Thanks Oleg,
that was very helpful. i can work with that. read the rest of this if you are curious where your hints took me. you are right, I need to make the functional dependency explicit: | class Table t c | t -> c where | toLists :: t -> [[c]] | fromLists :: [[c]] -> t | | instance Table [[c]] c where | toLists = id | fromLists = id | | instance (Table t c, Show c) => Show t where | showsPrec p t = showParen (p > 10) $ showString "fromLists " . shows (head . head $ toLists t) this compiles, and 'show' prints the first cell of each table. i also understand now why i can't just print all of them: [[Int]] is one of the types where the instances overlap. | instance Show Int | instance Show a => Show [a] vs. | instance Table [[Int]] Int | instance (Table [[Int]] Int, Show Int) => Show [[Int]] the advanced overlap code you are referencing below is fascinating, but isn't it a different problem? i don't want to distinguish different the cases "Show c" and "Typeable c", but i want to use whatever instance of "Show c" is available to implement "Show t". i can't make your solution work for this, because one of the two overlapping instances (namely "Show a => Show [a]") is already provided by the surrounding code that i cannot outfit with the advanced overlap trick. or am i missing something here? i think what i will do is to instantiate all table types individually: | instance Show c => Show (SimpleTable c) where | showsPrec p t = showParen (p > 10) $ showString "FastTable " . | shows (toLists t) cheers, matthias On Wed, May 09, 2012 at 06:41:00AM -0000, o...@okmij.org wrote: > Date: 9 May 2012 06:41:00 -0000 > From: o...@okmij.org > To: f...@etc-network.de > CC: haskell-cafe@haskell.org > Subject: Re: [Q] multiparam class undecidable types > > > > | instance (Table a c, Show c) => Show a where > > I would have thought that there is on overlap: the instance in my code > > above defines how to show a table if the cell is showable; > > No, the instance defines how to show values of any type; that type > must be an instance of Table. There is no `if' here: instances are > selected regardless of the context such as (Table a c, Show c) > above. The constraints in the context apply after the selection, not > during. > > Please see > ``Choosing a type-class instance based on the context'' > http://okmij.org/ftp/Haskell/TypeClass.html#class-based-overloading > > for the explanation and the solution to essentially the same problem. > > There are other problems with the instance: > > | instance (Table a c, Show c) => Show a where > > For example, there is no information for the type checker to determine > the type c. Presumably there could be instances > Table [[Int]] Int > Table [[Int]] Bool > So, when the a in (Table a c) is instantiated to [[Int]] > there could be two possibilities for c: Int and Bool. You can argue > that the type of the table uniquely determines the type of its > cells. You should tell the type checker of that, using functional > dependencies or associated types: > > class Table table where > type Cell table :: * > toLists :: table -> [[Cell table]] > > > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
