against the idioms and best strengths of
> the language. It can be hard to think of everything as having a recursive
> solution, especially since it can often have non-obvious implementation
> when you first start. Frequently, new functional programmers find
> themselves repeating the habits they've been taught in imperative
> languages, and those can be hard to unlearn.
>
> And thank you for your kind words! I'm honored, but I'm just an avid
> learner with a master's; universities wouldn't accept me. I think I'll
> pursue a PhD soon, but I have to finish my obligations with my current line
> of work first. Also, Dr. Sleator's fix of your function clearly illustrates
> what I've said above while also nudging you away from using references.
> Thanks Dr. Sleator.
>
>
>
> On Sunday, March 27, 2016 10:21 PM, "[email protected]
> [ocaml_beginners]" <[email protected]> wrote:
>
>
>
> Part of the problem is that the "reference solution" is incomprehensible.
> Here's how I did it:
>
> let pack li =
>   let rec grab_block li ac =
>     (* Grab a block of equal elements from li and put them into ac.
>        Return the pair (li,ac). *)
>     match li with
>       | [] -> ([],ac)
>       | x::tail ->
>     if ac=[] || x = List.hd ac then grab_block tail (x::ac)
>     else (li, ac)
>   in
>   let rec loop li ac = if li = [] then List.rev ac else
>       let (rest,bl) = grab_block li [] in
>       loop rest (bl::ac)
>   in
>   loop li []
>
> I think this is much easier to understand, and just as efficient.
>
> ---DS
>
>
> 
>