Le 29 juil. 09 à 01:28, Jon Harrop <j...@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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