[Haskell-cafe] Type inference
Cale Gibbard
cgibbard at gmail.com
Wed Feb 8 23:14:23 EST 2006
On 08/02/06, Brian Hulley <brianh at metamilk.com> wrote:
> Fred Hosch wrote:
> > Is type inferencing in Haskell essentially the same as in SML? Thanks.
>
> Well, that depends on what you mean by "essentially the same" ;-)
>
> Both languages are based on the same Hindley-Milner type inference
> algorithm, so both suffer from the same problem that a function such as
>
> f g x y = (g x, g y)
>
> can't be given a very satisfactory type (ie there exist perfectly good
> programs that will be rejected by both SML and Haskell due to limitations
> inherent in the kinds (excuse the pun) of type system that can be used with
> HM type inference)
>
> However, Haskell has a lot of advanced features that are bolted on to this
> foundation, which SML doesn't. One such feature is arbitrary rank
> polymorphism, which allows you to use a function argument in more than one
> way within a function, for example (compile with ghc -fglasgow-exts):
>
> h :: (forall a. a->a) -> b -> c -> (b,c)
> h g x y = (g x, g y) -- the argument g is used polymorphically
>
> This function can't be written in SML. Note that although h is similar to f,
> there would not exist a type for h if g could be an arbitrary function ie
> a->d instead of a->a.
The type forall a d. a -> d isn't terribly useful, since there just
aren't many functions of that type. You can give a type signature
like:
f :: (forall a b. a -> b) -> c -> d -> (e,f)
f g x y = (g x, g y)
but good luck actually applying the function in a useful way :) The
only thing you can reasonably pass as g (barring the existence of
functions which completely break the type system) is bottom.
So what is it that you're looking for here? I'm not sure I understand.
- Cale
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