Hi Adrian
You can use Data.Derive to do this for you:
http://www-users.cs.york.ac.uk/~ndm/derive/
Or DrIFT: http://repetae.net/~john/computer/haskell/DrIFT/
Thanks
Neil
On 7/7/07, Adrian Hey <[EMAIL PROTECTED]> wrote:
Adrian Hey wrote:
> Hello,
>
> I'm trying to make the type (ListGT map k a) an instance of Typeable,
> where map is kind (* -> *).
>
> data ListGT map k a
> = Empt
> | BraF ![k] a !(map (ListGT map k a))
> | BraE ![k] !(map (ListGT map k a))
>
> I thought I'd cracked it with something like this..
>
> instance (Typeable (map (ListGT map k a)), Typeable k, Typeable a) =>
> Typeable (ListGT map k a) where
> typeOf lgt = mkTyConApp (mkTyCon "Data.Trie.General.ListGT")
> [mTypeRep, kTypeRep, aTypeRep]
> where BraF [k] a m = lgt -- This is just to get types for k a m !!
> kTypeRep = typeOf k
> aTypeRep = typeOf a
> mTypeRep = typeOf m
>
> However, showing the resulting TypRep gives a stack overflow. I wasn't
> too surprised about this, so I tried replacing the last line with..
> mTypeRep = mkTyConApp (typeRepTyCon (typeOf m)) []
> ..thinking that this would make it terminate. But it doesn't.
>
> Could someone explain how to do this?
(Answering my own question) this seems to do the trick..
instance (Typeable1 map, Typeable k, Typeable a) =>
Typeable (ListGT map k a) where
typeOf lgt = mkTyConApp (mkTyCon "Data.Trie.General.ListGT")
[mTypeRep, kTypeRep, aTypeRep]
where BraF [k] a m = lgt -- This is just to get types for k a m !!
kTypeRep = typeOf k
aTypeRep = typeOf a
mTypeRep = typeOf1 m
Regards
--
Adrian Hey
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