Some mostly unrelated thoughts:
An instance head has the form `T a_1 ... a_n`, and the constraint can only
apply to the `a_i`s. Consider the Show instance for
pairs.
instance (Show a, Show b) => Show (a, b) -- Show ((,) a b)
The constraints only act on the parameters of the type.
It looks like you're taking the constraint to mean "whenever I have a Showable
`f String`, this is how to define a Show instance", but a constraint
actually means "use this rule to make a Show instance for any `A f`, and
it is an error if a Show instance for `f String` is not in scope".
In the second error, you are making the strong claim that your Show
instance for `A f` holds for any `f` and `a`. Even if you could trick
the compiler into allowing that, I don't think it would actually express
the constraint that you want it to.
Is there something a Show instance gets you that a pretty-print function wouldn't?
Dmitriy Matrosov
Hi.
Is there a way to avoid `UndecidableInstances` in following code:
data A f = A {_a1 :: f String}
instance Show (f String) => Show (A f) where
it does not compile with
1.hs:4:10: error: • The constraint ‘Show (f String)’ is no smaller than the instance head (Use UndecidableInstances to permit this) • In the instance declaration for ‘Show (A f)’
Though, initially, this was
{-# LANGUAGE RankNTypes #-}
data A f = A {_a1 :: f String}
instance forall f a. Show (f a) => Show (A f) where
which also does not compile with
1.hs:5:10: error: • Variable ‘a’ occurs more often in the constraint ‘Show (f a)’ than in the instance head (Use UndecidableInstances to permit this) • In the instance declaration for ‘Show (A f)’
The error is different and i don't sure, that this two cases are related.
I want these instances to make a type with many records parametrized by `Alternative` type, e.g.
data Volume t = Volume { _volName :: t Name , _volSize :: t Size , _volPath :: t Path , _pool :: t Pool }
When i try to make instances, which require `*` type, i will end with above cases.
-- Jack