[Haskell-beginners] Type of function with constant pattern

[j.romildo at gmail.com]

On Tue, Apr 10, 2012 at 05:30:37PM -0400, Mike Meyer wrote:
> On Tue, 10 Apr 2012 16:57:38 -0300
> j.romildo at gmail.com wrote:
> 
> > Hello.
> > 
> > Given the following function definitions
> > 
> >    f 0 = True
> > 
> >    g False = True
> > 
> > ghc infers the following types for the functions:
> > 
> >    f :: (Eq a, Num a) => a -> Bool
> >    g :: Bool -> Bool
> > 
> > Why f has "Eq a" in the context in ts type, and g does not?
> 
> Bool is an instance of Eq, so there's no need to say that your
> (non-existent) type variable has that constraint.
> 
> Using a numeric constants means you get a type variable with the Num
> constraint. Since Num doesn't imply Eq, that constraint is (as Tim
> pointed out) required so the guard can be checked.

Then consider the following type and function definitions:

   data T = A | B

   h A = True
   h B = False

The type T is not an instance of Eq. The inferred type by ghc for the
function h is

   h :: T -> Bool

which does not require T to be an instance of the Eq class.

Therefore pattern matching on the constant data constuctors A and B is
not related to the Eq class, which provides the equality function (==).

This example contradicts the explanation given above by Mike.

And the question remains: why pattern matching on numeric constants,
differently from other constants, requires the Eq class?

Romildo