Hello,

From: Khudyakov Alexey Hello

However I run into problem with them. It's possible to create instance for type class which doesn't have superclass. If it does have one compiler complains that it could not deduce context. All my attempts to provide context fail.

What have you tried?

Is that accidental limitation or because of Some Good Reason? Or just due to lack of understanding on my side?

Below is simplest example.

...

{-# LANGUAGE FlexibleInstances #-} {-# LANGUAGE TemplateHaskell #-} import Language.Haskell.TH

-- OK but require FlexibleInstances makeEq :: Name -> Q [Dec] makeEq name =     [d| instance Eq $(conT name) where           (==) = undefined      |]

-- Could not deduce Eq context makeOrd :: Name -> Q [Dec] makeOrd name =     [d| instance Ord $(conT name) where           compare = undefined      |]

And GHC output:

test.hs:14:17:    Could not deduce (Eq t) from the context ()      arising from the superclasses of an instance declaration                   at test.hs:14:17-32    Possible fix: add (Eq t) to the context of the instance declaration    In the instance declaration for `Ord t_aS5'    In the Template Haskell quotation      [d|          instance Ord $(conT name) where              { compare = undefined } |]    In the expression:        [d|            instance Ord $(conT name) where                { compare = undefined } |]

I think this will work if you add the Eq constraint in the instance definition: makeOrd :: Name -> Q [Dec] makeOrd name =     [d| instance Eq $(conT name) => Ord $(conT name) where           compare = undefined      |] The problem is that, when the (Ord t) instance is compiled, the compiler needs to know where to find the (Eq t) dictionary because Eq is a superclass of Ord. If 't' is a concrete type, e.g. Int, the compiler will look for the Eq instance and use it directly. If 't' is a type variable, as in your code, then the compiler needs to have the Eq constraint so that it can find the dictionary. In your case you provide an Eq instance for any type 't'. In theory I think the compiler could examine the instance, see that it's valid for any 't', and use that in your definition, but I don't think it's a good idea. This would be an extra branch that would only apply to a class that is instantiated for all types. Such an instance couldn't do anything useful; as an example your Eq instance is undefined. That's why GHC doesn't look for instances like the one you provided when determining type class constraints; they can't provide the necessary functionality. I suspect this feature could be added and then your code would compile, but you couldn't use it for anything so there's no point to doing so. Cheers, John