[Haskell-beginners] Type of function with constant pattern
j.romildo at gmail.com
j.romildo at gmail.com
Wed Apr 11 02:40:43 CEST 2012
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
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