Yet Another Haskell Tutorial/Language advanced/Solutions
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Sections and Infix Operators
editLocal Declarations
editPartial Application
editFunction func3
cannot be converted into point-free style. The
others look something like:
func1 x = map (*x) func2 f g = filter f . map g func4 = map (+2) . filter (`elem` [1..10]) . (5:) func5 = flip foldr 0 . flip . curry
You might have been tempted to try to write func2
as filter f
. map
, trying to eta-reduce off the g
. In this case, this isn't
possible. This is because the function composition operator (.
)
has type (b -> c) -> (a -> b) -> (a -> c)
. In this
case, we're trying to use map
as the second argument. But
map
takes two arguments, while (.)
expects a function which
takes only one.
Pattern Matching
editGuards
editInstance Declarations
editThe Eq Class
editThe Show Class
editOther Important Classes
editThe Ord Class
editThe Enum Class
editThe Num Class
editThe Read Class
editClass Contexts
editDeriving Classes
editDatatypes Revisited
editNamed Fields
editMore Lists
editStandard List Functions
editList Comprehensions
editArrays
editFinite Maps
editLayout
editThe Final Word on Lists
editWe can start out with a recursive definition:
and [] = True and (x:xs) = x && and xs
From here, we can clearly rewrite this as:
and = foldr (&&) True
We can write this recursively as:
concatMap f [] = [] concatMap f (x:xs) = f x ++ concatMap f xs
This hints that we can write this as:
concatMap f = foldr (\a b -> f a ++ b) []
Now, we can do point elimination to get:
foldr (\a b -> f a ++ b) [] ==> foldr (\a b -> (++) (f a) b) [] ==> foldr (\a -> (++) (f a)) [] ==> foldr (\a -> ((++) . f) a) [] ==> foldr ((++) . f) []