Re: [Haskell-cafe] Defining methods generically for a class
That's probably the best thing to do, yes. The purpose of doing so was for Data.List.delete, but I see now there's a Data.List.deleteBy, so I can use the regionEquals function as my equality predicate. On Thu, Jan 8, 2009 at 12:26 PM, Martijn van Steenbergen wrote: > Hi Jeff, > > Jeff Heard wrote: >> >> instance Region a => Eq a where >> regiona == regionb = all $ zipWith (==) (bounds regiona) (bounds >> regionb) > > If you want to be Haskell98 compliant, why not define regionEquals :: Region > a => a -> a -> Bool as above and use that everywhere instead of (==)? > > If you insist on using the overloaded (==), then in Haskell98 you will need > to define individual Eq instances for all your custom region types. However, > you can simply define (==) = regionEquals for all those types. > > Hope this helps, > > Martijn. > ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Fwd: [Haskell-cafe] Defining methods generically for a class
-- Forwarded message -- From: Jeff Heard Date: Thu, Jan 8, 2009 at 12:26 PM Subject: Re: [Haskell-cafe] Defining methods generically for a class To: Cristiano Paris Not really... I'm not testing if each of the items a are equal, but rather that in the context of them being a Region, they are equal.As long "dim" (which is in the class) can be defined, then equality is defined over all types Region. On Thu, Jan 8, 2009 at 12:16 PM, Cristiano Paris wrote: > On Thu, Jan 8, 2009 at 6:04 PM, Jeff Heard > wrote: >> ... >> How do I declare all Regions to be Eqs without putting it in the class >> body (since I define a function over all Regions that is independent >> of datatype that is an instance of Region)? > > Would this be a solution? > > class Eq a => Region a where > ... > > Cristiano > ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Defining methods generically for a class
Hi Jeff, Jeff Heard wrote: instance Region a => Eq a where regiona == regionb = all $ zipWith (==) (bounds regiona) (bounds regionb) If you want to be Haskell98 compliant, why not define regionEquals :: Region a => a -> a -> Bool as above and use that everywhere instead of (==)? If you insist on using the overloaded (==), then in Haskell98 you will need to define individual Eq instances for all your custom region types. However, you can simply define (==) = regionEquals for all those types. Hope this helps, Martijn. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Defining methods generically for a class
On Thu, Jan 8, 2009 at 6:04 PM, Jeff Heard wrote: > ... > How do I declare all Regions to be Eqs without putting it in the class > body (since I define a function over all Regions that is independent > of datatype that is an instance of Region)? Would this be a solution? class Eq a => Region a where ... Cristiano ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Defining methods generically for a class
I have the following class: class Region a where numDimensions :: a -> Int dim :: Int -> a -> (Double,Double) merge :: a -> a -> a and several ancillary methods defined, the most importance of which is: bounds :: Region a => a -> [(Double,Double)] bounds r = take (numDimensions r) . map dim . iterate (+1) $ 0 Let's say that I want all Regions to also be of class Eq, where regiona == regionb = all $ zipWith (==) (bounds regiona) (bounds regionb) How do I declare all Regions to be Eqs without putting it in the class body (since I define a function over all Regions that is independent of datatype that is an instance of Region)? Is it merely: instance Region a => Eq a where regiona == regionb = all $ zipWith (==) (bounds regiona) (bounds regionb) -- Jeff ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
