2008/12/26 Oscar Picasso
Hi,
I can write: *Main> let yes = not . not *Main> :t yes yes :: Bool -> Bool
But not: *Main> let isNotEqual = not . (==)
The definition of (.): f . g = \x -> f (g x) So: not . (==) = \x -> not ((==) x) But (==) x is a function (of type a -> Bool, which returns whether its argument is equal to x), not a Bool as not is expecting. Composition like that usually only works for single argument functions. It gets uglier quickly for multi arg functions: isNotEqual = (not .) . (==) You can define your own operator to help: (.:) = (.) . (.) Which is a bit of silly definition. I typically don't like pointfree style for any but the most transparent of uses, so in real life (tm), I'd probably write: (f .: g) x y = f (g x y) But be prepared for others to disagree with me. Luke
<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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