Le 29 juil. 09 à 01:28, Jon Harrop <jon@ffconsultancy.com> a écrit :

> On Tuesday 28 July 2009 22:47:25 Aaron Bohannon wrote:
>> let f (x : 'a) : 'a = x in (f true, f 3);;
>
> Err, good question. :-)
>
> # let f : _ = fun x -> x in f true, f 3;;
> - : bool * int = (true, 3)
The type of f here is something like 'a. 'a -> 'a. So you can use it  
on bool or int. I understand it like [], like in the expression (3 ::  
[], true :: []) where it has type 'a. 'a list (unfortunately it is not  
synctacally correct to say
let f : 'a. 'a -> 'a = fun x -> x, I don't know why.)

>
> # let f : 'a = fun x -> x in f true, f 3;;
> Error: This expression has type int but is here used with type bool
Here you specify that there exists a type named 'a, which is the  
return type of f and which should unify with both bool and int. Hence  
the typing error.


>
>
> I'm guessing the scope of the 'a is not what you'd expect but I have  
> no idea
> why. I'd have thought the latter would be a harmless type  
> annotation...

My guess is that the problem here is not about scope but about  
quantifiers : the type of f is quantified universally by default but  
the annotation constrains the type of f to a fixed (yet unknown) type  
in the subsequent expression (in that sense, yes, the scope of f plays  
a crucial role).

HTH,

Ph.

>
>
> Any takers?
>
> -- 
> Dr Jon Harrop, Flying Frog Consultancy Ltd.
> http://www.ffconsultancy.com/?e
>
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