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

I just start ghci from shell and do nothing else. In fact, i really donot know `Monad ((->) a) ` . Would you mind expplain it ? Yusaku Hashimoto wrote:

Did you import the module includes the instance of Monad ((->) e) somewhere in your code loaded in ghci?

I tried this on a fresh ghci 6.12, but I got "No instance" error.

-nwn

On Sat, Mar 27, 2010 at 9:20 AM, zaxis wrote:

...

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!

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