I guess a canonical example of the reason behind this restriction would be
the following:

let put = let r = ref [] in fun x -> r := x::!r

OCaml will tell you that it has type '_a -> unit. It would be unsound
(exercise!) to decide it has type 'a -> unit.

Of course, your example is perfectly sound, but the typing algorithm
doesn't know that.

--
Arnaud Spiwack

On 10 January 2012 16:45, Romain Bardou <bardou@lsv.ens-cachan.fr> wrote:

> Le 10/01/2012 16:29, Dario Teixeira a écrit :
>
>  Hi,
>>
>> Consider functions foobar1 and foobar2:
>>
>>
>> type 'a t = {id: int; x: 'a}
>>
>> let foobar1: 'a ->  'a t =
>>         fun x ->  {id = 0; x}
>>
>> let foobar2: 'a ->  'a t =
>>         let ctr = ref 0 in
>>         fun x ->  incr ctr; {id = !ctr; x}
>>
>>
>>
>> I would expect them to have the same type, because foobar2's
>> use of a reference cell is kept private.  However, they don't.
>> In fact, foobar2 is not really polymorphic:
>>
>>
>> type 'a t = { id : int; x : 'a; }
>> val foobar1 : 'a ->  'a t
>> val foobar2 : '_a ->  '_a t
>>
>>
>> It's easy to get around this issue by putting the reference cell
>> outside of foobar2.  Function foobar3 does just this, and behaves
>> as expected:
>>
>>
>> let next =
>>         let ctr = ref 0 in
>>         fun () ->  incr ctr; !ctr
>>
>> let foobar3: 'a ->  'a t =
>>         fun x ->  {id = next (); x}
>>
>>
>>
>> Could someone point me to a good explanation of what's going on?
>> (I have the feeling I've read about this restriction before.)
>>
>> Best regards,
>> Dario Teixeira
>>
>>
>>
>>
> Hello,
>
> This is, basically, the value restriction: you cannot let-generalize
> something which is not *syntactically* a value. Function foobar1 is
> syntactically a function, and a function is a value. Function foobar2 is
> not: it starts with a let-binding. It computes something before returning a
> function. It cannot be generalized.
>
> Cheers,
>
> --
> Romain Bardou
>
>
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