[Haskell-cafe] Why is it so different between 6.12.1 and 6.10.4_1
?
Yusaku Hashimoto
nonowarn at gmail.com
Sat Mar 27 16:20:54 EDT 2010
Hmm, When a ghci was started, there should be the only loaded module
(Prelude.) And in both 6.10 and 6.12, such instance is not defined or
exported in its Prelude. So please try `ghci -ignore-dot-ghci`. It
invokes ghci without reading ~/.ghci and ./.ghci.
And `((->) a)` is known as the Reader Monad, `a` can be regarded as
the environment. My typical usage of that is like following:
import Control.Monad
data Vec = Vec { x :: Int, y :: Int }
absolute :: Vec -> Double
absolute = sqrt . fromIntegral . liftM2 (+) (square . x) (square . y)
where
square a = a * a
The definition of `absolute` above can be rewritten as
absolute p = sqrt . fromIntegral $ square (x p) + square (y p)
where
square a = a * a
How `square . x` and `square . y` share the argument? Because `Monad
((->) a)` is defined as
instance Monad ((->) a) where
return x = \a -> x
m >>= f = \a -> f (m a) a
Note `(>>=)` propagates `a` into both of its arguments. That's why the
functions read same argument.
HTH
-nwn
On Sat, Mar 27, 2010 at 3:31 PM, zaxis <z_axis at 163.com> wrote:
>
> 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 <z_axis at 163.com> 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!
>>>
>>>
>>> -----
>>> fac n = let { f = foldr (*) 1 [1..n] } in f
>>> --
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>
>
> -----
> 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---tp28049329p28050535.html
> Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
>
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