[Haskell-beginners] Type of function with constant pattern

[Brent Yorgey]

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.

This is not really the reason; no instance of Eq for Bool is required.
Note that you can even pattern-match like this on a type which is not
an instance of Eq:

  data Foo = Bar | Baz

  f Bar = True

Here the type of f is inferred as  Foo -> Bool, even though there is
no instance of Eq for Foo.  Pattern-matching is more fundamental than
equality testing.

The reason Eq is needed for the definition  f 0 = True   is that 0 is
not actually a constructor.  Numeric literal patterns are provided as
a convenience, but they desugar into a guard of the form

  f x | x == 0 = ...

Hence, an instance of Eq is needed.

-Brent