文件夾返回Haskell中的列表

Question

我正在閱讀本指南，並了解返回數字時文件foldr工作方式。

sum' :: (Num a) => [a] -> a  
sum' xs = foldl (\acc x -> acc + x) 0 xs 

ghci> sum' [3,5,2,1]  
11

現在，我需要使用foldr返回列表，並且無法運行此代碼。

map' :: (a -> b) -> [a] -> [b]  
map' f xs = foldr (\x acc -> f x : acc) [] xs 

我不知道f應該是什么。

Answer 1

也許這可以幫助：

map' succ [1,2,3]
= foldr (\x acc -> succ x : acc) [] [1,2,3]
= foldr (\x acc -> succ x : acc) [] [1,2,3]
= (\x acc -> succ x : acc) 1 (foldr (\x acc -> succ x : acc) [] [2,3])
= succ x : acc  where x=1 ; acc=foldr (\x acc -> succ x : acc) [] [2,3]
= succ 1 : foldr (\x acc -> succ x : acc) [] [2,3]
= succ 1 : (\x acc -> succ x : acc) 2 (foldr (\x acc -> succ x : acc) [] [3])
= succ 1 : succ 2 : foldr (\x acc -> succ x : acc) [] [3]
= succ 1 : succ 2 : (\x acc -> succ x : acc) 3 (foldr (\x acc -> succ x : acc) [] [])
= succ 1 : succ 2 : succ 3 : foldr (\x acc -> succ x : acc) [] []
= succ 1 : succ 2 : succ 3 : []
= 2 : 3 : 4 : []
= [2,3,4]

Answer 2

看一下類型：

map' :: (a -> b) -> [a] -> [b] 
map'       f        xs  =  ys   where ys = ...

f是類型a -> b的函數：

f :: a -> b
f    x =  y   where y = ...

意思是，給定類型a的x ，它會產生類型b的y 。 因此map' f的類型為[a] -> [b] ，

map' :: (a -> b) -> [a] -> [b] 
map'       f        xs  =  ys   where ys = ...
map'       f     :: [a] -> [b]

也就是說，給定類型為[a]的列表xs ，它將產生類型為[b]的列表ys 。

所以，無論f你使用，它必須與列表xs與您打算使用map' f ：它必須接受的元素xs作為參數。