Indeed, the types foo :: forall a . (Num a) => a -> (Int, Float) and foo :: (forall a . (Num a) => a) -> (Int, Float) are quite different.
The first one say, I (foo) can handle any kind of numeric 'a' you (the caller) can pick. You (the caller) get to choose exactly what type you give me. The second one says, I (foo) require you (the caller) to give me an numeric 'a' that I can use any way I want. You (the caller) don't get to choose what type you give me, you have to give me a polymorphic one. -- Lennart On Thu, Oct 8, 2009 at 5:35 PM, Bulat Ziganshin <bulat.zigans...@gmail.com> wrote: > Hello Cristiano, > > Thursday, October 8, 2009, 7:14:20 PM, you wrote: > >> Could you explain why, under NoMonomorphismRestriction, this typechecks: > >> let a = 1 in (a + (1 :: Int),a + (1 :: Float)) > >> while this not: > >> foo :: Num a => a -> (Int,Float) >> foo k = (k + (1 :: Int), k + (1.0 :: Float)) > > i think it's because type is different: > > foo :: (forall a. (Num a) => a) -> (Int,Float) > > in first equation it probably inferred correctly > > > > -- > Best regards, > Bulat mailto:bulat.zigans...@gmail.com > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
