Haskell does not play as well with overloading as one would do it in C++; every name used must be fully qualified. Indeed, if we try something like Indeed, if we try something like data A = A Int deriving (Show, Eq) test = A 3 unA (A i) = i class Group a where (+) :: a -> a -> a instance Group A where (+) x y = A $ unA x + unA y we will get Ambiguous occurrence `+' It could refer to either `Main.+', defined at ****.hs:7:1 or `Prelude.+', imported from Prelude Failed, modules loaded: none. Haskell has its own brand of 'overloading': type classes. Every (+) sign used assumes that the operands are of the Num typeclass in particular. In order to define (+) on something else you will need to instance the Num typeclass over your A type. I am not sure what you mean by "the stuff defined in class Num is meanless to A." Strictly speaking nothing needs to be defined in a typeclass declaration other than the required type signatures. To instance the Num typeclass with A, though, assuming that A constructors take something that works with Num, you would do something similar to what Miguel posted: data A = A Int deriving (Show, Eq) test = A 3 unA (A i) = i instance Num A where (+) x y = A $ (unA x) + (unA y) (-) x y = A $ (unA x) - (unA y) (*) x y = A $ (unA x) * (unA y) abs x = A $ (unA $ abs x) signum y = A $ (unA $ signum y) fromInteger i = A (fromInteger i) Look at fromInteger, which must take Integer as as argument. That may be inconvenient for you. The Awesome Prelude, referenced in Chris's post, is a way of defining less specific version of basic types like Bool so that you have more choices in defining things like fromInteger in the Num typeclass (which must take an Integer; it is 'sad' if that Integer refers to a grounded, specific type). Still, if not every one of the Num operations make sense for your A type, you can leave them blank and get a warning. On Sun, Nov 21, 2010 at 10:48 PM, Magicloud Magiclouds < magicloud.magiclouds@gmail.com> wrote:
Hi, For example, I have a data A defined. Then I want to add (+) and (-) operators to it, as a sugar (compared to addA/minusA). But * or other stuff defined in class Num is meanless to A. So I just do: (+) :: A -> A -> A (+) a b = A (elem1 a + elem1 b) (elem2 a + elem2 b) -- I got errors here, for the (+) is ambiguous.
So, just wondering, does this way work in Haskell? -- 竹密岂妨流水过 山高哪阻野云飞 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe