I understand simple foldr statements like
foldr (+) 0 [1,2,3]
However, I'm having trouble with more complicated foldr statements, namely ones that take 2 parameters in the function, and with / and - computations. Could anyone explain the steps that occur to get these answers?
foldr (\x y -> (x+y)*2) 2 [1,3] = 22
foldr (/) 2 [8,12,24,4] = 8.0
Thanks.
The function argument of a fold always takes two arguments. (+)
and (/)
are binary functions just like the one in your second example.
Prelude> :t (+)
(+) :: Num a => a -> a -> a
If we rephrase the second example as
foldr f 2 [1,3]
where
f x y = (x+y)*2
we can expand the right fold using exactly the same scheme we would use for (+)
:
foldr f 2 [1,3]
foldr f 2 (1 : 3 : [])
1 `f` foldr f 2 (3 : [])
1 `f` (3 `f` foldr f 2 [])
1 `f` (3 `f` 2)
1 `f` 10
22
It is worth noting that foldr
is right-associative, which shows in how the parentheses nest. Conversely, foldl
, and its useful cousin foldl'
, are left-associative.
The foldr
function is defined as follows:
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr _ a [] = a
foldr f a (x:xs) = f x (foldr f a xs)
Now consider the following expression:
foldr (\x y -> (x + y) * 2) 2 [1,3]
We'll give the lambda a name:
f x y = (x + y) * 2
Hence:
foldr f 2 [1,3]
-- is
f 1 (foldr f 2 [3])
-- is
f 1 (f 3 (foldr f 2 []))
-- is
f 1 (f 3 2)
-- is
f 1 10
-- is
22
Similarly:
foldr (/) 2 [8,12,24,4]
-- is
8 / (foldr (/) 2 [12,24,4])
-- is
8 / (12 / (foldr (/) 2 [24,4]))
-- is
8 / (12 / (24 / (foldr (/) 2 [4])))
-- is
8 / (12 / (24 / (4 / (foldr (/) 2 []))))
-- is
8 / (12 / (24 / (4 / 2)))
-- is
8 / (12 / (24 / 2.0))
-- is
8 / (12 / 12.0)
-- is
8 / 1.0
-- is
8.0
Hope that helped.
You can think of foldr
, maybe
, and either
as functions that replace the data constructors of their respective types with functions and/or values of your choice:
data Maybe a = Nothing | Just a
maybe :: b -> (a -> b) -> Maybe a -> b
maybe nothing _just Nothing = nothing
maybe _nothing just (Just a) = just a
data Either a b = Left a | Right b
either :: (a -> c) -> (b -> c) -> Either a b -> c
either left _right (Left a) = left a
either _left right (Right b) = right b
data List a = Cons a (List a) | Empty
foldr :: (a -> b -> b) -> b -> List a -> b
foldr cons empty = loop
where loop (Cons a as) = cons a (loop as)
loop Empty = empty
So in general, you don't really have to think about the recursion involved, just think of it as replacing the data constructors:
foldr f nil (1 : (2 : (3 : []))) == (1 `f` (2 `f` (3 `f` nil)))
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.