Hi Alfonso!
"Alfonso Acosta"
type FSet a = Show a => (a -> Double)
type only works for redefinitions (i.e. adding the |Show a| constraint makes FSet a different type to (a -> Double)).
Yes, I know. What I mean, if type is a type macro, why cannot it expand type constraints too? Quoting the GHC manual
7.4.1.3. Liberalised type synonyms
Type synonyms are like macros at the type level, and GHC does validity checking on types only after expanding type synonyms.
Seemed to me like a syntactical restriction, however that parses OK, but the typer complains. I guess this is due to types like
type A a = Show a => a type B a = Show a => a
so if you do
f :: A a -> B b it should get translated to f :: (Show a => a) -> (Show b => b)
Which is not valid. I wonder if regrouping all the contexts in the left side should work:
f :: (Show a, Show b) => a -> b
In addition you seem to be trying to pack a dictionary with the type (something you cannot do in Haskell without existentials). This is
Why? My first guess it that there's no way to translate that to system F, but if we look at 'type' as a macro, it could make sense.
just a guess, but it seems that a definition using existentials is what you're looking for.
data FSet a = forall a. Show a => FSet (a -> Double)
That would be
data FSet = forall a. Show a => FSet (a -> Double)
not? The above is like
data FSet b = forall a. Show a => FSet (a -> Double)
I'm not looking yet for existential types, as then my functions will have to hold for every a, which turns out to be very inconvenient. The reason I don't want to use data, which works ok, is because I'd like to use FSet as a function, but not to write the type constraint in every function using as it is now.
type FSet a = a -> Double a :: Show a => FSet a ...
Thanks for your answer, Emilio pd: Greetings from Angel too :)