On oct 15 2006 at 05:29, Denis Bueno wrote:
> ,----
> | val of_array: ?getter : ('b -> 'a) -> 'b array -> 'a t
> |   (** [of_array xs] returns a set containing the elements in  
> [xs]. *)
> `----
>
> I get a type error when trying to compile this. I don't know how to
> fix it, or even if it's possible without wrecking my nice of_array
> signature.

Your very nice of_array signature... I'm afraid you're signature is  
awful. Basically you're trying to get a 'a -> 'b type, which is  
impossible without using Obj.magic, and awful anyway. See, if you do

# let my_array = [| (1,'a') ; (2,'b') |];;
val my_array : (int * char) array = [|(1, 'a'); (2, 'b')|]
# let set = of_array ~getter:snd my_array;;
val set : char list = ['a'; 'b']

This is okay, but if you decide not to give the optional argument:

# let set2 = of_array my_array

Then what type should be inferred for set2? of course the real type  
is (int * char) t, but with the signature you gave for of_array you  
cannot infer it, since they're is no way to know that the default  
value for the optional argument has the type 'a -> 'a.

By implementing of_array with Obj.magic, you get:

# let set2 = of_array my_array;;
val set2 : 'a list = [<poly>; <poly>]

and then you can do funny (and ugly) things:

# add 4 set2;;
- : int list = [4; 354636; 354630]
# add 'c' set2;;
- : char list = ['c'; '\
36'; '\
30']
# add (3,'c') set2;;
- : (int * char) list = [(1, 'a'); (2, 'b'); (3, 'c')]

I implemented of_array this way:
# let of_array ?(getter = fun x -> Obj.magic x) xs =
   Array.fold_right (fun x set -> add (getter x) set) xs empty;;
   val of_array : ?getter:('a -> 'b) -> 'a array -> 'b list = <fun>

Regards,

Etienne