[Haskell-cafe] Function composition

[Tony Morris]

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Check the type of (.)

Prelude> :t (.)
(.) :: (b -> c) -> (a -> b) -> a -> c

Then the type of (.) not

Prelude> :t (.) not
(.) not :: (a -> Bool) -> a -> Bool

Now try to apply (==)

Prelude> :t (.) not (==) -- not going to happen

Won't happen. What do you want to happen?


Oscar Picasso wrote:
> Hi,
>
> I can write:
> *Main> let yes = not . not
> *Main> :t yes
> yes :: Bool -> Bool
>
> But not:
> *Main> let isNotEqual = not . (==)
>
> <interactive>:1:23:
>     Couldn't match expected type `Bool'
>            against inferred type `a -> Bool'
>     Probable cause: `==' is applied to too few arguments
>     In the second argument of `(.)', namely `(==)'
>     In the expression: not . (==)
>
> Why?
>
> Oscar
>
>
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- --
Tony Morris
http://tmorris.net/

S, K and I ought to be enough for anybody.

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