In 6.12.1 under archlinux
let f x y z = x + y + z :t f f :: (Num a) => a -> a -> a -> a
:t (>>=) . f (>>=) . f :: (Num a) => a -> ((a -> a) -> a -> b) -> a -> b ((>>=) . f) 1 (\f x -> f x) 2 5
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 Sincerely! ----- fac n = let { f = foldr (*) 1 [1..n] } in f -- View this message in context: http://old.nabble.com/Why-is-it-so-different-between-6.12.1-and-6.10.4_1---t... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.