[Haskell-cafe] Re: Newcomers question

[Alexander Dunlap]

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.

Alex

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.
>
> Thanks.
>
>
> 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
>>
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