On Tue, Apr 10, 2012 at 05:30:37PM -0400, Mike Meyer wrote:
On Tue, 10 Apr 2012 16:57:38 -0300 j.romildo@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