zaxis
In 6.10.4_1 under freebsd
let f x y z = x + y + z *Money> :t f f :: (Num a) => a -> a -> a -> a
:t (>>=) . f (>>=) . f :: (Monad ((->) a), Num a) => a -> ((a -> a) -> a -> b) -> a -> b ((>>=) . f) 1 (\f x -> f x) 2
<interactive>:1:1: No instance for (Monad ((->) a)) arising from a use of `>>=' at <interactive>:1:1-5 Possible fix: add an instance declaration for (Monad ((->) a)) In the first argument of `(.)', namely `(>>=)' In the expression: ((>>=) . f) 1 (\ f x -> f x) 2 In the definition of `it': it = ((>>=) . f) 1 (\ f x -> f x) 2
Some definitions and exports got changed, so in 6.12 the (-> a) Monad instance is exported whereas in 6.10 it isn't.
fac n = let { f = foldr (*) 1 [1..n] } in f
Why do you bother with the interior definition of f in there? fac = product . enumFromTo 1 -- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com IvanMiljenovic.wordpress.com