Mateusz Kowalczyk | 10 May 15:58 2013
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Typeclass with an ‘or’ restriction.


Greetings,

We can currently do something like
> class (Num a, Eq a) => Foo a where bar :: a -> a -> Bool bar =
> (==)

This means that our `a' has to be an instance of Num and Eq. Apologies
for a bit of an artificial example.

Is there a way however to do something along the lines of:
> class Eq a => Foo a where bar :: a -> a -> Bool bar = (==)
> 
> class Num a => Foo a where bar :: a -> a -> Bool bar _ _ = False
This would allow us to make an instance of Num be an instance of Foo
or an instance of Eq to be an instance of Foo.

The compiler currently complains about multiple declarations. Is there
currently a way to achieve this?

The main issue I can see with this is that given an instance of both,
Num and Eq, it wouldn't be possible to pick the correct default
implementation.

Purely a theoretical question.

--

-- 
Mateusz K.
Adam Gundry | 10 May 17:10 2013
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Re: Typeclass with an ‘or’ restriction.

Hi Mateusz,

It's not directly possible to write a class with a choice of
superclasses; as you point out, it's not really clear what that would
mean. One workaround, though it might not be sensible in practice, is
the following.

> {-# LANGUAGE ConstraintKinds, GADTs #-}

First, reify the constraints we are interested in as types that pack up
the corresponding dictionary. Thanks to ConstraintKinds, it's possible
to do this once and for all.

> data Dict c where
>   Dict :: c => Dict c

Now we can describe types with either Num or Eq dictionaries (or both)
as a class.  The proxy argument makes it easy to specify the type, in
the absence of explicit type application.

> class NumOrEq a where
>   numOrEq :: proxy a -> Either (Dict (Num a)) (Dict (Eq a))

Something like your Foo class can then be defined like this:

> class NumOrEq a => Foo a where
>   bar :: a -> a -> Bool
>   bar x y = case numOrEq [x] of
>                 Left Dict   -> False
>                 Right Dict  -> x == y
(Continue reading)


Gmane