Maybe it'd be more intuitive if written backwards:
AppEq f a <= (Applicative f, Eq a)
or even
AppEq f a => (Applicative f, Eq a)
On 27 Nov 2008, at 00:39, Ryan Ingram wrote:
A common mistake (and a confusing bit about typeclasses) is that
whether or not the constraints on an instance apply are irrelevant.
Specifically, the code "instance (Applicative f, Eq a) => AppEq f a"
means that, given any types f and a, I can tell you how to make them
an instance of AppEq. But I also ask you to please add the
constraints "Applicative f" and "Eq a". That is to say, only the
stuff on the right of the => apply when determining whether an
instance applies.
If you take out the overlapping specific instance for Interval, the
compiler will give you a different error:
"No instance for Applicative Interval". You can see what happened
here: the compiler wants an instance for AppEq Interval Integer. It
sees the instance "AppEq f a" and adds the constraints "Ord Integer"
and "Applicative Interval". Ord Integer is already fulfilled, but it
can't discharge the constraint on Applicative, so the compile fails.
Similarily, in your case, the compiler can't decide whether to apply
the "Ord a => AppEq Interval a" instance, or the "Applicative f, Eq a
=> AppEq f a" instance; the right hand sides of the instance
declarations both match (and add different constraints to the left
hand side).
You can use -XOverlappingInstances, but beware, dragons lie in that
direction.
I think this is a fundamental weakness of the typeclass system, but I
haven't seen a design that avoids it for code as complicated as this.
On Wed, Nov 26, 2008 at 12:05 PM, Paul Johnson <[EMAIL PROTECTED]>
wrote:
Hi,
I'm trying to set up some operators for applicative versions of
prelude
types. For instance:
-- | Applicative Equality.
class (Eq a) => AppEq f a where
(.==.), (./=.) :: f a -> f a -> f Bool
instance (Applicative f, Eq a) => AppEq f a where
(.==.) = liftA2 (==)
(./=.) = liftA2 (/=)
Hopefully the intention is fairly straightforward: if "f" is an
instance of
Applicative then the lifted implementation of the underlying type.
Otherwise I can just give my own instance, which is useful for
things that
"wrap" prelude types but where "fmap" doesn't work. For instance:
data (Ord a) => Interval a = Interval a a
instance (Ord a) => AppEq Interval a where
i1@(Interval _ u1) .==. i2@(Interval _ u2)
| isSingleton i1 && isSingleton i2 && u1 == u2 = Interval True
True
| has i1 u2 || has i2 u1 = Interval False
True
| otherwise = Interval False
False
i1 ./=. i2 = let Interval b1 b2 = (i1 .==. i2) in Interval (not b2)
(not
b1)
isSingleton :: (Ord a) => Interval a -> Bool
isSingleton (Interval lower upper) = lower == upper
has :: (Ord a) => Interval a -> a -> Bool
has (Interval lower upper) v = v >= lower && v <= upper
You can't (easily) define fmap for Interval because the function
given as an
argument might not be monotonic. So instead you have to write custom
implementations for each lifted function, as shown here for (.==.)
and
(./=.) . The same principle works for AppOrd, AppNum etc, but I'm
trying to
solve the problem for just AppEq for now.
This compiles, but when I try to use it I get this in ghci:
*Interval> let i1 = Interval 4 5
*Interval> let i2 = Interval 4 6
*Interval> i1 .==. i2
<interactive>:1:0:
Overlapping instances for AppEq Interval Integer
arising from a use of `.==.' at <interactive>:1:0-9
Matching instances:
instance (Ord a) => AppEq Interval a
-- Defined at Interval.hs:(22,0)-(27,78)
instance (Control.Applicative.Applicative f, Eq a) => AppEq f a
-- Defined at AppPrelude.hs:(32,0)-(34,23)
In the expression: i1 .==. i2
In the definition of `it': it = i1 .==. i2
I'm puzzled, because Interval is not an instance of Applicative, so
the
second instance doesn't apply. Can anyone help me out?
I'm using ghc 6.8.3, so its possible that this was a bug fixed in
6.10.
Paul.
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