Hi, class Processor a where ready :: (forall b c. a → [b → c]) {- instance Processor (b → c) where ready = repeat instance Processor [b → c] where ready = id-} doSth :: (Show p, Processor p) ⇒ p → IO () doSth p = print p ------------------------------- Why can I not declare the above instances and always get: message.hs:229:10: Couldn't match expected type `b' against inferred type `b1' `b' is a rigid type variable bound by the instance declaration at message.hs:228:20 `b1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:19 Expected type: b -> c Inferred type: b1 -> c1 In the expression: repeat In the definition of `ready': ready = repeat message.hs:229:10: Couldn't match expected type `c' against inferred type `c1' `c' is a rigid type variable bound by the instance declaration at message.hs:228:24 `c1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:21 Expected type: b -> c Inferred type: b1 -> c1 In the expression: repeat In the definition of `ready': ready = repeat message.hs:232:10: Couldn't match expected type `b1' against inferred type `b' `b1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:19 `b' is a rigid type variable bound by the instance declaration at message.hs:231:20 Expected type: [b1 -> c] Inferred type: [b -> c1] In the expression: id In the definition of `ready': ready = id message.hs:232:10: Couldn't match expected type `c1' against inferred type `c' `c1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:21 `c' is a rigid type variable bound by the instance declaration at message.hs:231:24 Expected type: [b -> c1] Inferred type: [b1 -> c] In the expression: id In the definition of `ready': ready = id Is there a way around this? Regards, CS