
Let's try to see what is going on using GHCi 6.10.1: Prelude> :t lookup lookup :: (Eq a) => a -> [(a, b)] -> Maybe b Prelude> :t flip lookup flip lookup :: (Eq a) => [(a, b)] -> a -> Maybe b Everything ok. Lets try binding: Prelude> let f = flip lookup Prelude> :t f f :: [((), b)] -> () -> Maybe b Woops. Whenever you see a type of a function becoming less polymorphic than you expected, try adding a parameter: Prelude> let g x = flip lookup x Prelude> :t g g :: (Eq a) => [(a, b)] -> a -> Maybe b Nice! Let's try again without the monomorphism restriction[1] Prelude> :s -XNoMonomorphismRestriction Prelude> let h = flip lookup Prelude> :t h h :: (Eq a) => [(a, b)] -> a -> Maybe b Prelude> :s -XMonomorphismRestriction So, that's where the culprit lies! As we declared 'f' without a type signature and without explicit arguments, the monomorphism restriction didn't let us overload our function with the 'Eq' class. Note that if we activated all warnings, GHC would tell us that something is wrong: Prelude> :s -Wall Prelude> let j = flip lookup <interactive>:1:13: Warning: Defaulting the following constraint(s) to type `()' `Eq a' arising from a use of `lookup' at <interactive>:1:13-18 In the first argument of `flip', namely `lookup' In the expression: flip lookup In the definition of `j': j = flip lookup We can't just remove the 'Eq' constraint, if we want to refine the type of 'lookup' then we need to supply something that implements 'Eq'. As GHC doesn't know what you want to do, it just defaults to (). If, for example, you used the new function in the same let-binding, GHC would infer another type and show no warnings: Prelude> let k = flip lookup; l = k [] (1 :: Int) Prelude> :t k k :: [(Int, b)] -> Int -> Maybe b But note that even if you use the function where you defined it, it can't be polymorphic Prelude> let m = flip lookup; n = m [] (1 :: Int); o = m [] (1 :: Double) <interactive>:1:52: Couldn't match expected type `Int' against inferred type `Double' In the second argument of `m', namely `(1 :: Double)' In the expression: m [] (1 :: Double) In the definition of `o': o = m [] (1 :: Double) In short: add an explicit parameter or use a type signature. =) Hope that helps! [1] http://www.haskell.org/haskellwiki/Monomorphism_restriction -- Felipe.