Re: [Haskell-cafe] type inference question

The reason a gets a single type is the monomorphism restriction (read the report). Using NoMonomorphismRestriction your example with a works fine. On Thu, Oct 8, 2009 at 12:29 PM, Cristiano Paris wrote:

On Thu, Oct 8, 2009 at 11:04 AM, minh thu wrote:

...

Hi,

I'd like to know what are the typing rules used in Haskell (98 is ok).

Specifically, I'd like to know what makes

let i = \x -> x in (i True, i 1)

legal, and not

let a = 1 in (a + (1 :: Int), a + (1.0 :: Float))

Is it correct that polymorphic functions can be used polymorphically (in multiple places) while non-functions receive a monomorphic type ?

First, "1" IS a constant function so it's in no way special and is a value like any other.

That said, the type of 1 is (Num t) => t, hence polymorphic. But, when used in the first element of the tuple, a is assigned a more concrete type (Int) which mismatches with the second element of the tuple, which is a Float.

If you swap the tuple, you'll find that the error reported by ghci is the very same as before, except that the two types are swapped.

Is it possible to rewrite the expression so as to work? The answer is yes, using existential quantification (and Rank2 polymorphism).

Here you are: {-# LANGUAGE ExistentialQuantification, Rank2Types #-} foo :: (forall a. (Num a) => a) -> (Int,Float) foo = \x -> (x + (1 :: Int), x + (1 :: Float))

Hence:

foo 1 --> (2,2.0)

Bye,

Cristiano _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe