OK, I understand that now but I've got a supplimentary question. If I put: instance Eq b => Eq (a -> b) where (==) = liftA2 (Prelude.==) to do the Eq part I get another error: Couldn't match expected type `Bool' against inferred type `a -> Bool' In the expression: liftA2 (==) In the definition of `==': == = liftA2 (==) In the instance declaration for `Eq (a -> b)' Now can someone please explain this? I'm hoping that when I've understood this stuff I'll have made a small step to understanding Haskell. Thanks. On 31 Oct, 23:38, Daniel Peebles wrote:

For some reason, Show and Eq are superclasses of Num (despite Num not actually using any of their methods), meaning that the compiler forces you to write instances of Eq and Show before it even lets you write a Num instance. I don't think anybody likes this, but I think we're stuck with it for the foreseeable future.

On Sat, Oct 31, 2009 at 7:31 PM, b1g3ar5 wrote:

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I'm trying:

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instance Num b => Num (a -> b) where fromInteger = pure . Prelude.fromInteger negate = fmap Prelude.negate (+) = liftA2 (Prelude.+) (*) = liftA2 (Prelude.*) abs = fmap Prelude.abs signum = fmap Prelude.signum

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but the compiler rejects it with:

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src\Main.hs:24:9:    Could not deduce (Show (a -> b), Eq (a -> b))      from the context (Num b)      arising from the superclasses of an instance declaration                   at src\Main.hs:24:9-29    Possible fix:      add (Show (a -> b), Eq (a -> b)) to the context of        the instance declaration      or add an instance declaration for (Show (a -> b), Eq (a -> b))    In the instance declaration for `Num (a -> b)'

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Could someone please explain this to me?

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I thought that it might be that it couldn't work out the functions necessary for (a->b) to be in the classes Show and Eq - so I tried adding definitions for == ans show, but it made no difference.

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Thanks

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On Sun, Nov 1, 2009 at 11:09 AM, b1g3ar5 wrote:

OK, I understand that now but I've got a supplimentary question.

If I put:

instance Eq b => Eq (a -> b) where    (==) = liftA2 (Prelude.==)

You don't need the "Prelude." here.

to do the Eq part I get another error:

   Couldn't match expected type `Bool'           against inferred type `a -> Bool'    In the expression: liftA2 (==)    In the definition of `==': == = liftA2 (==)    In the instance declaration for `Eq (a -> b)'

Now can someone please explain this?

The type of liftA2 (==) is forall f b. Applicative f => f b -> f b -> f Bool. In your case, f is (->) a, so liftA2 (==) :: (a -> b) -> (a -> b) -> (a -> Bool), which can't be unified with the type for (==). Personally, I recommend against trying to make (a -> b) an instance of Num. If you really want to do arithmetic on functions, it's easier to just do something along these lines: (.+.),(.-.),(.*.) :: (Applicative f, Num a) => f a -> f a -> f a (.+.) = liftA2 (+) (.-.) = liftA2 (-) (.*.) = liftA2 (*) -- Dave Menendez http://www.eyrie.org/~zednenem/