CODEWITHSUNDEEP
haskell
Rohit Sharma
Rohit Sharma

Reputation: 6500

Function composition does not work unless used with dot operator

Why does the first case works OK but not the second? I believe . operator chains the function and since f returns Boolean i could "not" it, but why does it not work in the second case?

*Main> let filterNot f xs = filter (not . f) xs
*Main> let filterNot f xs = filter (not (f)) xs

Upvotes: 0

Views: 152

Answers (1)

bheklilr
bheklilr

Reputation: 54068

This is because there is a difference between not . f and not f. If you want to do it without using ., you'll have to use a lambda:

filterNot f xs = filter (\x -> not (f x)) xs

But this is precisely the definition of .!

(.) :: (b -> c) -> (a -> b) -> (a -> c)
f . g = \x -> f (g x)

Just because f returns a Bool does not mean that you can not it. You can only not a Bool itself, not a function that returns one. There is a large difference between a -> Bool and just Bool. The definition of not is

not :: Bool -> Bool
not True = False
not False = True

So if f itself is not True or False exactly, then you can't apply not to it.

(Whew, that's a not of negatives in this explanation)

Upvotes: 9

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