[Haskell] Swapping parameters and type classes

[Mads Lindstrøm]

Hi Bas

Thank you for the answer.

I tried to "fill in some blanks" in the example you gave. And mostly got
a lot of context reduction stack overflows :(

Here is my example (a little closer to what I actually need):

data Foo a b = Foo { first :: a, second :: b }

class Bar (x :: * -> *) where
    foo :: x a -> a

instance Bar (Foo a) where
    foo x = second x

type family BarB a b :: * -> *
type instance BarB a b = Foo b

instance Bar (BarB a b) where
    foo x = second x     -- this unexpectedly works! 
--  foo x = first x      -- This unexpectedly gives context reduction stack overflow

What surprises me is that I still need to look at `second`, even though
I use BarB. I thought I was swapping the parameters. Whats more changing
the line:

type instance BarB a b = Foo b

to

type instance BarB a b = Foo a  -- the last letter changed

has no effect.


Greetings,

Mads Lindstrøm

P.s. Why can we not just have the option of being explicit about which type parameters are applied? Something like:

"instance Bar (apply a. Foo a b)" which would apply a and be identical to "instance Bar (Foo a)"
"instance Bar (apply b. Foo a b)" which would apply b and be what I am trying to achieve.

It would seem a lot more natural to me. But maybe there are other reasons why type families are a better solution?

I do not know if I use the right terminology when saying "apply". Please correct if there is more correct terms.


Bas van Dijk:
> On 9/16/07, Mads Lindstrøm <mads_lindstroem at yahoo.dk> wrote:
> > But what if I want to "apply" the 'b' ? How do I do that ?
> 
> The following uses type families (functions) and compiles under GHC HEAD:
> 
> {-# OPTIONS_GHC -XTypeFamilies -XEmptyDataDecls -XTypeSynonymInstances #-}
> 
> data Foo a b
> 
> class Bar (x :: * -> *)
> 
> instance Bar (Foo a)
> 
> type family BarB a b :: * -> *
> type instance BarB a b = Foo b
> 
> instance Bar (BarB a b)
> 
> 
> regards,
> 
> Bas van Dijk