[Haskell-cafe] Re: Newcomers question
alexander.dunlap at gmail.com
Sun Nov 1 11:49:57 EST 2009
The type of liftA2 :: Applicative f =>(a -> b -> c) -> f a -> f b -> f
c. Thus, the type of liftA2 (==) :: (Eq b, Applicative f) => f b -> f
b -> f Bool. In your case, f :: a -> b, so liftA2 (==) :: (Eq b) => (a
-> b) -> (a -> b) -> (a -> Bool). (==) takes two arguments, so you're
left with the type (liftA2 (==)) x y :: a -> Bool. This contradicts
the class definition of Eq, which says that the type of (==) after
giving it two arguments must be Bool, not a -> Bool.
I hope that's clear enough. Busting out GHCi and using :t to find the
types of a lot of these expressions can be really helpful.
On Sun, Nov 1, 2009 at 8:09 AM, b1g3ar5 <nick.straw at gmail.com> 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.==)
> 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.
> On 31 Oct, 23:38, Daniel Peebles <pumpkin... at gmail.com> 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 <nick.st... at gmail.com> wrote:
>> > I'm trying:
>> > 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
>> > but the compiler rejects it with:
>> > 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)'
>> > Could someone please explain this to me?
>> > 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.
>> > Thanks
>> > _______________________________________________
>> > Haskell-Cafe mailing list
>> > Haskell-C... at haskell.org
>> Haskell-Cafe mailing list
>> Haskell-C... at haskell.orghttp://www.haskell.org/mailman/listinfo/haskell-cafe
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
More information about the Haskell-Cafe