The problem in the class is that it doesn't necessarily know that (allows opts) and (crit a allows opts) necessarily are working on the exact same Flt instance. (allows opts) could return [Foo], and but crit is constrainted by the argument passed to it which is the a in question. The fact that there is only one possible instance right now does not change the fact that there could be more in the future. denied is already constrained because its return value is used as an argument of crit. To fix it {-# LANGUAGE ScopedTypeVariables #-} class Flt a where .... where allowed = if null $ (allows opts :: [a]) then True else a `crit` (allows opts) denied = if null $ (denies opts :: [a]) then False else (a :: a) `crit` (denies opts) When you see an error Flt a1 does not match Flt a, that's a classic sign that it doesn't know a1 and a are the same type. As for the instance it has the exact same problem. If you were to pull allowed into its own function outside the class you could constrain both functions at the same time, at the cost of some verbosity. {-# LANGUAGE ScopedTypeVariables #-} class Flt a where ... flt opts a = allowed2 opts a && not denied instance Flt MyType where ... flt opts a = allowed2 opts a && not denied allowed2 :: forall a. Flt a => FltOpts -> a -> Bool allowed2 opts a = if null $ (allows opts :: [a]) then True else a `crit` (allows opts) On Mon, Sep 25, 2017 at 6:06 AM, Baa wrote:

Hello, everyone.

Considering, I have a class:

class Flt a where allows :: FltOpts -> [a] denies :: FltOpts -> [a] crit :: a -> [a] -> Bool flt :: FltOpts -> a -> Bool flt opts a = allowed && not denied where allowed = if null $ allows opts then True else a `crit` (allows opts) denied = if null $ denies opts then False else a `crit` (denies opts)

I get error here:

• Could not deduce (Flt a1) arising from a use of ‘allows’ from the context: Flt a bound by the class declaration for ‘Flt’ at .../.stack-work/intero/intero5319V42.hs:(31,1)-(38,97) The type variable ‘a1’ is ambiguous These potential instance exist: instance Flt MyType -- Defined at ... • In the second argument of ‘($)’, namely ‘allows opts’ ....................................................

As I understand, GHC can not deduce type if it's a return's value (contraposition?). OK, but it knows its type: it is `[a]`! What is the problem to keep `flt` method as a generic, i.e. without concreate type, but only `[a]` ?

Second, I implemented instance:

instance Flt MyType where allows = ... denies = ... flt opts a = allowed && not denied where allowed = if null $ (allows opts::[MyType]) then True else a `crit` (allows opts) denied = if null $ (denies opts::[MyType]) then False else a `crit` (denies opts)

and without this explicite type annotation of `allows opts` I get again ambigous error. But why? GHC knows that `allows` returns `[a]` and `a` is `MyType`, so `[a]` is `[MyType]`. Why I need to write it explicitly? May be I need some extension here?

=== Best regards, Paul _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners