Re: [Haskell-cafe] Why is it so different between 6.12.1 and 6.10.4_1 ?

On Fri, Mar 26, 2010 at 8:59 PM, Ivan Lazar Miljenovic wrote:

zaxis writes:

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In 6.10.4_1 under freebsd

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let f x y z = x + y + z *Money> :t f f :: (Num a) => a -> a -> a -> a

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: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.

What? From where? I thought the whole reason the Monad ((->) a) instance was in Control.Monad.Instances (instead of Prelude) was to retain compatibility with the library report. -- Dave Menendez http://www.eyrie.org/~zednenem/