On 5/14/07, Roberto Zunino
Also, using only rank-1:
polyf :: Int -> a -> Int polyf x y = if x==0 then 0 else if x==1 then polyf (x-1) (\z->z) else polyf (x-2) 3
Here passing both 3 and (\z->z) as y confuses the type inference.
Actually, I tried this in ghci... Should this work? polyf.hs: polyf x y = if x==0 then 0 else if x==1 then polyf (x-1) (\z->z) else polyf (x-2) 3 NOTE: no type signature Prelude> :l polyf [1 of 1] Compiling Main ( polyf.hs, interpreted ) Ok, modules loaded: Main. *Main> :t polyf polyf :: forall a t1 t. (Num (t1 -> t1), Num a, Num t) => a -> (t1 -> t1) -> t The inference assigns y the type (t1 -> t1) even though it is assigned the value 3? Regards, Chris